Graduate Microeconomic Theory
Scott Cunningham, Baylor University
September 1, 2016
1 / 526
Outline
Part 1: Math introduction
• Calculus
• Quasi-concavity
• Convex sets
• Probability
Part 2: Consumer theory
• Utility maximization and expenditure minimization
• Derived demand
2 / 526
Outline cont...
Part 3: Producer theory
• Profit maximization and cost minimization
• Derived input demand
Part 4: Game theory
• Nash equilibrium.
• Mixed strategies.
• Subgame perfection.
• Bayesian equilibrium.
3 / 526
This is not a pipe? Then what is it
4 / 526
This is not a heart? Then what is it
5 / 526
Characteristics of economic models I
• All models are wrong but some are useful. (George Box)
• If you want literal realism, look at the world around you; if
you want understanding, look at theories. (Dorfman, 1964)
• We would, of course, dismiss the rigorous proof as being
superfluous: if a theorem is geometrically obvious why prove
it? This was exactly the attitude taken in the eighteenth
century. The result, in the nineteenth century, was chaos and
confusion: for intuition, unsupported by logic, habitually
assumes that everything is much nicer behaved than it really
is. (Steward, 1975)
Characteristics of economic models II
• Ceteris paribus – holding all else constant, causal models
• Mapping of real items to mathematical notation for the
purposes of analysis, explanation and prediction
• Positive/normative distinction
7 / 526
Maximization of a function of one variable
π = f (x)
0
−20
−40
−4
−2
0
2
4
x
8 / 526
Derivatives I
• Manager may not have an accurate picture of the market and
try varying x to see where a maximum profit is obtained
• She may start at low levels of x, observe profits, then increase
x and see if profits rise or fall
• The commonsense idea that profits have increased in response
to an increase in x can be stated formally as
π2 − π1
∆π
> 0, or
>0
x2 − x1
∆x
• The limit of ∆π
∆x for small changes in x is called the derivative
of the function, π = f (x) and is denoted by
dπ
dx
or
df
dx
or f 0 (x)
dπ
f (x1 + h) − f (x1 )
= limh→0
dx
h
9 / 526
Derivatives II
• Value of the derivative at a point: sometimes we wish to note
explicitly the point at which the derivative is to be evaluated,
or
dπ
dx |x=x1
• At other times, we are interested in the value of dπ
dx for all
possible values of x and not a particular point
• In the previous figure, on the rising part of the function,
dπ
>0
dx |q=q0
whereas on the falling part of the function
dπ
<0
dx |q=q0
10 / 526
Derivatives III
• First order conditions for a maximum
df
=0
dx |x=x ∗
• Second-order conditions:
• Notice that the derivative equals zero for both functions below
π = f (x)
60
40
20
0
−20
−40
−60
−4
−2
0
2
4
x
11 / 526
Derivatives IV
• Second derivatives: the derivative of a derivative is called a
second derivative and is denoted by
d 2π
d 2f
,
or
, or f 00 (x)
dx 2
dx 2
• The additional condition for x ∗ to represent a local maximum
is therefore
d 2π
<0
dx 2 |x=x ∗
where the notation reminds us that this second derivative
must be evaluated at the optima, x ∗
• Hence the necessary condition for a optima is
dπ
=0
dx
and the sufficient condition for a maximum (minimum) is
d 2π
< (>)0
dx 2 |x=x ∗
12 / 526
Rules for derivatives
da
dx = 0
then d[afdx(x)] = af 0 (x)
a
a−1
then dx
dx = ax
1
If a is a constant then
2
If a is a constant,
3
If a is a constant,
4
dlnx
dx
5
= x1 where ln signifies the logarithm to the base e
(= 2.71828)
dax
dx
= ax lna for any constant a
13 / 526
Rules for derivatives II
Now suppose that f (x) and g (x) are two functions of x and f 0 (x)
and g 0 (x) both exist. Then:
6
7
8
d[f (x)+g (x)]
= f 0 (x) + g 0 (x)
dx
d[f (x)·g (x)]
= f 0 (x)g (x) + g 0 (x)f (x)
dx
0
df (x)/g (x)
(x)g 0 (x)
= f (x)g (x)−f
provided
dx
[g (x)]2
g (x) 6= 0
14 / 526
Rules for derivatives III
9
If y = f (x) and x = g (z) and if both f 0 (x) and g 0 (x) exist,
dy dx
dg
df
then dy
dz = dx · dz = dx · dz . This is called the “chain rule”
d(ax)
de ax
ax
d(ax) · dx = e
d[ln(ax)]
d(ax)
11
= d[ln(ax)]
dx
d(ax) · dx
10
12
de ax
dx
=
d[ln(x 2 )]
dx
=
d[ln(x 2 )]
d(x 2 )
·
d(x 2 )
dx
· a = ae ax
=
1
ax
· a = x1 .
=
1
x2
· 2x =
2
x
15 / 526
Example
Suppose that the relationship between profits (π) and quantity
produced (x) is given by
π(x) = 1, 000x − 5x 2
Use the first and second order conditions that we previously
discussed to find the amount of production that maximizes profits
16 / 526
Solution
Profit function: π(x) = 1, 000x − 5x 2 .
First order condition: f 0 (x) = 0
dπ
= 1, 000 − 10x = 0
dx
∗
x = 100
Second order condition: f 00 (x) < 0
d 2π
dx 2
= −10 < 0
17 / 526
Functions of several variables
Economic problems seldom involve functions of only a single
variable – utility, for instance, is a function of all goods consumed.
In these circumstances, the dependence of one variable (y ) on a
series of of other variables (x1 , x2 , . . . , xn ) is denoted by
y = f (x1 , x2 , . . . , xn ).
∂y
• Partial derivative of y with respect to x1 is denoted by ∂x
or
1
∂f
∂x1
or f1
• Formal definition of the partial derivative
∂f
f (x1 + h, x¯2 , . . . , x¯n ) − f (x1 , x¯2 , . . . , x¯n )
= limh→0
∂x1 |x¯2 ,...,x¯n
h
where x¯2 , . . . , x¯n are held constant at preassigned values.
18 / 526
Functions of several variables II
Let y = f (x1 , x2 ) = ax12 + bx1 x2 + cx22 . Then
∂f
= f1 = 2ax1 + bx2 ,
∂x1
and
∂f
= f2 = bx1 + 2cx2
∂x2
19 / 526
Functions of several variables III
Let y = f (x1 , x2 ) = a ln x1 + b ln x2 . Then
a
∂f
= f1 = ,
∂x1
x1
and
b
∂f
= f2 =
∂x2
x2
20 / 526
Functions of several variables IV
• Ceteris paribus
• Partial derivatives are the mathematical means by which we
invoke the ceteris paribus assumption because we are
augmenting the value of one variable while we hold constant
all the others
• Units of measurement
21 / 526
Functions of several variables V
Let y = f (x1 , x2 ) = ax12 + bx1 x2 + cx22 . Then
1
f11 = 2a
2
f12 = b
3
f21 = b
4
f22 = 2c
Let y = f (x1 , x2 ) = alnx1 + blnx2 .
1
f11 = −ax1−2
2
f12 = 0
3
f21 = 0
4
f22 = −bx2−2
22 / 526
Functions of several variables VI
• Young’s theorem:
• These examples illustrate the mathematical result that, under
general conditions, the order in which partial differentiation is
conducted to evaluate second-order partial derivatives does not
matter. That is
fij = fji
for any pair of variables xi , xj .
• Cross-partial effects are symmetric
• Chain rule:
∂f dx1
∂f dx2
∂f dx3
dy
=
·
+
·
+
·
da
∂x1 da
∂x2 da
∂x3 da
23 / 526
Implicit functions I
• If the value of a function is held constant, an implicit
relationship is created among the independent variables that
enter into the function.
• That is, the independent variables can no longer take on any
values, but rather must take on only that set of values that
result in the function’s retaining the required value
• The most useful result provided by this approach is the ability
it affords the user to quantify the trade-offs inherent in an
economic model
• Consider the function y = f (x1 , x2 ). If we hold y constant, we
have created an implicit relationship between the x’s showing
how changes in them must be related to keep the value of the
function constant
24 / 526
Implicit functions II
• Given y = f (x1 , x2 ), set y = 0
y = 0 = f (x1 , x2 )
• Let x2 = g (x1 ) and use the chain rule to differentiate this
function with respect to x1
0 = f1 + f2 ·
dg (x1 )
dx1
d0
• Recall that dx
=0
1
• Thus we have shown that the partial derivatives of the
function f can be used to derive an explicit expression for the
trade-offs between x1 and x2
25 / 526
Implicit functions III
• Example: Let the production possibility frontier for two goods
have the following form:
200 = x 2 + 0.25y 2
• Notice that this function is set to a constant (e.g., 200). We
can therefore study the relationship between the variables by
using the implicit function result:
−fx
−2x
−4x
dy
=
=
=
dx
fy
0.5y
y
26 / 526
Unconstrained optimization I
• Using partial derivatives we can find the optimal value for a
function of several variables
• Let’s consider first a simple one-variable case – imagine an
economic agent varying output, x, by a small amount (dx)
and observing the change in profits, y , dy ; this change is
given by
dy = f 0 (x)dx
• This identity records the fact that the change in y is equal to
the change in x times the slope of the function, or the
point-slope formula used for linear equations in basic algebra
• The necessary condition for a maximum (minimum) is that
dy = 0 for small changes in x around the optimal point –
otherwise, y could be increased by suitable changes in output,
x.
• But dx does not necessarily equal 0 – setting dy = 0 implies
that at the desired point, f 0 (x) = 0 – we call this the first
27 / 526
order conditions
Unconstrained optimization II
• First order conditions for a maximum when there are several
variables is an extension of the simpler case
• The necessary condition for a point to be a local maximum is
that at this point
f1 = f2 = · · · = fn = 0
• A point at which this holds is called the critical point of the
function
• Recall that first order conditions are only necessary conditions,
not sufficient conditions – we have to investigate the second
order conditions to determine whether the critical point has
maximized or minimized the function
28 / 526
Unconstrained optimization III
• Suppose that y is a function of x1 and x2 given by
y = −x12 + 2x1 − x22 + 4x2 + 5
• Taking the partial derivatives with respect to x1 and x2 and
applying the necessary conditions yields
∂y
∂x1
∂y
∂x2
= −2x1 + 2 = 0,
= −2x2 + 4 = 0
• Therefore the function is at a critical point when x1 = 1 and
x2 = 2. At that point, y = 10 and that is the optimized value
for the function
• To know if this is a max or a min, we have to investigate the
second order conditions. Here we introduce matrix algebra to
facilitate this
29 / 526
Matrix algebra I
1
An n × k matrix, A, is a rectangular
form:

a11 a12
a21 a22

A = [aij ] =  .
..
 ..
.
an1 an2
array of terms of the

· · · a1k
· · · a2k 

.. 
..
.
. 
· · · ank
Here i = 1, n; j = 1, k. Matrices can be added, subtracted, or
multiplied providing their dimensions are conformable
2
If n = k, then A is a square matrix. A square matrix is
symmetric if aij = aji . The identity matrix, In , is an n × n
square matrix where aij = 1 if i = j and aij = 0 if i 6= j
30 / 526
Matrix algebra II
3
The determinant of a square matrix denoted |A| is a scalar
(i.e., a single term) found by suitably multiplying together all
the terms in the matrix. If A is 2 × 2,
|A| = a11 a22 − a21 a12
1 3
Example: If A =
then |A| = 2 − 15 = −13
5 2
The inverse of an n × n square matrix, A, is another n × n
matrix A−1 such that
4
A · A−1 = In
5
Not every square matrix has an inverse. A necessary and
sufficient condition for the existence of A−1 is that |A| =
6 0
The leading principal minors of an n × n square matrix, A, are
the series of determinants of the first p rows and columns of
A where p = 1, n. If A is a 2 × 2 matrix, then the first leading
principal minor is a11 and the second is a11 a22 − a12 a21
31 / 526
Matrix algebra III
6
7
An n × n square matrix, A, is positive definite if all its leading
principal minors are positive. The matrix is negative definite if
its principal minors alternate in sign starting with a minus. If
some of the determinants in this definition are 0, then the
matrix is said to be positive semidefinite or negative
semidefinite
A particularly useful symmetric matrix is the Hessian matrix
formed by all the second-order partial derivatives of a
function. If f is a continuous and twice differentiable function
of n variables, then its Hessian is given by:


f11 f12 · · · f1n
f21 f22 · · · f2n 


H(f ) =  .
.. . .
.. 
 ..
.
. 
.
fn1 fn2 · · · fnn
32 / 526
Matrix algebra IV
• A concave function is one that is always below (or on) any
tangent to it
• A convex function is one that is always above (or on) any
tangent to it.
• The concavity or convexity of a function is determined by its
second derivative(s)
• Concavity requires that the Hessian matrix be negative definite
• Convexity requires that the Hessian be positive definite
• If f (x1 , x2 ) is a function of two variables, the Hessian is given
f11 f12
. This is negative definite if f11 < 0 and
f21 f22
f11 f22 − f21 f12 > 0
by
33 / 526
Maximization I
• The first order conditions for an unconstrained maximization
of a function of many variables requires finding a point at
which the partial derivatives are zero
• If the function is concave it will be below its tangent plane at
this point
• Note, this will be a local maximum if the function is concave
in a region and global if the function is concave everywhere
34 / 526
Maximization II
• Sometimes we restrict the values that the x’s can take on in
an maximization or minimization problem
• Matrix algebra provides a compact (even if it’s not intuitive)
way to denote the conditions where a constrained max (min)
is a max (min)
• Assume we wish to maximize
f (x1 , . . . , xn )
subject to the constraint
g (x1 , . . . , xn ) = 0
35 / 526
Maximization III
• The first order conditions for a maximum are of the form
fi + λgi = 0 where λ is the Lagrange multiplier for this
problem. Second order conditions for a maximum are based
on the augmented bordered Hessian.


0 g1 g2 · · · gn
g1 f11 f12 · · · f1n 


g2 f21 f22 · · · f2n 



 ..

.
gn fn1 fn2 · · · fnn
• For a maximum, (−1)Hb must be negative definite
• the leading principal minors of Hb must follow the pattern
− + − + − and so forth, starting with the second such minor
• notice that the first leading principal minor is of Hb is 0.
• For a minimum, the second order conditions require that
(−1)Hb be positive definite – that is, all the leading principal
minors of Hb (except the first) should be negative
36 / 526
Maximization IV
• Let y = f (x1 , x2 ). Take total differential
dy = f1 dx1 + fx dx2
• The differential of that function is given by
d 2 y = (f11 dx1 + f12 dx2 )dx1 + (f21 dx1 + f22 dx2 )dx2
or
d 2 y = f11 dx12 + f12 dx2 dx1 + f21 dx1 dx2 + f22 dx22
• Recall Young’s theorem (f12 = f21 ). Rearrange the terms
d 2 y = f11 dx12 + 2f12 dx1 dx2 + f22 dx22
37 / 526
Maximization V
• For the previous equation to be unambiguously negative for
any changes in the x 0 s, it is obviously necessary that f11 and
f22 be negative.
• If neither dx1 nor dx2 is 0, we then must consider the cross
partial f12 in deciding whether d 2 y is unambiguously negative
• It requires that the Hessian matrix
f11 f12
f21 f22
be negative definite
• This requires that the determinant of this Hessian matrix be
positive
38 / 526
Constrained maximization I
• Consider the problem of choosing x1 and x2 to maximize y = f (x1 , x2 ) subject
to a linear constraint
c = b1 x1 + b2 x2
where c, b1 and b2 are constant parameters
• The first order conditions for a maximum can be derived by setting up the
Lagrangean expression
L = f (x1 , x2 ) + λ(c − b1 x1 − b2 x2 )
• Partial differentiation with respect to x1 , x2 and λ yields three first order
conditions (FOC)
∂L
∂x1
∂L
∂x2
∂L
∂λ
= f1 − λb1 = 0
= f2 − λb2 = 0
= c − b1 x1 − b2 x2 = 0
• These three equations can be solved in general to find the optimal values of x1 ,
x2 and λ
39 / 526
Constrained maximization II
• To ensure that the critical point derived is a local maximum,
we must again examine movements away from the critical
points by using the second total differential
d 2 y = f11 dx12 + 2f12 dx1 dx2 + f22 dx22
• In this case, because of the linear constraint, not all possible
small changes in the x 0 s are permissible; only those values of
x1 and x2 that satisfy the constraint are legal alternatives to
the critical point
• To examine such changes, we have to calculate the total
differential of the constraint
−b1 dx1 − b2 dx2 = 0
or
b1
dx1
b2
• This equation shows the relative changes in x1 and x2 that are
legal in considering movements from the critical point
40 / 526
dx2 = −
Constrained maximization III
• To proceed further, use the first two FOC
f1
b1
=
f2
b2
• Combine with the last equation on the previous slide
f1
dx2 = − dx1
f2
• Substitute into the d 2 y equation and show that for d 2 y < 0
d 2y
2
f1
f1
dx1 + f22 − dx1
f2
f2
2
f1
f
= f11 dx12 − 2f12 dx12 + f22 12 dx12
f2
f2
= f11 dx12 + 2f12 dx1
−
• Combine terms and put over a common denominator
d 2y =
2
dx1
f11 f22 − 2f12 f1 f2 + f22 f12
f22
41 / 526
Constrained maximization IV
• Consequently, for d 2 y < 0, it must be the case that
d 2 y = f11 f22 − 2f12 f1 f2 + f22 f12 < 0
• Although that maybe looks like a bunch of weird symbols, the
condition characterizes a set of functions called
“quasi-concave functions”
• Quasi-concave functions have the property that the set of all
points for which such a function takes on a value greater than
any specific constant is a convex set
• That is, any two points in the set can be joined by a line
contained completely within the set
42 / 526
Homogenous functions I
• One important set of properties relates to how a function
behaves when all (or most) of its arguments are increased
proportionally
• For instance if you doubled all prices and income, what would
happen to demand?
• Specifically, a function f (x1 , x2 , . . . , xn ) is said to be
homogenous of degree k if
f (tx1 , tx2 , . . . , txn ) = t k f (x1 , x2 , . . . , xn )
• Most important cases are k = 1 or k = 0
• k = 1: doubling all arguments doubles the value of the
function itself (e.g., demand would double if prices and income
doubled)
• k = 0: doubling all arguments leaves the value of the function
unchanged (e.g., demand is unchanged if prices and income
double)
43 / 526
Homogenous functions II
• If a function is homogenous of degree k and can be
differentiated, the partial derivatives of the function are
homogenous of degree k − 1
Proof.
Differentiate the previous equation with respect to its first
argument
∂f (x1 , . . . , xn )
∂f (tx1 , . . . , txn )
· t = tk
∂x1
∂x1
or
f1 (tx1 , . . . , txn ) = t k−1 f1 (x1 , . . . , xn )
which shows that f1 meets the definition for homogeneity of degree
k −1
44 / 526
Homogenous functions III
• Another useful feature of homogenous functions can be shown
by differentiating the definition for homogeneity with respect
to the proportionality factor, t
• Differentiate the right side of the equation then the left side
so that we can see this
kt k−1 f (x1 , . . . , xn ) = x1 f1 (tx1 , . . . , txn )+· · ·+xn fn (tx1 , · · · , txn )
Let t = 1
kf (x1 , . . . , xn ) = x1 f1 (x1 , . . . , xn ) + · · · + xn fn (x1 , . . . , xn )
• This equation is called “Euler’s theorem” for homogenous
functions; it shows that for a homogenous function there is a
definite relationship between the values of the function and
the values of its partial derivatives
45 / 526
Homogenous functions IV
• Monotonic transformations preserve the order of the
relationship between the arguments of a function and the
value of that function
• If certain sets of x 0 s yield larger values for f , they will also
under a monotonic transformation of f
• Homothetic functions are formed by taking a monotonic
transformation of a homogenous function
• One kind of monotonic transformation is to multiply it by 1,
leaving the function unchanged
• All homogenous functions are therefore also homothetic
• Except in special cases, homothetic functions do not possess
the homogeneity properties of their underlying functions
• But they do preserve one feature – the implicit trade-offs
implied by the function depend only on the ratio of the two
variables being traded, not on their absolute levels
46 / 526
Homogenous functions V
• Recall that for a two-variable function of the form
y = f (x1 , x2 ),
dx2
f1
=−
dx1
f2
• If f is homogenous of degree k, its partial derivatives will be
homogenous of degree k − 1, and letting t = 1/x2 :
dx2
tk k−1 f1 (tx1 , tx2 )
f1 (tx1 , tx2 )
f1 (x1 /x2 , 1)
= − k−1
=−
=−
dx1
f2 (tx1 , tx2 )
f2 (x1 /x2 , 1)
tk
f2 (tx1 , tx2 )
which shows that the trade-offs implicit in f depend only on
the ratio of x1 to x2
• If we apply any monotonic transformation F (with F 0 > 0) to
the original homogenous function, the trade-offs are
unchanged:
dx2
F 0 f1 (x1 /x2 , 1)
f1 (x1 /x2 , 1)
=− 0
=−
dx1
F f2 (x1 /x2 , 1)
f2 (x1 /x2 , 1)
47 / 526
Cramer’s rule I
• Cramer’s rule is a method of solving a system of linear
equations through the use of determinants. Cramer’s rule is
given by the equation
|Aj |
xj =
|A|
where xj is the jth endogenous variable in a system of
equations, |A| is the determinant of the original A matrix, and
|Aj | is the determinant a special matrix formed as part of
Cramer’s rule.
• Note the determinant for a matrix must not equal 0. If |A| = 0
then there is no solution, or there are infinite solutions (from
dividing by zero). When A 6= 0 then a unique solution exists
48 / 526
Cramer’s rule II
• To use Cramer’s rule, two (or more) linear equations are
arranged in the matrix form Ax = d. For a two equation
model:
a11 a12 x1
d
= 1
a21 a22 x2
d2
• The first matrix corresponds to the number of equations in a
system (here two equations)
• The second matrix holds the number of endogenous variables
in the system (here two variables)
• Remember that the matrix must be square so the number of
equations must equal the same number of endogenous
variables
• Finally, the last matrix contains the exogenous terms of each
linear equation
49 / 526
Cramer’s rule III
• Using Cramer’s rule to solve for the unknowns in the following
linear equations
2x1 + 6x2 = 22
−x1 + 5x2 = 53
• Then,
2 6
−1 5
x1
22
=
x2
53
• The primary determinant of |A| = 10 − (−6) = 16
|A |
• We need to construct xj = |A|j for j = 1 and j = 2
50 / 526
Cramer’s rule IV
• The first special determinant |A1 | is found by replacing the
first column of the primary matrix with the d exogenous
column matrix
22 6
53 5
• |A1 | = (22 × 5) − (53 × 6) = −208
• The second special determinant |A2 | is found by replacing the
second column of the primary matrix with d
2 22
−1 53
• |A2 | = (2 × 53) − (−22) = 128
• Now with these three determinants, we can solve for x1 and x2
using xj =
|Aj |
|A| .
x1 = −13, x2 = 8
51 / 526
Mathematical statistics I
• A random variable describes in numerical form the outcomes
from an experiment that is subject to chance
• Let’s say we flip a coin and observe whether it lands heads or
tails
• If we call this random variable x, we can denote the possible
outcome of the variable as
(
1 if coin is heads,
x=
0 if coin is tails
• Notice that before the coin flip, x could’ve been either 0 or 1.
• Only after the uncertainty is resolved do we know the value of
x
52 / 526
Mathematical statistics II
• For any random variable, its probability density function (pdf)
shows the probability that each specific outcome will occur
• The expected value of a random variable is the numerical
value that the random variable might be expected to have on
average
• The variance of a random variable is a measure of its
dispersion and is defined as the expected squared deviation
from a random variable
Var (x) = σx2 = E [(x − E (x))2 ]
• The covariance between two variables seeks to measure the
direction of association between the variables
Cov (x, y ) = E [(x − E (x))(y − E (y ))]
53 / 526
Consumer theory: Preferences and
utility
54 / 526
Preferences
We start a theory of the consumer based on her preferences
• “A is preferred to B” means if both items have the same price
and are affordable, the agent would choose A.
55 / 526
Preference Notation
• We represent strict preferences with the symbol, as in
A B (i.e., “A is strictly preferred to B”)
• We represent “weak” preferences with the % symbol, as in
A % B (i.e., “A is weakly preferred to B” or “A is at least as
good as B”)
• We represent indifference with the ∼ symbol, as in A ∼ B
(i.e., “The agent is indifferent between A and B”)
56 / 526
Axioms of rational choice
• We make the following assumptions about her preferences:
1 Completeness - Given any two goods, say A and B, an
individual can always specify exactly one of the following
possibilities: (i) A ≺ B; (ii) A B; (iii) A ∼ B; (iii) A % B;
(iii) A - B
2
3
Transitivity: If A B and B C , then A C (i.e., internal
consistency)
4 Continuity: If A B, and C is “close to” B, then A C
57 / 526
Utility I
• Given complete, transitive and continuous preferences, we can
formally create a utility function that captures the preference
ordering
• If a person prefers A to B, we say that the utility she gets
from A is larger than the utility she gets from B
• Old idea going back to 19th century economist Jeremy
Bentham who assumed consumption yielded personal and
invisible “utils”
58 / 526
Utility II
• Utility turns complete, transitive and continuous preferences
into a utility number.
• The utility function U(bundle of goods) turns the preferences
the agent has about the bundle of goods into a number that
represents “how much” the person likes the bundle.
• So if someone has preferences where A B, then
U(A) > U(B)
59 / 526
Utility III
• Although we will stick with two goods most of the time, a
utility function can be created for any number of goods.
Utility = U(x1 , x2 , . . . , other things)
• Don’t get distracted by this language – anything can enter a
utility function
• Utility can be a function of friendship and material wealth,
guns and butter, leisure and consumption, and so forth
• It can even be a function of another person’s utility! (i.e.,
altruism)
60 / 526
Utility IV
• The same ordinal preferences can be represented by a number
of different utility functions.
• For example these two utility functions represent the same
preferences:
U(x, y ) = x 1/2 y 1/2
U(x, y ) = 5x 1/2 5y 1/2
• If a given utility function UDude (x, y ) represents the Dude’s
preferences then any monotonic transformation of UDude (x, y )
also represents the Dude’s preferences.
61 / 526
Economic goods
10
9
8
7
Quantity of y
6
5
4
ambiguous
preferred to x*,y*
3
y*
ambiguous
worse than x*,y*
1
0
0
1
x*
3
4
5
6
7
8
9
Quantity of x
62 / 526
Indifference and Substitution I
• The reason that the shaded reason is preferred is because we
assume that more is always better
• The reason that the area below x ∗ , y ∗ is worse than x ∗ , y ∗ is
because more is always better
• It’s not theoretically possible to say anything conclusively
about the ambiguous regions because they involve trade-offs –
a little more x and a little less than y
63 / 526
Indifference and Substitution II
• An indifference curve shows a set of consumption bundles
about which the individual is indifferent.
• The bundles all provide the same level of utility
• The negative of the slope of an indifference curve (U1 ) at
some point is termed the marginal rate of substitution. That is
MRS = −
dy
dx |U=U1
where the notation indicates that the slope is calculated along
the U1 indifference curve
64 / 526
Indifference curves and Utility Functions
• Everything along the indifference curve has the same level of
utility (e.g., U1 )
• The indifference curves is a ”contour curve” or ”level set” of a
utility function
• Example [LINK]
65 / 526
Indifference curves
• Indifference curves for utility functions that meet our basic
assumptions have the following traits:
• Convex to the origin.
• More is better – higher indifference curves yield higher levels of
utility.
• Transitivity implies that indifference curves never cross.
66 / 526
Convex to the Origin
10
9
8
7
Quantity of Y
6
5
y1
y2
U1
2
1
0
0
1
2
x1
3
x2
4
5
6
7
8
9
Quantity of X
67 / 526
More is Better
7
6
5
4
More preferred
Y Axis
Increasing utility
3
2
U3
1
U2
U1
0
0
1
2
3
4
X Axis
68 / 526
Transitivity
10
9
8
Quantity of Y
7
C
6
5
D
4
E
3
A
2
U1
1
B
U2
0
0
1
2
3
4
5
6
7
8
9
10
Quantity of X
69 / 526
Convexity of indifference curves I
10
9
8
7
Quantity of Y
6
5
y1
y2
U1
2
1
0
0
1
2
x1
3
x2
4
5
6
7
8
9
Quantity of X
• The slope of U1 and the MRS tells us something about the
trades this person will voluntarily make
70 / 526
Convexity of indifference curves II
10
9
8
7
Quantity of Y
6
5
y1
y2
U1
2
1
0
0
1
2
x1
3
x2
4
5
6
7
8
9
Quantity of X
• At x1 , y1 , the person has a lot of y and is willing to trade
away a significant amount to get one more x – the
indifference curve is relatively steep.
• At x2 , y2 , the slope is less steep; they are less willing to trade
y for the same amount of x as they had previously.
71 / 526
Convexity of indifference curves III
10
9
8
7
Quantity of Y
6
5
y1
y2
U1
2
1
0
0
1
2
x1
3
x2
4
5
6
7
8
9
Quantity of X
• We characterize a shape like the one shown in this figure as
diminishing MRS – the trades necessary to stay on the
indifference curve are getting larger and larger the less you
have of that good
72 / 526
Convexity of indifference curves IV
• An alternative way of stating the principle of diminishing MRS
is to speak in terms of a convex set
• A set of points is said to be convex if any two points within
the set can be joined by a straight line that is contained
completely within the set
73 / 526
Convexity of indifference curves V
10
U1
9
8
7
Quantity of y
y1
5
4
y*
3
y2
U1
1
0
0
1
x1
3
x*
4
5
6
x2
8
9
10
Quantity of x
74 / 526
Convexity of indifference curves VI
• The assumption of a diminishing MRS is the same as
assuming that all combinations of x and y that are preferred
or indifferent to a particular combination x ∗ , y ∗ form a convex
set .
• Convexity implies that consumers prefer moderation – if the
indifference curve is strictly convex.
• So given three bundles on the same indifference curve (x1 , y1 ),
(x ∗ , y ∗ ), and (x2 , y2 ), the consumer will prefer the average of
the first and last points to the the second point
75 / 526
Convexity of indifference curves VII
• If an indifference curve is convex then we know that the utility
function the IC belongs to is quasi-concave.
• If a utility function is quasi-concave then all of it’s indifference
curve’s are convex.
76 / 526
All of these things go together.
• Quasi-Concave Utility Function
• Convexity of Indifference Curves
• Convexity of the ”weakly preferred to set”.
• Diminishing MRS.
77 / 526
Convexity of indifference curves VIII
• It may be helpful to see a violation of diminishing MRS to
understand what it is:
78 / 526
Convexity of indifference curves IX
10
9
8
Quantity of y
7
6
5
4
3
2
1
0
0
1
2
3
4
5
6
7
8
9
10
Quantity of x
79 / 526
Example: Guns and Butter I
• Suppose a person’s ranking of guns (y ) and butter (x) could
be represented with the following utility function
u(x, y ) =
√
x ·y
• To find the indifference curve associated with this utility
function, we need only find a level surface associated with it
(e.g., set it to 10)
u(x, y ) = 10 =
√
x ·y
80 / 526
Example: Guns and Butter II
• That square root is going to be a little complicated, so let’s
get rid of it – remember that utility is ordinal so any
monotonic transformation preserves the preference orderings;
let’s square both sides (i.e., a monotonic transformation)
2
10
=
√
2
x ·y
100 = x · y
81 / 526
Example: Guns and Butter III
• Recall that MRS = − dy
dx , so rearrange the equation in terms
of y
y = 100/x
• Now take the derivative with respect to x
MRS = −dy /dx = 100/x 2
• This MRS is unique to the chosen indifference curve (i.e., the
level surface 10 associated with x · y , the transformed utility
function)
82 / 526
Example: Guns and Butter IV
• Now evaluate MRS at 5 guns and 20 units of butter:
MRS(5, 20) = 100/25 = 4.
• The person is willing to give up 4 units of butter for 1 unit of
guns
• Now evaluate MRS at 20 guns and 5 units of butter:
MRS(20, 5) = 100/400 = 0.25.
• She’s willing to give up a fourth of butter for 1 unit of guns
83 / 526
Example: Guns and Butter IV
• Convexity: to see that moderation is preferred, what if we
averaged 5 and 20 guns (butter) and gave her 12.5 of each
p
U(x, y ) = U(12.5, 12.5) = (12.5 · 12.5) = 12.5 utils
which is more than the utility that we assumed (e.g., 10)
84 / 526
More MRS I
• Let U = U(x, y ) and totally differentiate
dU = Ux dx + Uy dy
• Recall that indifference curves are level surfaces – that is,
along an indifference curve, utility does not change, so set
dU = 0
Ux dx + Uy dy = 0
• Solve for dy
dy
Ux
=−
dx |U(x,y )=k
Uy
85 / 526
More MRS II
• Recall that MRS = − dy
dx , therefore
MRS = −
dy
Ux
=
dx |U(x,y )=k
Uy
• Because the MRS is the ratio of two marginal utilities, the
units (e.g., utils) “drop out” in the computation; the MRS will
be unchanged no matter what specific utility ranking is used
86 / 526
Quasi-concavity of U(.) I
• For the function U(x, y ) to be quasi-concave, the following
condition must hold
Uxx Ux2 − 2Uxy Ux Uy + Uyy Uy2 < 0
• We tend to assume positive marginal utilities (Ux > 0 and
Uy > 0) because we are dealing with “goods” not “bads”
87 / 526
Quasi-concavity of U(.) II
• Notice that it is not sufficient that Uxx < 0 and Uyy < 0 to
ensure this because we still have to deal with the Uxy cross
partial effect
• If Uxy > 0, then the condition holds. If Uxy < 0 then we
require it not be “too negative”.
88 / 526
Convexity of ICs
• How do we show that indifference curves are convex? We
need to show that MRS is diminishing in x. Recall that
MRS(x, y ) = −
dy
Ux (x, y )
=
dx
Uy (x, y )
• Take the derivative with respect to x
Ux (x,y )
dU
dMRS(x, y )
y (x,y )
=
dx
dx
• Let’s look at some examples
89 / 526
Convexity of ICs: U(x, y ) = 3x + y I
• Let U(x, y ) = 3x + y . Find the MRS
MRS(x, y ) =
3
Ux
= =3
Uy
1
• To check whether the indifference curve is a convex set, MRS
must be diminishing in x
dMRS(x, y )
=0
dx
90 / 526
Convexity of ICs: U(x, y ) = 3x + y II
• So we can see here that this utility function does not have
convex indifference curves.
• Let’s see what the indifference curves associated with this
utility function look like.
• Set U = 9 and solve for y
y = 9 − 3x
91 / 526
Convexity of ICs: U(x, y ) = 3x + y III
10
9
8
Quantity of y
7
6
5
4
3
2
1
0
U=9
0
1
2
3
4
5
6
7
8
9
10
Quantity of x
92 / 526
Convexity of ICs: U(x, y ) =
• Let U(x, y ) =
√
√
xy I
xy . Find the MRS
MRS(x, y ) =
0.5x −0.5 y 0 .5
y
Ux
=
=
0.5
−0.5
Uy
0.5x y
x
• Check for diminishing MRS
dMRS(x, y )
y
=− 2 <0
dx
x
93 / 526
Convexity of ICs: U(x, y ) =
√
xy II
• So we can see here that this utility function has convex
indifference curves, meaning it must be a quasi-concave utility
function
• Let’s see what the indifference curves look like. Set U = 2,
square both sides and get
y=
4
x
94 / 526
Convexity of ICs: U(x, y ) =
√
xy III
10
9
8
quantity of y
7
6
5
4
3
2
1
0
U=2
0
1
2
3
4
5
6
7
8
9
10
quantity of x
95 / 526
Utility functions for specific preferences
• There’s four broad utility function types for specific kinds of
preferences that you should be familiar with
1
2
3
4
Cobb-Douglas Utility
Perfect substitutes
Perfect complements
CES utility
• Let’s look at them each separately
96 / 526
Cobb-Douglas utility
• The previous indifference curve can be generated using a
Cobb-Douglas utility function
U(x, y ) = x α y β
where α and β are both positive constants
• It is often convenient to normalize these parameters so that
α + β = 1.
97 / 526
Cobb-Douglas II
Cobb-Douglass Utility on Wolfram Alpha [LINK]
10
9
8
quantity of y
7
6
5
4
3
2
1
0
U=8
0
1
2
3
4
5
6
7
8
9
10
quantity of x
98 / 526
Perfect substitutes I
• The earlier linear indifference curve is is generated by a utility
function of the form
U(x, y ) = αx + βy
where again α and β are positive constants
• These generate linear indifference curves where MRS is
constant across all consumption bundles
99 / 526
Perfect substitutes II
• This means that the internal rates of exchange necessary to
maintain indifference is the same no matter how much of x
the consumer has
• Examples would be things like goods that are identical to
consumers
100 / 526
Perfect substitutes III
Perfect Substitutes on Wolfram Alpha [LINK]
10
9
8
7
quantity of y
6
5
4
3
2
1
0
U=8
0
1
2
3
4
5
6
7
8
9
10
quantity of x
101 / 526
Perfect complements I
• Some goods “go together” – peanut butter and jelly, left and
right shoes, hammer and nails
• The utility function describing perfect complements creates an
L-shaped indifference curve
U(x, y ) = min(αx, βy )
where the min operator means that utility is given by the
smaller of the two terms in the parentheses
102 / 526
Perfect complements II
• Example: if we let utility be a function of ounces of coffee (x)
and ounces of cream (y ), utility would be
U(x, y ) = min(x, 8y )
103 / 526
Perfect complements III
• Now 8 ounces of coffee and 1 ounce of cream provides 8 units
of utility. If we gave her 16 ounces of coffee and 1 ounce of
cream, she’d still only get 8 units of utility.
• Consumption will occur therefore at the vertices of the
indifference curves shown in the next figure
104 / 526
Perfect complements IV
Perfect Complements on Wolfram Alpha [LINK]
10
9
8
quantity of y
7
6
5
4
3
U=8
2
1
0
0
1
2
3
4
5
6
7
8
9
10
quantity of x
105 / 526
CES Utility I
• Each of the three specific utility functions we reviewed in the
previous slides are special cases of a more general CES
function
yδ
xδ
+
U(x, y ) =
δ
δ
where δ ≤ 1, δ 6= 0, and
U(x, y ) = lnx + lny
when δ = 0
• When δ = 1, the case of perfect substitutes
• When δ = 0, the Cobb-Douglas case
• When δ = ∞, the fixed proportions shape
106 / 526
CES Utility II
• Each of the three specific utility functions we reviewed in the
previous slides are special cases of a more general CES
function
yδ
xδ
+
U(x, y ) =
δ
δ
where δ ≤ 1, δ 6= 0, and
U(x, y ) = lnx + lny
when δ = 0
• When δ = 1, the case of perfect substitutes
• When δ = 0, the Cobb-Douglas case
• When δ = ∞, the fixed proportions shape
107 / 526
CES Utility III
CES Utility where δ 6= 0 on Wolfram Alpha [LINK]
CES Utility where δ = 0 on Wolfram Alpha [LINK]
108 / 526
Utility Maximization and
Expenditure Minimization
109 / 526
Utility maximization
• To maximize utility given a fixed amount of income to spend.
An individual will buy a bundle of goods that:
• Their internal rate of exchange between any two goods (their
MRS) is equal to the rate of exchange between the same two
goods in the market (the price ratio).
MRS =
px
py
• They spend all of their income.
110 / 526
• There are two distinct parts to the maximization problem:
1 Maximize the consumer’s utility function: U(x, y )
2 Subject to their budget constraint: I = px x + py y
111 / 526
• The slope of the budget constraint is ppx
y
• To see that solve the BC for y and rewrite the function in
“point slope” form: y = mx + b)
• We use the Lagrangian method to incorporate the budget
constraint into the utility maximization problem.
L = U(x, y ) + λ(I − px x − py y )
• This is why we call it “Constrained Optimization”
112 / 526
• The Lagrangian method finds the optimal values by
maximizing L by selecting values of x, y , and λ.
• These optimal values will still depend on prices and income
• The Lagrangian gives us the Uncompensated Demand
Functions x ∗ (px , py , I ), y ∗ (px , py , I ).
113 / 526
Step 1: First Order Conditions (FOCs)
• Take the derivative of L with respect to the three arguments
and set them equal to 0
(FOC 1)
(FOC 2)
(FOC 3)
dL
dx
dL
dy
dL
dλ
= Ux − λpx = 0
= Uy − λpy = 0
= I − px x − py y = 0
• This gives us a system of three equations and three unknown
variables.
• Notice that in the first two equations, we can isolate the MRS
114 / 526
Step 2: Solve The System of Equations
• In this two-good case, we can use simple substitution to solve
the system.1
• One approach to substitution is to solve each of the first two
FOCs for λ.
Ux
px
Uy
(FOC 20 ) λ =
py
(FOC 10 )
λ=
• Then since λ = λ we can set FOC1’ and FOC2’ equal to each
other.
1
Or you can use Cramer’s Rule.
115 / 526
Step 2: Solve The System of Equations
λ = λ
Uy
Ux
=
px
py
• Rearrange these terms and you can see the MRS=price ratio
equation.
Ux
px
=
Uy
py
116 / 526
Lagrangian Method and Tangency
• When the individual has maximized utility subject to a linear
constraint, there’s a tangency condition that occurs:
• the person has implicitly set equal the MRS (the slope of the
indifference curve) to the price ratio (the slope of the budget
constraint)
• Interpretation: Utility max implies setting equal the internal
rate of trade (MRS) to the external rate of trade (price ratio).
117 / 526
Tangency I
10
9
U3
8
U2
U1
7
Quantity of y
6
B
5
D
4
C
3
U3
A
2
U2
U1
1
0
I=PxX+PyY
0
1
2
3
4
5
6
7
8
9
10
Quantity of x
118 / 526
Tangency II
10
9
8
7
Quantity of y
6
MRS = Ux/Uy
5
4
3
y*
2
1
Px/Py
0
0
1
2 x*
3
4
5
6
7
8
9
10
Quantity of x
119 / 526
The Lagrange Multiplier λ
• What is the interpretation of the Lagrange multiplier itself?
λ=
•
Ux
px
Uy
Ux
=
px
py
is the “utility per dollar spent” for good x.
• At the utility-maximizing point, each good purchased should
yield the same marginal utility per dollar spent on that good.
120 / 526
The Lagrange Multiplier λ
• If this was not true, then the consumer could reallocate
spending patterns in the direction of the good with the higher
marginal utility per dollar (away from the one with the lower
marginal utility per dollar).
• We almost always have diminishing MU, this would push
down the marginal utility per dollar of the good being more
intensively consumed, and push up the marginal utility per
dollar of the good she’s moving away from.
121 / 526
Tangency Condition is Not Always Met
• For some utility functions the tangency condition is not
satisfied.
• Sometimes there is no tangency because of the shape of the
indifference curves (as with perfect substitutes and perfect
compliments).
• At other times the mathematical tangency point occurs at a
bundle that is “out of bounds” and has negative amounts of a
good in it.
122 / 526
Tangency Condition is Not Always Met:
Quasilinear Utility
10
9
U1
U2
U3
2
3
4
8
7
Quantity of y
6
5
4
3
2
1
0
E
0
1
5
6
7
8
9
10
Quantity of x
• This is an example of Quasilinear Utility
• The IC does bound a convex “weakly preferred-to set”.
• The tangency point would occur outside the possible set of
bundles.
123 / 526
Tangency Condition is Not Always Met:
Quasilinear Utility
10
9
U1
U2
U3
2
3
4
8
7
Quantity of y
6
5
4
3
2
1
0
E
0
1
5
6
7
8
9
10
Quantity of x
• Point E yields highest utility surface that is also affordable,
even though at that point, MRS 6=
px
py .
• This is a “corner solution” because the consumer is optimally
choosing y = 0
124 / 526
Second order condition I
10
9
8
7
Quantity of y
6
5
4
MRS=Px/Py
3
2
1
0
0
1
2
3
4
5
6
7
8
9
10
Quantity of x
• Not all points that satisfy the first order conditions (tangency)
will maximize utility.
• We need to check to make sure that the tangency has found a
max and not a min
125 / 526
Second order condition II
10
9
8
7
Quantity of y
6
5
4
MRS=Px/Py
3
2
1
0
0
1
2
3
4
5
6
7
8
9
10
Quantity of x
• We can check the second order condition to make sure the
Utility function is quasiconcave.
126 / 526
Second order condition II
10
9
8
7
Quantity of y
6
5
4
MRS=Px/Py
3
2
1
0
0
1
2
3
4
5
6
7
8
9
10
Quantity of x
• You can also check for the convexity of the IC or the “weakly
preferred to set” to ensure quasiconcavity of the utility
function.
127 / 526
Second order condition III
• We know that we have quasiconcavity if
Uxx Uy2 − 2Uxy Ux Uy + Uyy Ux2 < 0
• This formula checks the function in all directions to make sure
the curvature meets our requirements.
• We are checking the “principal leading minors of the bordered
Hessian”.
128 / 526
Second order condition IV
• The bordered Hessian comes from a rewriting and
manipulation of the FOC’s into matrix form.
• A max will be negative definite with alternating leading
minors starting with the second leading minor


0
−px −py
Hb = −px Uxx Uxy 
−py Uyx Uyy
• Leading minors are:
• First leading minor is zero
2
• Second leading minor is −p
x <0
p2
• Third leading minor is − Uy2
y
Uxx Uy2 − 2Uxy Ux Uy + Uyy Ux2
which is positive if the utility function is quasi concave
129 / 526
Example I: U(x, y ) = ln x + y (I)
• Assume a consumer maximizes
U(x, y ) = ln x + y
subject to a linear budget constraint
I = px x + py y
• The Lagrangian is:
L = ln x + y + λ(I − px x − py y )
130 / 526
Example I: U(x, y ) = ln x + y (II)
L = ln x + y + λ(I − px x − py y )
• First order conditions
(FOC 1)
(FOC 2)
(FOC 3)
dL
dx
dL
dy
dL
dλ
=
1
− λpx = 0
x
= 1 − λpy = 0
= I − px x − py y = 0
131 / 526
Example I: U(x, y ) = ln x + y (III)
• Solve (FOC1) and (FOC2) for λ
(FOC 10 ) λ = x −1 px−1
(FOC 20 ) λ = py−1
132 / 526
Example I: U(x, y ) = ln x + y (IV)
• Set (FOC1’) = (FOC2’) and solve:
λ = λ
x −1 px−1
xpx
∗
= py−1
= py
x (px , py , I ) = py px−1
(∗)
• Normally the x (or y) you find in this step still contains y (or
x). So it is not the uncompensated demand function.
• For this particular quasilinear case it is the uncompensated
demand because the optimal amount of x does not depend on
y.
133 / 526
Example I: U(x, y ) = ln x + y (V)
• To solve for y, substitute (*) into FOC3:
(FOC 3) I − px x − py y
= 0
− px (py px−1 )
I − px px−1 py
− py y
= 0
− py y
= 0
I
I − py
= py y
I − py
y ∗ (px , py , I ) =
py
134 / 526
Example I: U(x, y ) = ln x + y (V)
• To solve for λ, substitute x ∗ or y ∗ into FOC1 or FOC2. Pick
the one that looks easiest to solve.
• Because FOC2 does not have an x or a y , I pick that one to
“substitute into” and solve for λ.
1 − λpy
= 0
(FOC 2)
1 = λpy
λ = py−1
135 / 526
Example I: U(x, y ) = ln x + y (VI)
• To summarize, for:
L = ln x + y + λ(I − px x − py y )
• These are our demand functions x ∗ , y ∗ and the marginal
utility of income λ∗ .
y ∗ (px , py , I ) =
I − py
py
x ∗ (px , py , I ) = py px−1
λ∗ (px , py , I ) = py−1
136 / 526
Indirect utility I
• x ∗ (px , py , I ) and y ∗ (px , py , I ) tell us the optimal bundle for
the consumer.
• For any income and prices we can plug them in a find the
consumers optimal bundle.
• To determine the level of utility at the optimal bundle
substitute x ∗ (px , py , I ) and y ∗ (px , py , I ) into U(x, y ).
• The resulting expression is Indirect Utility.
137 / 526
Indirect utility II
• Indirect utility, V (px , py , I ), is the same as maximized utility
subject to the resource constraint.
• It is written this way to look at the way prices and income
indirectly affect the consumer’s utility.
138 / 526
Indirect utility III
• Using the demand functions from the previous example, we
can solve for indirect utility.
• So we have:
U(x, y ) = ln x + y
∗
(1)
x (px , py , I ) =
py px−1
(2)
y ∗ (px , py , I ) =
I − py
py
(3)
• Subbing ??, and ?? into ??:
V (px , py , I ) = ln(px py−1 ) +
I − py
py
• For V (px , py , I ), maximized utility is not a function of goods
consumed and income, but rather prices and income.
• Some situations will call for U(x,y) and others will call for
V (px , py , I ).
139 / 526
Indirect utility IV
• Indirect utility satisfies our FOC’s for maximizing. Look at the
Lagrangian with the optimal values plugged in.
V (px , py , I ) = U[x ∗ (px , py , I ), y ∗ (px , py , I )] +
λ(px , py , I ) I − px x ∗ (px , py , I ) − py y ∗ (px , py , I )
• The three first order conditions hold for this expression:
Ux − λpx = 0
Uy − λpy = 0
I − px x − py y = 0
140 / 526
Indirect Utility and Prices I
• Indirect utility is non-increasing in prices:
• Lets see why:
∂V (px , py , I )
∂px
∂V (px ,py ,I )
∂px
≤0
∂x
∂y
∂λ
+ Uy
+
(I − px x − py y )
∂px
∂px
∂px
∂x
∂y
−λx − λpx
− λpy
∂px
∂px
∂x
∂y
∂λ
= (Ux − λpx )
+ (Uy − λpy )
+ (I − px x − py y )
∂px
∂px
∂px
−λx ∗ (px , py , I )
= Ux
141 / 526
Indirect Utility and Prices II
∂V (px , py , I )
∂px
∂x
∂y
∂λ
= (Ux − λpx )
+ (U − λp )
+ (I − px x − py y )
| {z } ∂px | y {z y } ∂px |
{z
} ∂px
FOC 1
FOC 2
FOC 3
−λx ∗ (px , py , I )
• The three terms in parentheses are our first order conditions, each of which
equals zero when utility equals indirect utility
∂V (px , py , I )
= −λx ∗ (px , py , I ) ≤ 0
∂px
• If the constraint is binding, then λ > 0. If there are no corner solutions, then
x ∗ > 0, and therefore
∂V (px ,py ,I )
∂px
<0
• If the constraint isn’t binding or x ∗ = 0, then
∂V (px ,py ,I )
∂px
=0
142 / 526
Indirect Utility and Income I
• Indirect utility is non-decreasing in income:
∂V (px , py , I )
∂I
∂V (px ,py ,I )
∂I
≥0
∂x
∂y
∂λ
+ Uy
+
(I − px x − py y )
∂I
∂I
∂I
∂x
∂y
+λ(px , py , I ) − λpx
− λpy
∂I
∂I
∂x
∂y
∂λ
= (Ux − λpx )
+ (Uy − λpy )
+ (I − px x − py y )
∂I
∂I
∂I
+λ(px , py , I )
= Ux
143 / 526
Indirect Utility and Income II
∂V (px , py , I )
∂I
∂x
∂y
∂λ
= (Ux − λpx )
+ (Uy − λpy )
+ (I − px x − py y )
| {z } ∂I
| {z } ∂I
|
{z
} ∂I
FOC 1=0
FOC 2=0
FOC 3=0
+λ(px , py , I )
• The three terms in parentheses are our first order conditions equaling zero
when utility equals indirect utility
• We can now see that the reason indirect utility is non-decreasing in income is
∂V (px , py , I )
= λ(px , py , I ) ≥ 0
∂I
• And λ = 0 when the constraint is not binding (and therefore more income
doesn’t matter) but is otherwise positive
144 / 526
Roy’s Identity
∂V
x
• Roy’s identity: x ∗ (px , py , I ) = − ∂p
∂V
∂I
• This shows us we can derive the consumer’s uncompensated
demand by calculating how a consumer’s indirect utility
changes as price and income change.
145 / 526
Example II: U(x, y ) = (xy )1/2
• Consider a simplified version of the Cobb-Douglas utility
function.
1
U(x, y ) = (xy ) 2
• Solve for uncompensated demand functions:
∂L
∂x
∂L
∂y
∂L
∂λ
=
=
1 −1 1
x 2 y 2 − λpx = 0
2
1 1 −1
x 2 y 2 − λpy = 0
2
= I − px x − py y
• Demand functions are x ∗ (px , py , I ) = 2pI , y ∗ (px , py , I ) = 2pI .
x
y
146 / 526
Example II: U(x, y ) = (xy )1/2
• Demand functions are:
I
2px
I
2py
x ∗ (px , py , I ) =
y ∗ (px , py , I ) =
• Sub demand equations into U(x,y).
• Indirect utility is V (px , py , I ) =
I
1/2 1/2
py
2px
147 / 526
Indirect Utility is Homogeneous of Degree
0
• Indirect utility is homogenous of degree 0 in prices and
income.
• That means if you double all the inputs to the function the
output stays the same.
• So for Indirect Utility if we doubled income and prices, the
consumer’s utility would remain the same.
• To illustrate, let’s use our numerical example from earlier
where V (px , py , I ) =
I
2(px py )0.5
• Double prices and income and observe what happens to
indirect utility
V (2px , 2py , 2I ) =
2I
I
=
0.5
2(4px py )
2(px py )0.5
• This means that pure inflation doesn’t harm the consumer
148 / 526
Example III: V (px , py , I ) =
• Recall the last example:
I
2(px py )1/2
V (px , py , I ) =
I
2(px py )1/2
• Double prices and income and observe what happens to
indirect utility
V (2px , 2py , 2I ) =
2I
I
=
1/2
2(4px py )
2(px py )1/2
• One implication is that pure inflation (both prices and income
increase at the same rate) doesn’t harm the consumer.
149 / 526
Indirect Utility and Lump Sums
• Consumers will weakly prefer a lump sum tax to an equivalent
revenue generating sales tax. To simplify, let px = 1, py = 4
and I = 8. This yields x ∗ = 4, y ∗ = 1 and V = 2 utils
• Sales tax
• Tax x at $1 per unit. x ∗ =
V =
8
2·80.5
8
∗
4 = 2, y = 1 and
= 1.41 utils. Tax revenue = x · t = 2 · $1 = $2
• Lump sum income tax
• Tax I at $2. New income = $6, x =
V = 64 = 1.5 utils. Tax revenue $2
6
2
= 3, y =
6
8
= 0.75,
• More utility under lump sum income tax (1.5 utils) than the
sales tax (1.41 utils)
• This would also hold for a subsidy – generally people are
better off under cash transfers than price subsidies of the
same revenue level
150 / 526
Expenditure I
• Total Expenditure is E = px x + py y , that is the total amount
of money a consumer spends on their bundle of goods.
• Note: This is the same formula as the budget constraint
because I = E , consumers use all of their income.
151 / 526
Expenditure II
• Most constrained maximization problems have a “dual”
constrained minimization problem.
• Mathematically, consumers maximizing utility subject to a
fixed budget constraint, solves the same problem as minimizing
expenditure subject to a fixed utility function.
152 / 526
Expenditure III
• This “dual problem” to utility maximization is called
Expenditure Minimization.
• The consumer wants to minimize expenditure E = px x + py y ,
subject to a fixed level of utility Ū = U(x, y )
• When might we use expenditure minimization? Maybe your
family has decided that a certain vacation should produce a
certain amount of utils, and then sets out to find the cheapest
vacation that accomplishes that level of utility
153 / 526
Expenditure IV
• When might we use expenditure minimization?
• Maybe your family has decided that a certain vacation should
produce a certain amount of utils, and then sets out to find
the cheapest vacation that accomplishes that level of utility
154 / 526
Expenditure IV
• The Lagrangean method can also be used as before, with the
objective function and constraints reversed.
L = px x + py y + λ[Ū − U(x, y )]
• The three first order conditions are
(FOC 1)
(FOC 2)
(FOC 3)
∂L
∂x
∂L
∂y
∂L
∂λ
= px − λUx = 0
= py − λUy = 0
= Ū − U(x, y ) = 0
• Now we have a system of 3 equations and 3 unknowns, that
can be solved the same was as before.
• Also note, MRS = ppx , can be seen by combining the first to
y
FOCs
155 / 526
Expenditure V
10
9
8
B
7
quantity of y
6
5
A
4
3
C
2
Fixed utility surface
1
0
E1
0
1
2
3
4
5
E2
6
7
8
9
10
quantity of x
156 / 526
Expenditure Function I
• Solving the system of 3 equations and 3 unknowns will give us:
x ∗ (px , py , Ū)
y ∗ (px , py , Ū)
λ∗ (px , py , Ū)
• These are the compensated demand functions.
• We can substitute them back into the objective function (for
this dual case that is the BC) to get the “expenditure
function”.
157 / 526
Expenditure Function II
• The expenditure function tells us the minimum spending
necessary to achieve a fixed level of utility.
E (px , py , Ū) = px x ∗ (px , py , Ū) + py y ∗ (px , py , Ū)
• Notice that expenditure is a function of px , py and Ū not
income, because in the dual problem, there is no income
explicitly in the function.
158 / 526
Expenditure Function II
• Because expenditure is observed, unlike utility, the
expenditure function is widely used in applied economics; it’s
important therefore to understand some of its properties
• Expenditure functions are homogenous of degree one in all
prices.
• If you double prices then you must double your expenditure to
get the same level of utility (Ū).
159 / 526
Expenditure Function III
10
9
8
E/Py
quantity of y
6
5
4
3
U=10
2
1
Px/Py
0
0
1
2
3
4
5
6
E/Px
8
9
10
quantity of x
160 / 526
E(.) is Non-Decreasing in Prices I
∂E
• Expenditure functions are nondecreasing in prices: ∂p
≥ 0.
x
• In general the expenditure function is:
E (px , py , Ū) = px x ∗ (px , py , Ū) + py y ∗ (px , py , Ū)
+λ∗ (px , py , Ū)(Ū − U[x(px , py , Ū), y (px , py , Ū)])
161 / 526
E(.) is Non-Decreasing in Prices II
• Take the derivative of E () with respect to a price:
∂E
∂px
∂x
∂y
∂λ
∂x
∂y
+ py
+
Ū − U(x, y ) − λUx
− λUy
∂px
∂px
∂px
∂px
∂px
∂x
∂y
∂λ
= x + (px − λUx )
+ (py − λUy )
+ Ū − U(x, y )
∂px
∂py
∂px
= x + px
• Recall our first order conditions are:
(FOC 1)
(FOC 2)
(FOC 3)
∂L
∂x
∂L
∂y
∂L
∂λ
= px − λUx = 0
= py − λUy = 0
= Ū − U(x, y ) = 0
162 / 526
E(.) is Non-Decreasing in Prices III
• Now apply the FOCs to the derivative.
∂E
∂px
∂y
∂λ
∂x
= x + (px − λUx )
+ (p − λU )
+ Ū − U(x, y )
| {z } ∂px | y {z y } ∂py
∂px
|
{z
}
FOC 1=0
FOC 2=0
FOC 3=0
So :
= x ≥0
• So as prices increase, expenditure cannot go down.
∂E
• This is also called Shepherd’s Lemma: ∂p
= x ∗ (px , py , Ū)
x
• It can be very helpful in solving problems.
163 / 526
E(.) is Concave in Prices
2
• Expenditure functions are concave in prices: ∂∂pE2 < 0
x
• We will cover this more later.
∂E
• For now, notice Shepherd’s Lemma, ∂p
= x(px , py , Ū)
x
• Therefore
∂2E
∂px2
=
∂x(px ,py ,Ū)
∂px
which is non-positive by the law of
demand.
164 / 526
Income and Substitution Effects
165 / 526
Demand Functions
166 / 526
Basics of Demand Functions
• Constrained utility maximization gives us uncompensated
demand functions.
x ∗ (px , py , I )
y ∗ (px , py , I )
167 / 526
Properties of Demand Functions:
Homogeneity I
• Demand functions are homogenous of degree zero in all prices
and income.
x ∗ (px , py , I ) = x ∗ (2px , 2py , 2I ) = x ∗ (tpx , tpy , tI )
168 / 526
Properties of Demand Functions:
Homogeneity II
• Numerical example with a Cobb-Douglas uncompensated
demand.
x ∗ (px , py , I ) =
I
2px
• Multiply income and all prices by a positive constant t (here
2).
x(2px , 2py , 2I ) =
2I
I
=
2 · 2px
2px
• The optimal bundle is the same because the budget constraint
didn’t shift.
169 / 526
Properties of Demand Functions:
Homogeneity III
10
9
U1
8
7
Quantity of y
2I/2Py
5
4
y*
2
U1
1
0
2Px/2Py
0
1
2
x*(2Px,2Py,2I)
4
5
2I/2Px
7
8
9
10
Quantity of x
170 / 526
Demand Curves
• We can think of a demand function in the same way we think
about any demand curve.
• Demand curves relate the price of a good to the quantity
demanded for that good.
• The demand curve from Principals classes is the market
demand curve.
• In this case we have an individual demand curve.
• To see the demand curve for x we hold py and I constant.
x ∗ (px , p¯y , I¯)
• So now if you plug in any px for the function would return the
quantity demanded of x
171 / 526
Change in Qty. Demanded v/s Shifts in
Demand
• For the demand curve x ∗ (px , p¯y , I¯)
• Changes to px are movements along the demand curve.
• Changes to p¯y or I¯ shift the entire demand curve.
• Whether it shifts to the right or left will depend on they type
of good it is.
• For I¯: Normal or Inferior
• For p¯y : Substitutes or Compliments.
172 / 526
Normal Goods
• For normal goods, as income increases, quantity demanded
for the good increases.
∂x ∗ (px , py , I )
>0
∂I
or
∂y ∗ (px , py , I )
>0
∂I
• Examples: Just about everything :)
173 / 526
Normal Goods Graphically
10
9
8
U2
7
U1
Quantity of Y
6
I2
5
I1
4
3
2
U2
1
0
I2
U1
I1
0
1
x1 Normality x2
4
5
6
7
8
9
10
Quantity of X
174 / 526
Inferior Goods
• For inferior goods, as income increases, quantity demanded
for the good decreases.
∂x ∗ (px , py , I )
<0
∂I
or
∂y ∗ (px , py , I )
<0
∂I
• Examples: Ramen Noodles, SPAM, pretty much everything
you ate in your dorm freshman year of college.
175 / 526
Inferior Goods: Graphically
10
9
U2
8
7
U1
I2
Y Axis
6
5
I1
4
Normality
3
U2
2
1
0
I2
I1
0
1
2
Inferiority
3
4
5
6
U1
7
8
9
10
X Axis
176 / 526
Changes in a Good’s Own-Price
177 / 526
Uncompensated and Compensated
Demand Functions I
• Utility maximization gives us the uncompensated demand
functions:
x ∗ (px , py , I )
y ∗ (px , py , I )
• Expenditure minimization gives us the compensated
demand functions:
xc∗ (px , py , Ū)
yc∗ (px , py , Ū)
178 / 526
Uncompensated and Compensated
Demand Functions II
• To see one of the difference consider an increase in one of the
prices.
• Uncompensated Demand: demand falls (
so does utility
)
( ∂U(x,y
∂px
< 0), and
< 0).
• Compensated Demand: demand falls (
utility does not
∂x ∗ (px ,py ,I )
∂px
)
( ∂U(x,y
∂px
∂xc∗ (px ,py ,Ū)
∂px
< 0), but
= 0).
179 / 526
An Increase in px on x ∗ (px , py , I ),
(uncompensated demand).
10
9
U2
8
U1
7
Quantity of Y
I/Py
5
4
3
Px
/
2
U2
Py
1
U1
Px'/Py
0
0
1
2
I/Px'
4
5
I/Px
7
8
9
10
Quantity of X
180 / 526
An Increase in px on xc∗ (px , py , Ū),
(compensated demand).
10
9
U1
8
7
Quantity of Y
E/Py
5
4
3
2
U1
1
0
Px/Py
0
1
2
3
4
5
E/Px
7
8
9
10
Quantity of X
181 / 526
Compensated v/s Uncompensated
Demand I
• Compensated demand functions and uncompensated demand
functions capture different aspects of the relationship between
prices, income, quantity demanded and utility.
• Uncompensated demand functions are also called “Marshallian
demand functions” after Alfred Marshall.
• Marshall was an early 20th century British economist.
• He developed much of the utility maximization paradigm in
microeconomics.
• He is also to blame for our “flipped” axes in Supply and
Demand diagrams.
182 / 526
Compensated v/s Uncompensated
Demand II
• Compensated demand functions are also called “Hicksian
demand functions” after Sir John Hicks.
• Hicks was one of the most influential economists of the 20th
century (1972 Nobel Prize for his work on general equilibrium
theory and welfare analysis).
• He studied demand based on expenditure minimization.
183 / 526
A Price Increase with Uncompensated
Demand
• Uncompensated:
∂x ∗ (px ,py ,I )
∂px
captures changes in the quantity
demanded that result from two possible sources.
They consume less x ∗ and more y ∗ , because y ∗ is now cheaper
(relative to x ∗ ) than it was before.
2 They consume less x ∗ because the higher price of x ∗ means
they can by less stuff overall.
1
184 / 526
A Price Increase with Compensated
Demand
• Compensated:
∂xc∗ (px ,py ,Ū)
∂px
captures changes in the quantity
demanded that result from:
1
They consume less x ∗ and more y ∗ , because y ∗ is now cheaper
(relative to x ∗ ) than it was before.
185 / 526
Introduction to Income and Substitution Effects
186 / 526
Two Effects of a Price Change
• For both x ∗ (px , py , I ), and xc∗ (px , py , Ū) the increase in price
led to a decline in the quantity demanded.
• For Uncompensated/Marshallian demand we have two effects
that led to lower quantity demanded:
• Substitution Effect.
• Income Effect
187 / 526
Income and Substitution Effects
(Uncompensated Demand)
• For x ∗ (px , py , I ) and y ∗ (px , py , I ) we have two effects of a
price increase that led to lower quantity demanded:
• Substitution effect: One good you like is now cheaper than it
was before, so you buy less of the good that increased in price
and more of good whose price stayed the same (meaning it
became relatively cheaper).
• Income Effect: The same bundles now cost more with the
higher price. This is like a decrease in income so you buy less
stuff.
188 / 526
The Substitution Effect I
• Why does the substitution effect induce people to buy less of
the good whose price increased?
• Recall that when you have a quasiconcave utility function you
must have diminishing MRS.
• An increase in px means the slope of the BC ( ppx ) must get
y
steeper.
• At that point MRS 6=
px
py
.
189 / 526
The Substitution Effect II
• In equilibrium the consumer chooses where MRS = ppx .
y
• MRS is diminishing as we move along the indifference curve,
so the only way to get back to MRS = ppyx is to decrease the
amount of x ∗ in the consumer’s optimal bundle.
• So the consumer has to move to a bundle on a different
indifference
curve in oder to be back at a bundle where
0
MRS = ppyx .
190 / 526
The Income Effect
• Uncompensated/Marshallian demand also has an Income
Effect.
• The increase in the price of x is a loss of “real income”.
• You can only buy smaller bundles now which is the same as a
loss of income.
• This means the optimal bundle will be on a lower indifference
curve.
∂V
• Recall for indirect utility: ∂p
≤ 0.
x
• For normal goods the income and substitution effects reinforce
each other.
• The substitution from x to y and the decline in “real income”
both induce you to buy less x.
191 / 526
Thought Experiment Price Decrease
• A consumer’s uncompensated demands are x ∗ (px , py , I )
x ∗ (px , py , I ) and the price of x is pxH .
• The price falls to pxL < pxH so the optimal bundle the
consumer had chosen now costs less.
• The consumer now has to re-optimize to spend all their
money and maximize utility.
192 / 526
Thought Experiment Price Decrease
• The consumer now has to re-optimize to spend all their
money and maximize utility.
• Consider the two possible things the consumer might do with
this new “extra money”
The consumer could take all their new extra money and buy
more of x ∗ (px , py , I ).
2 The consumer could take all their new extra money and buy
more of y ∗ (px , py , I ).
1
193 / 526
Two Effects of a Price Decrease (Graph)
10
U1
9
8
7
Qty of Good Y
I/py
6
5
4
3
2
H xx+
I=p
1
p yy
0
0
1
2
3
4
I/pHx
5
6
7
8
9
10
Qty of Good X
Budget Constraint with original prices.
194 / 526
Two Effects of a Price Decrease (Graph)
10
U1
9
8
7
Qty of Good Y
I/py
6
5
4
3
2
H xx+
I=p
1
p yy
0
0
1
2
3
4
I/pHx
5
6
7
8
9
10
Qty of Good X
Optimal Bundle at original prices.
195 / 526
Two Effects of a Price Decrease (Graph)
10
U1
9
8
7
Qty of Good Y
I/py
6
I=p L
x x+
py
y
5
4
3
2
H xx+
I=p
1
p yy
0
0
1
2
3
4
I/pHx
5
6
7
Qty of Good X
8
9
I/pLx
10
Price Change: px falls to a lower price pxL .
196 / 526
Two Effects of a Price Decrease (Graph)
10
U2
U1
9
8
7
Qty of Good Y
I/py
6
I=p L
x x+
py
y
5
4
3
2
H xx+
I=p
1
p yy
0
0
1
2
3
4
I/pHx
5
6
Qty of Good X
7
8
9
I/pLx
10
Optimal Bundle at new prices. Note the total change in x.
197 / 526
Two Effects of a Price Decrease (Graph)
10
U2
U1
9
8
7
Qty of Good Y
I/py
6
I=p L
x x+
py
y
5
4
3
2
H xx+
I=p
1
p yy
I h=
pL
x x+
0
py
y
0
1
2
3
4
I/pHx
5
6
Qty of Good X
7
8
9
I/pLx
10
1) Find Hypothetical BC: Income to buy old level of utility at new
prices.
198 / 526
Two Effects of a Price Decrease (Graph)
10
U2
U1
9
8
7
Qty of Good Y
I/py
6
I=p L
x x+
py
y
5
4
3
2
H xx+
I=p
1
p yy
I h=
pL
x x+
0
py
y
0
1
2
3
4
I/pHx
5
6
Qty of Good X
7
8
9
I/pLx
10
2) Find the Hypothetical Bundle. Old IC and Hypothetical BC.
199 / 526
Two Effects of a Price Decrease (Graph)
10
U2
U1
9
8
7
Qty of Good Y
I/py
6
I=p L
x x+
py
y
5
4
3
2
H xx+
I=p
1
0
Inc
I h=
pL
x x+
p yy
Sub.
py
y
0
1
2
3
4
I/pHx
5
6
Qty of Good X
7
8
9
I/pLx
10
3) Decompose Income and Substitution Effect w/ Hypothetical
Bundle
200 / 526
Normal and Inferior Goods
201 / 526
The Law of Demand and Normal Goods
(Uncompensated) I
• For Marshallian/uncompensated demand the law of demand
states that:
∂x ∗ (p ,p ,I )
x y
• For a normal good:
>0
∂I
• An increase in price will lead to a decrease in quantity
demanded:
∂x ∗ (px ,py ,I )
∂px
≤0
• The income and substitution effects both push quantity
demanded down.
202 / 526
The Law of Demand and Inferior Goods
(Uncompensated) II
• For Marshallian/uncompensated demand the law of demand
states that:
∂x ∗ (p ,p ,I )
x y
• For an inferior good:
<0
∂I
• The income effect and substitution effect go in different
directions.
• So unless we know if a good is inferior or normal there is some
ambiguity to whether increases in price will lower
Marshallian/uncompensated demand.
203 / 526
Example: The Law of Demand
x ∗ (px , py , I ) = 2pI x
• For the uncompensated demand function, x ∗ (px , py , I ) = 2pI
x
• Normal or Inferior?
∂x
1
=
>0
∂I
2px
• It is normal so we should expect the law of demand to hold.
∂x
I
=−
<0
∂px
2px 2
204 / 526
The Law of Demand and Normal Goods
(Compensated) I
• For Hicksian/compensated demand the law of demand states
that:
∂x ∗ (p ,p ,Ū)
x y
• For all goods c ∂p
≤0
x
• When price rises, utility is fixed, and an indifference curve is
convex.
• So as the BC changes slope the optimal bundle will have less
of the good whose price increased.
• This causes a move along the same indifference curve that
corresponds to a decline in quantity demanded of the good
whose price increased.
205 / 526
Price changes IX
10
9
U1
8
7
Quantity of Y
E/Py
5
4
3
2
U1
1
0
Px/Py
0
1
2
3
4
5
E/Px
7
8
9
10
Quantity of X
206 / 526
The Law of Demand and Normal Goods
(Compensated) II
• One common exception to a downward sloping Hicksian
demand curve is compliments U(x, y ) = min(αx, βy ).
• You always consume at the vertices of the I.C., so a rotation
in the BC leaves demand unchanged
• Note though that while consumption is unchanged when price
rises for the L-shaped indifference curve, expenditure rises.
207 / 526
Price changes X
10
9
U1
8
E2
7
Quantity of Y
6
5
E1
4
3
U1
Yc(Px,Py,U1)
1
Px'/Py
0
0
1
Xc(Px,Py,U1)
3
Px/Py
4
5
6
7
8
9
10
Quantity of X
208 / 526
Comparative Statics
209 / 526
Comparative statics I
• In general, a Comparative Static looks at how equilibrium
values change when you change something about a system.
• Consider the equilibrium size of a simple cake.
• If you add one more egg what happens to the size of the cake?
210 / 526
Comparative statics II
• The “statics” part refers to the fact that a comparison is
made after all adjustments have occurred.
• The “comparative” term refers to a before and after
comparison of the equilibrium values generated by slightly
different inputs.
• Such before and after comparisons can provide testable
predictions and policy implications of economic models.
• At this point what we care about is the sign of the change.
• Example: We what to know if
∂x ∗ (px ,py ,I )
∂I
> 0 because it is a
test of the normality of a good.
211 / 526
Comparative statics III
• One approach:
• Find the optimal bundle.
• Take
the derivatives.
∂x ∗ (px ,py ,I ) ∂x ∗ (px ,py ,I )
,
, or
∂px
∂py
∂y ∗ (px ,py ,I ) ∂y ∗ (px ,py ,I )
,
,
∂px
∂py
or
∂x ∗ (px ,py ,I )
∂I
∂y ∗ (px ,py ,I )
∂I
• We will indeed do this when working with specific utility
functions.
• However, there is a more general approach that will work with
any constrained optimaization.
212 / 526
Comparative Statics IV
• Recall the Lagrangean for expenditure min:
L = px x + py y + λ[Ū − U(x, y )]
• The three first order conditions are:
(FOC 1)
(FOC 2)
(FOC 3)
∂L
∂x
∂L
∂y
∂L
∂λ
= px − λUx = 0
= py − λUy = 0
= Ū − U(x, y ) = 0
• Now we have a system of 3 equations and 3 unknowns, but we
have not yet solved for the compensated demand functions.
213 / 526
Comparative Statics V
(FOC 1)
(FOC 2)
(FOC 3)
∂L
∂x
∂L
∂y
∂L
∂λ
= px − λUx = 0
= py − λUy = 0
= Ū − U(x, y ) = 0
• This system is not solved, but it is a mathematically
equivalent representation of the demand functions.
• It is a statement of the consumer’s optimal/equilibrium choice.
• Instead of many different partial derivatives of each demand
function, we can totally differentiate the whole system at once.
214 / 526
Comparative Statics VI
• Take the system of 3 equations and 3 unknowns.
px − λUx
= 0
py − λUy
= 0
Ū − U(x, y ) = 0
• Totally differentiate each equation:
dpx − λUxx dx − λUxy dy − Ux dλ = 0
dpy − λUyx dx − λUyy dy − Uy dλ = 0
dU − Ux dx − Uy dy
= 0
• It will help to rewrite this system in matrix form.
215 / 526
Comparative Statics VII
dpx − λUxx dx − λUxy dy − Ux dλ = 0
dpy − λUyx dx − λUyy dy − Uy dλ = 0
dU − Ux dx − Uy dy
= 0
• Rewrite the system of totally differentiated FOCs in Ax = d matrix form.

−λUxx
−λUyx
−Ux
−λUxy
−λUyy
−Uy
  

−dpx
−Ux
dx
−Uy   dy  = −dpy 
0
dλ
−dU
• It is exactly the same system of equations, but we this form is easier to work
with.
• Now we can solve for any of the comparative statics we want to look at.
216 / 526
Comparative statics VIII
What about
∂xc∗ (px ,py ,Ū)
?
∂px
ceteris paribus?
• For this particular comparative static we can sub in some
things to the A matrix.
• None of the other variables will change so all the other
derivatives can be set to zero.
• We can also make things simpler by setting px = 1

−λUxx
−λUyx
−Ux
−λUxy
−λUyy
−Uy
   
−Ux
dx
−1
−Uy   0  =  0 
0
0
0
217 / 526
Comparative statics IX
• Now we can use Cramer’s Rule to solve for

−λUxx
A = −λUyx
−Ux
−λUxy
−λUyy
−Uy
∂xc∗ (px ,py ,Ū)
∂px

−Ux
−Uy 
0
• The primary determinant is:
|A| = λ(Uxx Uy2 − 2Uxy Ux Uy + Uyy Ux2 ) < 0
218 / 526
Comparative statics X
• Following the Cramer’s Rule method we substitute the
exogenous right hand side column vector into the first column
of the primary matrix so that we can calculate |A1 |


−1 λUxy −Ux
A1 =  0 λUyy −Uy 
0 −Uy
0
219 / 526
Comparative statics X
• You can use the “augment” method to find the determinant
by repeating column’s 1 and 2

−1 λUxy −Ux
 0 λUyy −Uy
0 −Uy
0

−1 λUxy
0 λUyy 
0 −Uy
|A1 | = (−1)(λUyy )(0) + (λUxy )(−Uy )(0) + (−Ux )(0)(−Uy ) −
(0)(0)(λUxy ) − (−Uy )(−Uy )(−1) − (0)(λUyy )(−Ux ) = Uy2
220 / 526
Comparative statics XI
• You can also use the ”expansion method” of finding the
derivative.
• If you expand by first row (-1,0,0) makes that relatively simple:
|A1 | = −1
λUyy
−Uy
−Uy
= Uy2 > 0
0
U2
∂xc
• To calculate ∂p
= |A|y = (+)
(−) < 0.
x
• As expected compensated demand curves slope downward (it
only has a substitution effect).
221 / 526
The Own Price Slutsky Equation
222 / 526
Decomposing a Price Change I
• For compensated demand its always the case that
∂xc∗ (px ,py ,Ū)
∂px
<0
• But for uncompensated
∂x ∗ (px ,py ,I )
∂px
R0
• This is because the income and substitution effects may go in
opposite directions.
• We would like a more precise and general way of decomposing
the income and substitution effects than the graphical or
“hypothetical bundle” approach.
223 / 526
Decomposing a Price Change II
• We can use three facts we already know to pull apart or
“decompose”
•
∂x ∗ (px ,py ,I )
∂px
xc∗ (px ,py ,Ū)
perfectly
∂px
∂x ∗ (px ,py ,I )
.
∂px
captures the substitution effect part of
• At the optimal bundle I = E (px , py , Ū) because they
(ex)spend all of their income.
• At the optimal bundle xc∗ (px , py , Ū) = x ∗ (px , py , I )
224 / 526
xc∗ (px , py , Ū) = x ∗ (px , py , I )
10
9
8
7
Price of X
Px'
5
4
Px''
3
Px'''
X(Px,Py,I)
1
Xc(Px,Py,U)
substitution effect
0
0
1
x'
2 xc'
income effect
x=xc
substitution effect
4
xc'''
x'''
income effect
6
7
8
9
Quantity of X
225 / 526
Deriving the Own Price Slutsky Equation I
• So lets start at that optimal bundle by setting our two types
of demand equal.
xc (px , py , Ū) = x(px , py , I )
• Which is only true when I = E (px , py , Ū)
xc (px , py , Ū) = x[px , py , E (px , py , Ū)]
• Take derivative with respect to px .
∂x(px , py , I )
∂x(px , py , I ) ∂E (px , py , Ū)
∂xc (px , Py , Ū)
=
+
∂px
∂px
∂px
∂E (px , py , Ū)
226 / 526
Deriving the Own Price Slutsky Equation
II
• Lets clean up the notation a bit.
∂xc (.)
∂x(.)
∂x(.) ∂E (.)
=
+
∂px
∂px
∂E (.) ∂px
∂E
• An identity called Shepherd’s lemma is ∂p
= xc∗ (.)
x
∂xc (.)
∂x(.)
∂x(.)
=
+
xc (.)
∂px
∂px
∂E (.)
• Use E (px , py , Ū) = I again.
∂xc (.)
∂x(.) ∂x(.)
=
+
xc (.)
∂px
∂px
∂I
227 / 526
Deriving the Own Price Slutsky Equation
III
∂xc (.)
∂x(.) ∂x(.)
=
+
xc (.)
∂px
∂px
∂I
• Again use xc (px , py , Ū) = x(px , py , I ).
∂xc (.)
∂x(.) ∂x(.)
=
+
x(.)
∂px
∂px
∂I
• Now solve for the derivative we care about to get the Slutsky
equation.
∂x(.)
∂xc (.) ∂x(.)
=
−
x(.)
∂px
∂px
∂I
228 / 526
Own Price Slutsky Equation I
• The Slutsky equation decomposes the total effect of a price
change into two effects: the substitution effect and the
income effect.
∂xc (.)
∂x(.)
∂x(.)
=
− x(.)
∂p
∂p
| {z∂I }
| {zx }
| {zx }
Total
Substitution
Income
• We can sign some of these terms already.
∂x(.)
∂xc (.)
∂x(.)
=
− x(.)
|{z} | ∂I
∂px
∂px
{z }
| {z }
+
−
?
229 / 526
Own Price Slutsky Equation II
• If ∂x
∂I > 0 (normal good)
∂x(.)
∂xc (.)
∂x(.)
=
− x(.)
<0
|{z} | ∂I
∂px
∂px
{z
}
| {z }
+
−
+
• Then we know the good follows the law of demand and is not
a Giffen Good.
230 / 526
Own Price Slutsky Equation III
• If ∂x(.)
∂I < 0 (inferior good)
∂x(.)
∂xc (.)
∂x(.)
=
Q0
− x(.)
|{z} | ∂I
∂px
∂p
{z
}
| {zx }
+
−
−
∂x
• We still don’t know the overall effect of ∂p
.
x
c
∂x(.)
∂x (.)
∂x(.)
∂x(.)
• If ∂I < 0 and x(.) ∂I < ∂p then ∂p < 0
x
x
• But if
∂x
∂I
< 0 and x(.) ∂x(.)
∂I >
∂x c (.)
∂px
then
∂x
∂px
>0
• So if a good is very inferior, that income effect can overpower
the substitution effect and we can see a violation of the law of
demand.
231 / 526
Giffen Goods
• Legend has it the 19th century English economist, Robert
Giffen, observed this paradox first hand when the price of
potatoes rose during the Irish potato famine and people
consumed more potatoes.
• But Sherwin Rosen showed that Giffen, assuming the legend is
even true in the first place, was wrong. See Rosen (1999)
• Evidence for the upward sloping demand curve would not be
produced until Jensen and Miller2 used a randomized clinical
trial in rural China with randomized subsidies for wheat and
rice to demonstrate Giffen Behavior.
2
Jensen, Robert T., and Nolan H. Miller. 2008. ”Giffen Behavior and
Subsistence Consumption.” American Economic Review, 98(4): 1553-77.
232 / 526
Elasticity
233 / 526
Why Elasticity I
• We have computed and looked at and even decomposed
several ways that prices and income can affect demand.
∂y (.)
• So far we can’t compare ∂x(.)
∂px and ∂py .
• Those derivatives report the change in demand in the units of
the good.
• We have no way to compare a consumer’s response to a
change in different goods.
• We need a unit free measure of the response of one value (like
Qty x) to the change in another value (like px ).
234 / 526
Why Elasticity II
• Example: Good x is gallons gasoline, Good y is liters of helium
gas.
• py = 3, px = 2, I = 1000, and x ∗ (px , py , I ) =
∂x(.)
∂px
I
3px
= − 3pI 2 = − 1000
12 = −83.33̄ Gallons
x
• If py = 3, px = 1, I = 1000 and y ∗ (px , py , I ) =
∂y (.)
∂py
2I
= − 3p
2 =
y
−2000
27
2I
3py
¯ Liters
= −74.07
• Which good was more responsive to the increase in it’s price?
• We have no way to know.
235 / 526
Why Elasticity II
• Converting our derivatives into ”elasticities” will give us a unit
free and comparable measure of the response of something
like quantity demanded to a goods price.
• Thankfully it is pretty easy construct an elasticity. You just
multiply the derivative by a particular fraction.
236 / 526
Demand Elasticities I
1
Price elasticity of Marshallian demand: ex,px =
2
Income elasticity of demand: ex,I =
3
Cross-price elasticity of demand:
∂x(px ,py ,I )
∂px
∂x(px ,py ,I ) I
·x
∂I
∂x(px ,py ,I )
ex,py =
∂py
·
·
px
x
py
x
237 / 526
Demand Elasticities II
• Example: Good x is gallons gasoline, Good y is liters of helium
gas.
• py = 3, px = 2, I = 1000 and
x ∗ (px , py , I ) =
I
3px
=
ex,px
1000
6
= 166.66̄
∂x(.) px
·
∂px
x
I
px
= − 2·
3px x
2
1000
·
= −
=1
12 166.6̄
=
• A 1% increase in px leads to a 1% decrease in the gallons of
gasoline demanded.
238 / 526
Demand Elasticities III
• Example: Good x is gallons gasoline, Good y is liters of helium
gas.
• py = 3, px = 2, I = 1000 and
y ∗ (px , py , I ) =
2I
3py
ey ,py
=
2000
9
= 222.2/bar 2
∂y (.) py
·
∂py
y
2I py
= − 2 ·
3py y
2000
3
·
= −
= 1%
27 222.22̄
=
• A 1% increase in py leads to a 1% decrease in the liters of
helium demanded.
239 / 526
Interpreting Elasticity
• ex,px < −1, then uncompensated demand is “elastic”.
• ex,px = −1, then uncompensated demand is “unit elastic”.
• ex,px > −1, then uncompensated demand is “inelastic”.
240 / 526
Spending elasticities I
• Total spending on x: S = px x ∗ (px , py , I )
• What happens to total spending when price goes up?
∂S
∂px
∂x
= x + px
∂px
px ∂x
= x 1+
x ∂px
= x 1 + ex,px
• Notice that under normal conditions, ex,px < 0, therefore the
interior of the brackets can be positive or negative.
241 / 526
Spending elasticities II
• When the good is “elastic”, |ex,px | > 1 an increase in price
reduces spending
• When the good is “inelastic”, 0 ≤ |ex,px | < 1, an increase in
price increases spending
• When the good is unit elastic, |ex,px | = 1, an increase in price
leaves spending unchanged
242 / 526
Compensated Demand Elasticity
• The calculation and interpretation for compensated demand
are similar.
Compensated own-price elasticity of demand:
∂x (px ,py ,Ū) px
exc ,px = c ∂p
· xc
x
2 Compensated cross-price elasticity of demand:
∂x (px ,py ,Ū) py
exc ,py = c ∂p
· xc
y
1
243 / 526
Slutsky in Elasticity Form I
∂x(.)
∂xc (.) ∂x(.)
=
−
x(.)
∂px
∂px
∂I
• Multiply both sides by pxx
∂x(.) px
∂xc (.) px
∂x(.)
px
=
−
x(.)
∂px x
∂px x
∂I
x
244 / 526
Slutsky in Elasticity Form II
∂x(.) px
∂xc (.) px
∂x(.)
px
=
−
x(.)
∂px x
∂px x
∂I
x
• Multiply right-hand-side second term by II
∂x(.) px
∂xc (.) px
∂x(.)
px I
=
−
x(.)
∂px x
∂px x
∂I
x I
245 / 526
Slutsky in Elasticity Form III
∂x(.) px
∂xc (.) px
∂x(.)
px I
=
−
x(.)
∂px x
∂px x
∂I
xI
• Share of income a consumer spends on x: sx = pxI x
∂x(.) px
∂xc (.) px
∂x(.) I
=
−
sx
∂px x
∂px x
∂I
x
246 / 526
Slutsky in Elasticity Form IV
∂x(.) px
∂xc (.) px
∂x(.) I
=
−
sx
∂px x
∂px x
∂I x
• Now substitute in all the elasticities.
ex,px = exc ,px − sx ex,I
• This is the Slutsky Equation in elasticity form.
247 / 526
Relationships Between Elasticities
248 / 526
Demand Homogeneity and Elasticity
• Homogeneity: ex,px + ex,py + ex,I = 0
• Because demand is homogeneous of degree zero3 .
• All three elasticities of uncompensated demand must sum to 0.
3 ∗
x (tpx , tpy , tI ) = t k=0 x ∗ (px , py , I )
249 / 526
Deriving Demand Homogeneity and
Elasticity
• For uncompensated demand: x(tpx , tpy , tI ) = t 0 x(px , py , I )
• Differentiate with respect to t
∂x(tpx , tpy , tI )
∂t
∂x ∂tpx
∂x ∂tpy
∂x ∂tI
·
+
·
+
·
∂px
∂t
∂py
∂t
∂I ∂t
∂x
∂x
∂x
· px +
· py +
·I
∂px
∂py
∂I
=
∂t 0 x(px , py , I )
∂t
= 0
= 0
250 / 526
Deriving Demand Homogeneity and
Elasticity
∂x
∂x
∂x
· px +
· py +
·I =0
∂px
∂py
∂I
• Multiply both sides through by x1
∂x px
∂x py
∂x I
·
+
·
+
·
∂px x
∂py x
∂I x
ex,px + ex,py + ex,I
= 0
= 0
251 / 526
Engle Aggregation
• Engel Aggregation: sx ex,I + sy ey ,I = 1
• If you weight the share of income you spend on each good, by
the income elasticity for that good, then the sum of those
values will be 1.
• If we have the elasticity for one of the goods we can easily
estimate the elasticity for the other good.
252 / 526
Deriving Engle Aggregation
• Differentiate the budget constraint with respect to I
∂px x ∗ (px , py , I ) ∂py y ∗ (px , py , I )
+
∂I
∂I
∂y (.)
∂x(.)
+ py
1 = px
∂I
∂I
∂I
∂I
=
yI
• Multiply the first term by xI
xI and second term by yI
xI ∂x(.)
yI ∂y (.)
+ py
xI ∂I
yI ∂I
rearrange
x ∂x(.) I
y ∂y (.) I
1 = px
+ py
I ∂I x
I ∂I y
py y ∂y (.) I
px x ∂x(.) I
1 =
+
I
∂I x
I
∂I y
1 = sx ex,I + sy ey ,I
1 = px
253 / 526
Cournot Aggregation:
• sx ex,px + sy ey ,px = −sx
• sy ey ,py + sx ex,py = −sy
• Lets sign what we can assuming x is not a Giffen Good.
sx ex,px + sy ey ,px = − sx
|{z}
|{z}
|{z} |{z} |{z}
+
+
+
−
?
| {z }
| {z }
−
−
• If ex,py is negative (complementary goods) or zero the equation
will hold and be fine.
• If ex,py is positive (substitutable goods) the elasticity cannot
be so large as to violate the equation.
254 / 526
Deriving Cournot Aggregation I
• Start with the budget constraint with uncompensated demand
functions plugged in.
• This is not the same thing as the Expenditure Function.
• The expenditure function E() is the BC with the compensated
demand functions plugged in.
I = px x ∗ (px , py , I ) + py y ∗ (px , py , I )
• Differentiate the budget constraint with respect to px
∂px x ∗ (px , py , I ) ∂py y ∗ (px , py , I )
+
∂px
∂px
product rule
∂x ∗ (px , py , I )
0 = 1 · x ∗ (px , py , I ) +
· px
∂px
∂y ∗ (px , py , I )
+ 0 · y ∗ (px , py , I ) +
· py
∂px
∂I
∂px
=
255 / 526
Deriving Cournot Aggregation II
0 = x ∗ (.) +
∂y ∗ (.)
∂x ∗ (.)
px +
py
∂px
∂px
• Multiply this equation by pIx
0·
px
px
∂x ∗ (.)
px
∂y ∗ (.)
px
= x ∗ (.) ·
+
px ·
+
py ·
I
I
∂px
I
∂px
I
• Use xx and yy to get elasticities.
0 = x ∗ (.) ·
px
∂x ∗ (.)
px x
∂y ∗ (.)
px y
+
px ·
· +
py ·
·
I
∂px
I x
∂px
I y
• Rearrange to get:
0=
px x ∗ (.) ∂x ∗ (.) px xpx
∂y ∗ (.) px py y
+
·
·
+
·
·
I
∂px
x
I
∂px
y
I
256 / 526
Deriving Cournot Aggregation III
0=
xpx ∂x ∗ (.) px
px x ∗ (.) py y ∂y ∗ (.) px
+
+
I ∂px x
I
I ∂px y
• Substitutions:
0 = sx ex,px + sx + sy ey ,px
257 / 526
Consumer Welfare
258 / 526
Consumer Welfare and Expenditure
• We can use the expenditure function to look at several
measures of consumer welfare.
• The concept is called ”Compensating Variation”
• It asks “When the price of a good changes, how much money
do I have to give you (or take away from you) so you will have
the same utility as before the price change?”
259 / 526
Consumer Welfare and Expenditure
CV = E (px1 , py , U¯0 ) − E (px0 , py , U¯0 )
• px0 is the original price of x, px1 is the new price of x.
• U0 is the level of utility the consumer received at the original
prices.
260 / 526
Demand Relationships Among Goods
261 / 526
Cross Price Effects
262 / 526
From Own-Price to Cross-Price Effects
• Most of our analysis of our demand functions thus far has
been looking at “own-price” effects.
• We can apply a similar approach to study cross-price effects.
263 / 526
Decomposing a Cross Price Change
• We can use three facts we already know to pull apart or
“decompose”
1
∂x ∗ (px ,py ,I )
∂py
xc∗ (px ,py ,Ū)
perfectly
∂py
∂x ∗ (px ,py ,I )
.
∂py
captures the substitution effect part of
At the optimal bundle I = E (px , py , Ū) because they
(ex)spend all of their income.
3 At the optimal bundle xc∗ (px , py , Ū) = x ∗ (px , py , I )
2
264 / 526
Deriving the Cross Price Slutsky Equation
I
• So lets start at that optimal bundle by setting our two types
of demand equal.
xc (px , py , Ū) = x(px , py , I )
• Which is only true when I = E (px , py , Ū)
xc (px , py , Ū) = x[px , py , E (px , py , Ū)]
• Take derivative with respect to py .
∂x(px , py , I )
∂x(px , py , I ) ∂E (px , py , Ū)
∂xc (px , Py , Ū)
=
+
∂py
∂py
∂py
∂E (px , py , Ū)
265 / 526
Deriving the Cross Price Slutsky Equation
II
• Lets clean up the notation a bit.
∂x(.)
∂x(.) ∂E (.)
∂xc (.)
=
+
∂py
∂py
∂E (.) ∂py
∂E
• An identity called Shepherd’s lemma is ∂p
= yc (.)
y
∂xc (.)
∂x(.)
∂x(.)
=
+
yc (.)
∂py
∂py
∂E (.)
• Use E (px , py , Ū) = I again.
∂xc (.)
∂x(.) ∂x(.)
=
+
xc (.)
∂py
∂py
∂I
266 / 526
Deriving the Cross Price Slutsky Equation
III
∂xc (.)
∂x(.) ∂x(.)
=
+
xc (.)
∂py
∂py
∂I
• Again use xc (px , py , Ū) = x(px , py , I ).
∂xc (.)
∂x(.) ∂x(.)
=
+
x(.)
∂py
∂py
∂I
• Now solve for the derivative we care about to get the Slutsky
equation.
∂xc (.) ∂x(.)
∂x(.)
x(.)
=
−
∂py
∂py
∂I
267 / 526
Cross Price Slutsky Equation
∂x ∗ (px , py , I )
∂x ∗ (px , py , Ū)
∂x
= c
− y ∗ (px , py , I ) ·
∂py
∂py
∂I
• The first term on the right is the substitution effect and the
second term is the income effect the same as it was for the
own-price Slutsky.
268 / 526
Cross Price Elasticity Slutsky Equation I
• Just as we did with the own-price Slutsky, we can express this
as an elasticity.
ex,py = exc ,py − sy ex,I
• It would be a good exercise for you to derive the the elasticity
form the derivative form just as we did with the own-price
Slutsky.
269 / 526
Cross Price Elasticity Slutsky Equation II
ex,py = exc ,py − sy ex,I
• Note: The size of the income effect is determined by the share
of income spent on good y .
• The impact of a change in py on x ∗ () is determined by how
important y is to this person
• If we have convex indifference curves, the cross-price
substitution effect will be positive.
• Draw a simple IC graph with a Cobb-Douglas looking IC.
Change the price of y and see if you can see why this must be
the case.
270 / 526
Substitutes and Compliments
271 / 526
Gross Substitutes and Compliments I
• Cross price effects tell us if two goods are substitutes,
compliments or neither.
• Two goods, x and y are:
• Gross substitutes if
• Gross complements
∂x ∗ (px ,py ,I )
>0
∂py
∂x ∗ (px ,py ,I )
<
if
∂py
0
• Careful though these “Gross” definitions are not symmetric.
• The sign of
∂x ∗ (px ,py ,I )
,
∂py
and
∂y ∗ (px ,py ,I )
∂px
can be different.
272 / 526
Gross Substitutes and Compliments II
• Consider the case of U(x, y ) = ln x + y .
• The uncompensated demand functions are:
y ∗ (px , py , I ) =
x ∗ (px , py , I ) =
I − py
py
py
px
273 / 526
Gross Substitutes and Compliments III
• Lets check to see if x and y are gross complements or
substitutes?
∂x ∗ (px , py , I )
∂py
∗
∂y (px , py , I )
∂px
=
1
> 0 (Gross Substitutes)
px
= 0
• Asymmetry is an artifact of the uncompensated demand
function.
274 / 526
Net Substitutes and Compliments I
• Two goods, x and y are:
• Net substitutes if
• Net complements
∂xc∗ (px ,py ,I )
>0
∂py
∂xc∗ (px ,py ,I )
if
<
∂py
0
• This definition is symmetric:
∂xc∗ (px , py , I )
∂y ∗ (px , py , I )
= c
∂py
∂px
275 / 526
Net Substitutes and Compliments II
1
1
2
Ūp 2
• Ex: yc (px , py , Ū) = Ūp1x and xc (px , py , Ū) = 1y
py2
px2
∂yc
∂px
=
∂xc
∂py
=
Ū
1
2(px py ) 2
Ū
1
2(px py ) 2
276 / 526
Net Substitutes and Compliments III
∂xc
∂yc
=
∂py
∂px
• Can also be seen using Young’s Theorem (fxy = fyx ):
Exy
∂ ∂E∂x(.)
∂y
(Apply
∂xc∗ (.)
∂y
=
=
Shephard 0 s
=
Eyx
∂ ∂E∂y(.)
∂x
Lemma)
∂yc∗ (.)
∂x
277 / 526
Substitutability with Many Goods
278 / 526
More Than Two Goods I
• For many things in Micro Theory we can get by with only 2
goods.
• So far we have been using:
Bundle
: < x, y >
Prices
: < px , py >
Utility
:
U(x, y )
Uncomp. Demand
:
x ∗ (px , py , I ); y ∗ (px , py , I )
Comp. Demand
: xc∗ (px , py , I ); yc∗ (px , py , I )
• Occasionally we need to look at more than 2 goods.
279 / 526
More Than Two Goods II
• Bundle of n goods where n > 2:
X =< x1 , x2 , ..., xn >
P =< p1 , p2 , ..., pn >
• So xi ∈ X and xj ∈ X would be two different goods.
• For the agent:
U(x1 , x2 , ..., xn )
∗
xi (p1 , p2 , ..., pn ; I )
xic (p1 , p2 , ..., pn ; Ū)
or
U(X)
or
xi∗ (P; I )
or
xic (P; Ū)
280 / 526
More Than Two Goods III
• Now we can look at many goods at once.
• A wide variety of demand patterns become possible with more
than two goods.
• We can ask new questions about the relationships between
many goods.
281 / 526
Relationships Between Many Goods
• We often implicitly think of most goods as being somewhat
substitutable for each other.
• Example: A price increase in one market tends to increase
demand in most other markets.
• We can use theory to investigate our intuition.
282 / 526
Hicks’ Second Law of Demand
• John Hicks studied the issue of substitutability closely.
• He was able to demonstrate that in fact most goods must
have some degree of substitutability.
• The result is called Hicks’ Second Law.4
4
Back in my day we called it Hicks’ Third Law, but I guess this one got an
upgrade :)
283 / 526
Hicks’ Second Law of Demand I
• A consumer has compensated demand for good i of:5
xic (p1 , p2 , . . . , pn ; Ū)
• Compensated demand functions are homogeneous of degree
zero in prices.
• We can see this because, doubling prices while holding utility
constant does not change the optimal bundle.
5
Notice that the c for “compensated” is now the superscript.
284 / 526
Hicks’ Second Law of Demand II
• Euler’s theorem for a function f (.) that is homogeneous of
degree k with arguments < a1 , a2 , ..., an >.
∂f (a1 , a2 , ..., an )
∂a1
∂f (a1 , a2 , ..., an )
+ a2 ·
∂a2
+ ···
∂f (a1 , a2 , ..., an )
+ an ·
∂an
k · f (a1 , a2 , ..., an ) = a1 ·
285 / 526
Hicks’ Second Law of Demand III
• A consumer has compensated demand for good i of:6
xic (p1 , p2 , . . . , pn ; Ū)
• Lets apply Euler’s Theorem:
6
Notice that the c for “compensated” is now the superscript.
286 / 526
Hicks’ Second Law of Demand IV
• Euler’s theorem for xic (p1 , p2 , . . . , pn ; Ū) which has k = 0
∂xic (p1 , p2 , . . . , pn ; Ū)
∂p1
c
∂x (p1 , p2 , . . . , pn ; Ū)
+ p2 · i
∂p2
+ ···
∂x c (p1 , p2 , . . . , pn ; Ū)
+ pn · i
∂pn
0 · xic (p1 , p2 , . . . , pn ; Ū) = p1 ·
287 / 526
Hicks’ Second Law of Demand IV
• LHS = 0 and simplify the notation:
∂xic (P; Ū)
∂p1
∂xic (P; Ū)
+ p2 ·
∂p2
+ ···
∂x c (P; Ū)
+ pn · i
∂pn
0 = p1 ·
288 / 526
Hicks’ Second Law of Demand V
• Divide both sides by xic (P; Ū)
0
c
xi (P; Ū)
=
"
∂xic (P; Ū)
1
p
·
1
∂p1
xic (P; Ū)
+ p2 ·
∂xic (P; Ū)
∂p2
+ ···
∂x c (P; Ū)
+ pn · i
∂pn
#
289 / 526
Hicks’ Second Law of Demand VI
• Divide both sides by xic (P; Ū)
0
c
xi (P; Ū)
=
∂xic (P; Ū)
p1
·
∂p1
xic (P; Ū)
p2
xic (P; Ū)
+ ···
pn
+
c
xi (P; Ū)
+
·
∂xic (P; Ū)
∂p2
·
∂xic (P; Ū)
∂pn
290 / 526
Hicks’ Second Law of Demand VII
• Convert to elasticity:
c
0 = ei1
c
+ ei2
+ ···
c
+ ei3
291 / 526
Hicks’ Second Law of Demand VIII
c
c
c
ei1
+ ei2
+ · · · + ein
=0
c ≤ 0 by Hicks first law, so therefore
• We know that ei1
X
eijc ≥ 0
j6=i
• “Most goods are substitutes”
292 / 526
Composite Commodities
• Sometimes, when there are multiple goods, we may only care
about one of the goods, x1
• If the other goods’ prices typically move together (e.g., rise by
50%), then it may make sense to combine them
• We will define the total expenditure on all the other goods as
follows:
y = p20 x2 + p30 x3 + · · · + pn0 xn
• This person’s initial budget constraint is given by:
I = p1 x1 + p20 x2 + p30 x3 + · · · + pn0 xn = p1 x1 + y
• By assumption, all the prices p2 , . . . , pn must move in unison.
Assume they change by a factor of t (t > 0).
I = p1 x1 + tp20 x2 + tp30 x3 + · · · + tpn0 xn = p1 x1 + ty
293 / 526
Composite Commodities II
Composite Commodity
A composite commodity is a group of goods for which all prices
move together. These goods can be treated as a single
“commodity” in that the individual behaves as though he or she
were choosing between other goods and total spending on the
entire composite group.
294 / 526
Alchian and Allen theorem
“Why are Washington apples in local markets so small and
old-looking? The dried-up stems might seem they were taken out
of cold storage from some gathered last year. Recently, some
apple-picking friends brought some apples they had just picked,
and they were at least four times the size of those available for sale
here. Where do these big Delicious apples go? Are they shipped to
Europe, to the East or can they be bought here in Seattle?” –
M.W.P.’s letter to the “Troubleshooter” column of the
Seattle Times (October 19, 1975).
295 / 526
Alchian and Allen theorem II
“Regarding MWP’s complaint that all the good apples were being
shipped to the East, you might be interested to know that
‘shipping the good apples out’ has been a favorite classroom and
exam question in the economics department at UW for many years.
It is a real phenomenon, easily explained: Suppose for example a
‘good’ apple costs 10 cents and a ‘poor’ apple 5 cents locally. Then
since the decision to eat one good apple costs the same as eating
two poor apples, we can say that a good apple in essence ‘costs’
two poor apples. Two good apples cost four poor apples. Suppose
now that it costs 5 cents per apple (any apple) to ship apples East.
Then, in the East, good apples will cost 15 cents each and poor
ones 10 cents each. But now eating two good apples will cost
three – not four poor apples. Though both prices are higher, good
apples have become relatively cheaper, and a higher percentage of
good apples will be consumed in the East than here. It is no
conspiracy – just the laws of supply and demand.” – Eugene
Silberberg’s letter to the Seattle Times (October 28, 1975).
296 / 526
Alchian and Allen theorem III
• This phenomena was first discovered by Armen Alchian and
William Allen in the book Exchange and Production (1983).
• It states that when the prices of two substitute goods, such as
high and low grades of the same product, are both increased
by a fixed per-unit amount such as a transportation cost or a
lump-sum tax, consumption will shift toward the higher-grade
product.
• Examples:
• Difficulty finding high-quality apples to buy in Washington
state or good fresh oranges in Florida
• People with significant baby-sitting expenses are more likely to
have meals out at expensive (rather than cheap) restaurants
• Individuals are more likely to search for bargains for expensive
items than cheap ones
297 / 526
Alchian and Allen theorem IV
• Assume three goods
x1 = x1c (p1 , p2 , p3 , Ū)
x2 = x2c (p1 , p2 , p3 , Ū)
x3 = x3c (p1 , p2 , p3 , Ū)
• We will hereafter suppress the superscript, c, for notational
simplicity
• Let x3 be a Hicksian composite commodity representing all
other goods. Let x1 and x2 be the higher and lower grades of
some good (e.g., good and poor apples) such that the market
determines p1 > p2 .
• Other examples: x1 → high quality coffee beans; x2 → lower
quality coffee beans
298 / 526
Alchian and Allen theorem V
• Suppose that a transportation cost, t, per item is added to
the prices of x1 and x2 such that the new prices (in some
distant location) are p1 + t and p2 + t, respectively.
• The Alchian and Allen theorem states that:
∂(x1 /x2 )
>0
∂t
∂x2
1
• That is, while ∂x
∂t < 0 and ∂t < 0 by Hicks’ first law of
demand, the consumer shifts her consumption towards x1
relatively speaking
• Assumption: we are assuming x1 and x2 share qualities that
make them substitutes. These are different grades of one class
of commodities and the market selects one as more expensive
as a result.
299 / 526
Alchian and Allen theorem VI
• Use the quotient rule on the derivative
∂(x1 /x2 )
1
∂x1
∂x2
= 2 x2
− x1
∂t
x
∂t
∂t
• An increase in t is equivalent to increasing p1 and p2 by the
same amount. Thus
∂xi
∂xi
∂xi
=
+
for i = 1, 2
∂t
∂p1 ∂p2
• Let
∂xi
sij =
∂pj
• And rewrite
∂(x1 /x2 )
1
∂x1
∂x1
∂x2
∂x2
=
[
+
]x2 − [
+
x1
∂t
x 2 ∂p1 ∂p2
∂p1 ∂p2
1
=
[s11 + s12 ]x2 − [s21 + s22 ]x1
x2
300 / 526
Alchian and Allen theorem VII
• Make an algebraic simplification by distributing x1 , pull out
2
x1 , and do some other stuff :)
x1 [s11 + s12 ] [s21 + s22 ]
∂(x1 /x2 )
=
−
∂t
x2
x1
x2
x1 s11 s12 s21 s22
+
−
−
=
x2 x1
x1
x2
x2
x1 s11 p1 s12 p2 s21 p1 s22 p2
=
·
+
·
−
·
−
·
x2 x1 p1
x1 p2
x2 p1
x2 p2
p
∂xi
• Recall the definition of elasticity, eij = ∂p
· xij
j
∂(x1 /x2 )
x1
=
∂t
x2
e11 e12 e21 e22
+
−
−
p1
p2
p1
p2
301 / 526
Alchian and Allen theorem VIII
• Recall Hicks’ second law
e11 + e12 + e13 = 0 → e12 = −e13 − e11
e21 + e22 + e23 = 0 → e22 = −e23 − e21
• Substitute for e12 and e22 and rearrange
∂(x1 /x2 )
x1
=
∂t
x2
e11 e11 e13 e21 e21 e23
−
−
−
+
+
p1
p2
p2
p1
p2
p2
• And finally
∂(x1 /x2 )
x1
1
1
1
=
(e11 − e21 )( − ) + (e23 − e13 )( )
∂t
x2
p1 p2
p2
302 / 526
Alchian and Allen theorem IX
• What can we say about this expression:
∂(x1 /x2 )
x1
1
1
1
=
(e11 − e21 )( − ) + (e23 − e13 )( )
∂t
x2
p1 p2
p2
• We know that xx1 > 0
2
• We know that if p1 > p2 , then p1 < p1 . Assume p1 = 2 and
1
2
p2 = 1. Then 21 − 1 = − 21 . Therefore the second part of the
first term is clearly negative.
• We also know that the first part of the first term is negative.
Hicks’ first law states e11 < 0, and by assumption these two
goods are substitutes, therefore e21 > 0, and e11 − e21 < 0
• Thus the first part of the interior expression is positive.
• We also know that p2 > 0. So the validity of the Alchian and
1 /x2 )
Allen theorem (i.e., ∂(x∂t
> 0) comes down to e23 − e13 .
• This is mathematically indeterminate
• But if they are close substitutes, then it stands to reason that
e13 ≈ e23
303 / 526
Other Sections
• Composite commodities (p. 193-195)
• Household production model (p. 197-200)
• Shipping the good apples out (p. 203)
304 / 526
Producer Theory:
Production Functions
305 / 526
Marginal Product and Average Product
306 / 526
Everything is Different:
Everything is the Same
• We switch gears to talk about the other side of the market –
firms!
• We are modeling a different decision making process.
• Firms will aim to maximize profits/minimize costs.
• Thankfully, the mathematics is more or less the same.
307 / 526
What’s a firm?
• The basic model of the firm is pretty straightforward
• Firms turn inputs into outputs.
• If they’re in the business of making money, then their goal is to
maximize profits.
• If they’re not in the business of making money, then their goal
is minimize costs.
• Maximizing Profits and Minimizing Costs are “dual” problems.
• In IO there are more sophisticated models of firms.
308 / 526
Markets or Strategic Interaction?
• We start with a firm in a competitive market.
• Firms are price takers and interact with consumers via prices.
• The firms don’t directly take other firms into account.
• In a few weeks
• We will use game theory to model strategic firm interaction.
• Regardless, all firms have “production function”, and we’re
going to start there.
309 / 526
Production functions
• The firm’s production function for a particular good, q, is:
q = f (l, k)
• q is the amount of the good produced.
• f(.) is just the functional form that turns the inputs (l and k)
into outputs (q).
• It is analogous to U(x,y)
• k is capital.
• Machines, warehouses, etc..
• l is labor.
• People.
310 / 526
Why only k and l
• k and l are just “stand ins” for any two inputs.
• We can do almost everything we need to do with a two input
model.
• This is similar how we use U(x1 , x2 ) rather than
U(x1 , x2 , ..., xn )
311 / 526
A Quick Note of Notation
• The book uses the generic notation f(l,k) for production
functions.
• However, it draws graphs and as we will see calculates slopes
assuming that l is on the horizontal axis and k is on the
vertical axis.
• So to be consistent with the way we did things in consumer
theory I will use f(l,k).
312 / 526
Marginal (Physical) Product I
• The marginal physical product (MP) of an input is the
additional output that can be produced by using one more
unit of that input while holding all other inputs constant.
• Example:
• 50 laborers on a farm produce 100 bushels of wheat per year.
• 51 laborers on the same farm with the same equipment
produce produce 102.
• The MP of labor is 2.
• Just like MU, MP is found using ceteris paribus.
313 / 526
Marginal (Physical) Product II
∂q
∂f (l, k)
=
= fk
∂k
∂k
∂q
∂f (l, k)
MPl =
=
= fl
∂l
∂l
MPk =
• Looks familiar!
314 / 526
Diminishing Marginal Product
• Marginal product is not usually constant.
• One cook in the kitchen make 6 pizzas per hour
• Two cooks in the kitchen make 10 pizzas per hour.
• Three cooks in the kitchen make 12 pizzas per hour.
• First proposed by Malthus in the 19th century
315 / 526
Diminishing Marginal Product
• We can see this in the second (own) partial derivatives:
∂MPk
∂k
∂MPl
∂l
=
=
∂ 2 f (l, k)
= fkk < 0
∂k 2
∂ 2 f (l, k)
= fll < 0
∂l 2
316 / 526
Cross Partial Effects
• What about:
∂MPl
= flk
∂k
∂MPk
= fkl
∂l
• In most cases, flk > 0 and fkl > 0.
317 / 526
Average Physical Product I
• The common use of “labor productivity” is usually referring to
what we call average labor productivity
• “Rising productivity”, usually means that output per unit of
labor input has increased.
318 / 526
Average Physical Product II
q
f (l, k)
output
= =
labor input
l
l
output
q
f (l, k)
APk =
= =
capital input
k
k
APl =
• Notice that average productivity of one input depends on the
level of the other input used.
319 / 526
Example: Fixed level of Capital (k̄)
• Firms can vary l more easily than k.
• At times we treat k as fixed or constant: k̄.
• A firm producing flyswatters:
q = f (l, k̄) = 600k̄ 2 l 2 − k̄ 3 l 3
• If k = 10
q = 60, 000l 2 − 1, 000l 3
2
• Marginal product: MPl = ∂q
∂l = 120, 000l − 3, 000l q
• Average product: APl = ql = 60, 000l − 1, 000l 2
320 / 526
Isoquants
321 / 526
Isoquants I
• An isoquant is the set of k and l that satisfies f (l, k) = q0
• Where q0 is a given level of output.
• They are similar to indifference curves in many ways:
• They are the “level set” or “contour curve” of a 3-dimensional
function.
• There are infinitely many isoquants for a f (l, k).
• As we move in a northeasterly direction the Isoquants
represent higher levels of output.
322 / 526
Isoquants II
10
9
8
7
k per period
6
5
4
A
Ka
q=30
2
q=20
B
Kb
0
0
1
2
La
4
5
6
Lb
q=10
8
9
10
l per period
323 / 526
MRTS I
• The marginal rate of technical substitution (MRTS) is similar
to MRS
• MRTS tells us the rate at which labor can be substituted for
capital while holding output constant along an isoquant.
• It is the negative of the slope of the isoquant.
MRTS(l for k) = −
dk
dl |q=q0
• (l for k) just means l is on the horizontal axis.
• Note: Our MRS is could similarly be written MRS(x for y).
324 / 526
MRTS II
• Differentiate to find MRTS
q0 = f (l, k)
dq0 = fk dk + fl dl
fl
dk
=
−
dl
fk
dk
fl
= −
dl
fk
• dq0 = 0 because q0 is a constant.
• The MRTS is the ratio of the inputs’ marginal products.
325 / 526
Diminishing MRTS I
• To see MRTS is diminishing in l, rewrite as
MRTS =
fl (l, k)
fk (l, k)
• The level of k depends on l and vice versa, so lets make that
explicit for a moment:
MRTS =
fl [k(l), l(k)]
fk [k(l), l(k)]
326 / 526
Diminishing MRTS II
• Differentiate with respect to l
dMRTS
dl
=
dk
(flk dk
dl + fll )fk − (fkk dl + fkl )fl
fk2
=
fk2 fll − 2fl fk fkl + fl 2 fkk
fk3
Hint: Use the Quotient rule and substitute
dk
dl
= − ffkl
• Positive marginal products guarantees a positive denominator
• So dMRTS
< 0 if the numerator is negative.
dl
327 / 526
Diminishing MRTS III
fk2 fll − 2fl fk fkl + fl 2 fkk
fk3
• Checking the Numerator
• fll < 0 and fkk < 0 (diminishing marginal products)
• If fkl > 0, then the numerator is negative
• If fkl < 0, then the numerator is negative if this value is
relatively small.
• Functions for which the numerator is negative are called
quasi-concave functions.
328 / 526
Returns to Scale
329 / 526
Returns to Scale I
q = f (l, k)
• Multiply inputs by t (where t > 1).
• The returns to scale of the production function are:
Table: Returns to scale
Effect on output
f (tk, tl) = tf (l, k) = tq
f (tk, tl) < tf (l, k) = tq
f (tk, tl) > tf (l, k) = tq
Returns to scale
Constant
Decreasing
Increasing
330 / 526
Returns to Scale and Homogeneity
• When a production function exhibits constant returns to scale,
it meets the definition of “homogeneity”
• Homogenous of degree 1 in its inputs
f (tk, tl) = t 1 f (l, k) = tq
• That means its derivatives must be homogenous of degree
k −1
• Constant returns to scale implies homothetic production
functions
• MRTS for any constant returns to scale production function
will depend only on the ratio of the inputs, not on their
absolute levels
• its isoquants will be radial expansions of one another
331 / 526
Returns to Scale and Homotheticity
• Constant returns to scale implies homothetic production
functions
• MRTS for any constant returns to scale production function
will depend only on the ratio of the inputs, not on their
absolute levels
• its isoquants will be radial expansions of one another
332 / 526
Returns to Scale III
10
9
8
7
k per period
6
5
4
3
q=3
2
q=2
1
0
q=1
0
1
2
3
4
5
6
7
8
9
10
l per period
333 / 526
Elasticity of Substitution I
• How easy is it to substitute k for l or vice versa?
• Along one isoquant, the rate of technical substitution will
decrease as the capital-labor ratio decreases (i.e., as
decreases)
k
l
• Elasticity of substitution (σ) measures this degree of
responsiveness.
334 / 526
Elasticity of Substitution II
• Elasticity of substitution (σ)
σ =
=
=
percent ∆(k/l)
percent ∆MRTS
d(k/l) MRTS
·
dMRTS
k/l
d ln (k/l)
d ln MRTS
335 / 526
Four common production functions I
• Linear (σ = ∞)
q = f (l, k) = αk + βl
• Fixed proportions (σ = 0)
q = min (αk, βl), α, β > 0
• Cobb-Douglas (σ = 1)
q = f (l, k) = Ak α l β
f (tk, tl) = A(tk)α (tl)β = At α+β k α l β
= t α+β Ak α l β
= t α+β f (l, k)
hence if α + β = 1, then constant returns
336 / 526
Four common production functions II
• Constant Elasticity of Substitution (CES)
γ
q = f (l, k) = [k ρ + l ρ ] ρ
1
• σ = 1−ρ
• For γ > 1 gives increasing returns.
• For γ < 1 gives decreasing returns.
• Incorporates all three previous cases.
• Linear: ρ = 1
• Fixed Proportions: ρ = −∞
• Cobb-Douglas: ρ = 0
337 / 526
Cost Functions
338 / 526
Cost Minimization
339 / 526
What do we mean by ”cost”
• The economic cost of an input is the remuneration the input
would receive in its best alternative employment.
total cost = C = wl + vk
• We’ll start by looking at cost minimization.
• Then we will move to profit maximization in later sections.
340 / 526
Types of Costs
• Labor costs: expenditures on labor.
• Capital costs: what someone else would be willing to pay for
the use of a machine the firm is using.
• The cost of one machine-hour is the rental rate for that
machine in its best alternative use.
• This incorporates both explicit costs and opportunity costs.
• Entrepreneurial costs: The owner’s opportunity cost of
starting and/or running a firm is also an important cost.
341 / 526
A Quick Note on Notation I
• The book uses the generic notation f(k,l) for production
functions.
• Graphs are drawn with l is on the horizontal axis and k is on
the vertical axis.
• MRTS (l for k) is calculated as − dk
dl and
MUl
MUk .
• Which implies l is on the horizontal and k on the vertical.
342 / 526
A Quick Note on Notation II
• To be consistent with the way we did things in consumer
theory (horizontal, vertical) I will use f(l,k).
• Likewise, in generic notation prices will always go in the order
(price of horizontal input, price of vertical input) or (w , v ).
343 / 526
Cost minimization I
• Our objective function here is the Total Cost.
• The Constraint is the level of output: q0 = f (l, k):
L(l, k, λ) = wl + vk + λ q0 − f (l, k)
∂L
∂l
∂L
FOC 2
∂k
∂L
FOC 3
∂λ
FOC 1
= w − λfl = 0
= v − λfk = 0
= q0 − f (l, k) = 0
• System of 3 equations and 3 unknowns.
• Solve it for l c (w , v , q0 ), k c (w , v , q0 ) and λc (w , v , q0 )
344 / 526
Cost minimization II
• We can look at some general properties by dividing the first
two FOCs:
∂f /∂l
w
=
= MRTS of l for k
v
∂f /∂k
• Notice that marginal product per dollar spent on inputs is
equal across all inputs:
fk
fl
=
v
w
• Marginal cost
w
v
=
=λ
fl
fk
345 / 526
Cost minimization II
10
9
8
q=1 unit
7
k per period
6
5
4
Kc
2
q=1 unit
1
C1
0
0
1
Lc
3
4
C2
C3
5
6
7
8
9
10
l per period
346 / 526
Contingent Demand
• l c (w , v , q¯0 ) and k c (w , v , q¯0 ) are the contingent demand
functions for the inputs.
• It tells you the optimal bundle of inputs you should use to
produce q¯0 of the good, given prices w , and v .
• They are called “contingent demand” because the optimal
bundle is contingent on the level of output.
347 / 526
Expansion Paths
• Firm’s expansion path is the set of all cost minimizing
tangency’s as we increase only the level of output.
• Normal Input: If the firm increases output (q), then they
must also increase their demand for the input.
• Inferior Input: If the firm increases output (q), then they must
decrease their demand for the input.
348 / 526
Expansion Path: Both Normal Inputs
10
9
Expansion path
8
7
k per period
6
5
4
q=2 units
3
K1
2
1
q=1 unit
0
C1
0
1
L1 2
3
C2
4
5
6
7
8
9
10
l per period
349 / 526
Expansion Path: Labor is Inferior Input
Expansion path
10
9
8
7
k per period
6
q=2 units
5
4
3
2
1
q=1 unit
0
0
1
2
Input inferiority
3
4
5
6
7
8
9
10
l per period
350 / 526
Cost Function
351 / 526
Total Cost Functions
• The Total Cost formula is:
C = wl + rk
• The cost function is found by substituting in:
C (w , v , q0 ) = wl c (w , v , q¯0 ) + rk c (w , v , q¯0 )
• The total cost function tells us the total minimum cost of
producing a given level of output q0 , given input prices w and
v.
352 / 526
Related Cost functions
• Average cost (AC ):
AC (w , v , q) =
C (w , v , q)
q
• Marginal cost (MC ):
MC (w , v , q) =
∂C (w , v , q)
∂q
353 / 526
Related Cost Functions II
• MC is the important thing that ultimately drives the decision
making process.
• Looking at changes in things like average costs will be useful
too.
• Example: The shape of ATC is determined by marginal cost.
354 / 526
Related Cost Functions III
10
MC
9
e>1
8
7
ATC
Cost $$
MC=ATC
e=1
5
4
e<1
3
2
1
0
0
1
2
3
4
q*
6
7
8
9
10
Quantity (Q)
355 / 526
Quantity Elasticity of Cost
• If we increase costs, what is the size of the change in costs?
• When MC Q ATC:
MC
∂C
∂q
∂C
·q
∂q
∂C q
·
∂q C
eC ,q
Q ATC
C
Q
q
Q C
Q 1
Q 1
356 / 526
Minimum Marginal Cost: λc (w , v , q) I
• The value of λc (w , v , q) has a nice interpretation.
• It is the ”Minimum Marginal Costs” of another unit of output
q.
• Substitute contingent input demand functions back into
Lagrangean
C (w , v , q) = vk c (w , v , q) + wl c (w , v , q)
+λ(w , v , q) q0 − f [k c (w , v , q), l c (w , v , q)]
357 / 526
Minimum Marginal Cost: λc (w , v , q) II
C (w , v , q) = vk c (w , v , q) + wl c (w , v , q)
c
c
+λ(w , v , q) q0 − f [k (w , v , q), l (w , v , q)]
(.)
• Now find MC = ∂C
∂q
∂C (.)
∂q
∂k c (.)
∂l c (.) ∂λ(.)
+w
+
(q̄ − f (l, k))
∂q
∂q
∂q
∂k c (.)
∂l c (.)
−λfk
− λfl
+λ
∂q
∂q
= v
358 / 526
Minimum Marginal Cost: λc (w , v , q) III
∂C (.)
∂q
∂k c (.)
∂l c (.) ∂λ(.)
+w
+
(q̄ − f (l, k))
∂q
∂q
∂q
∂k c (.)
∂l c (.)
−λfk
− λfl
+λ
∂q
∂q
= v
• Now factor out the derivatives.
359 / 526
Minimum Marginal Cost: λc (w , v , q) IV
∂C (.)
∂q
∂k
∂l
∂λ
= (v − λfk )
+ (w − λfl )
+ q̄ − f (l, k)
+λ
∂q
∂q
∂q
= λc (w , v , q)
360 / 526
Properties of Input Demand and Cost Functions
361 / 526
Shephard’s Lemma
• Shepherd’s lemma works in a similar was as with Expenditure.
∂C (w , v , q)
= l c (w , v , q)
∂w
∂C (w , v , q)
= k c (w , v , q)
∂v
362 / 526
The Law of Demand for Inputs
• Downward sloping contingent input demand:
∂ 2 C (w , v , q)
∂k c (w , v , q)
=
≤0
2
∂v
∂v
• There is no Giffen phenomena.
• Notice that this also tells us that total cost is concave in input
prices.
• Cost increases at a decreasing rate as input prices increase.
363 / 526
Input Substitution
2
∂ C
∂k
• Input substitution: Cv ,w = ∂v
∂w = ∂w ≥ 0
• Curvature of the isoquant ensures this
364 / 526
Input Substitution
10
q=1 unit
9
w/v
8
7
quantity of K
K*
5
4
w/v
3
w/v'
2
K'
q=1 unit
w/v'
1
0
0
1
L*
3
4
5
6
L'
8
9
10
quantity of L
365 / 526
Homogeneity of Cost Functions
• The cost function is homogenous of degree 1 (H 1 ) in input
prices
C (tv , tw , q) = t 1 C (v , w , q) = tC (v , w , q)
• Average and marginal costs are also homogenous of degree 1
in input prices.
366 / 526
Homogeneity of Input Demand
• Input demand functions are homogenous of degree 0 (H 0 ) in
input prices
l c (tv , tw , q) = t 0 l c (w , v , q) = l c (w , v , q)
• It must be the case, because cost minimization subject to
fixed isoquant (q cannot change).
• Again we see that it is relative prices that matter.
• If you double the prices of all goods, and still want to produce
the same amount of the good then the firm will still use the
same mix of inputs.
367 / 526
Homogeneity of Input Demand (Graph)
10
9
tC/tv
quantity of K
7
6
5
K(tv,tw,q)
4
3
2
output = 1 unit
1
tv/tw
0
0
1
2
L(tv,tw,q)
4
5
6
tC/tw
8
9
10
quantity of L
368 / 526
Input Demand Slopes Down I
∂L
• We can show that ∂w
< 0 using Cramer’s rule, as well as
relying on the graphical proof. Recall the first order conditions
w − λfL = 0
v − λfK
= 0
q − f (l, k) = 0
• Totally differentiate the three equations
dw − λfLL dL − λfLK dK − fL dλ = 0
dv − λfKL dL − λfKK dK − fK dλ = 0
dq − fK dK − fL dL = 0
• And construct the Ab=c matrices

−λfLL −λfLK
−λfKL −λfKK
−fL
−fK
   

−fL
dL
−dw
−fK  · dK  =  −dv 
0
dλ
−dq
369 / 526
Input Demand Slopes Down II
• First, calculate the effect of wages on labor demand,
∂L
∂w
=
|A11 |
|A|
• Find the determinant of the coefficient matrix, |A|
−λfLL −fL −λfKL −λfKK + fK |A| = −fL −λfKL −fK −fL
−fK = λ(fKK fL2 − 2fKL fK fL + fLL fK2 ) < 0
• Find the determinant of the transformed matrix, |A11 |
−1 −λfLK
|A11 | = 0 −λfKK
0
−fK
−fL −λfKK
−fK = −1 −fK
0 −fK = fk2 > 0
0 fK2
∂L
11 |
• ∂w
= |A|A|
= |A|
<0
370 / 526
Input Demand Slopes Down III
∂L
12 |
• Next, calculate the substitution effect, ∂v
= |A|A|
• Find the determinant |A12 |
0 −λfLK
|A12 | = −1 −λfKK
0
−fK
−fL −λfLK
−fK = 1 −fK
0 −fL = −fL fL < 0
0 ∂L
12 |
K fL
• ∂v
= |A|A|
= −f|A|
>0
371 / 526
Changes in Marginal/Average cost
• The effects of changes in q, v and w on average and marginal
costs are sometimes ambiguous.
∂λ
• For instance, ∂MC
∂v = ∂v =?
∂λc (.)
∂v
=
=
=
=
∂(∂C (.)/∂q)
∂v
2
∂ C (.)
(rewriting )
∂q∂v
∂ 2 C (.)
(rewriting )
∂v ∂q
∂k c (.)
∂q
• If capital is a normal input, an increase in v will raise MC .
• If capital is an inferior input, an increase in v will reduce MC .
372 / 526
Inferior Inputs and Cramer’s Rule
∂λ
• Examine ∂w
using Cramer’s rule to better understand the role
of inferior inputs in determining the shape of the marginal
cost curve
−λfLL −λfLK −1
−λfKL −λfKK |A31 | = −λfKL −λfKK 0 = −1 −f
−f
L
K
−fL
−fK
0
= λ(fKK fL − fKL fK )
• The sign of |A31 | is ambiguous because fKL is ambiguous
• Proposition: A necessary condition for an increase in the
price of labor (capital) to decrease marginal cost is that
capital and labor are rivals in production (i.e., fKL < 0)
• The intuition for this prediction escapes your professor,
though the math is true and easy enough to follow
373 / 526
Input substitution
• Recall the formula for the elasticity of substiution
σ=
dln(k/l)
∂(k/l) MRTS
·
=
∂MRTS
k/l
dln(MRTS)
• But cost minimization says that when the firm optimizes
MRTS =
w
v
• Therefore we can substitute w /v for MRTS and get a new
expression for the elasticity of substitution
s=
∂(k/l) (w /v )
dln(k/l)
·
=
∂(w /v ) (k/l)
dln(w /v )
374 / 526
Input substitution II
• These two parameters are similar, but they are NOT the same
thing.
1
2
σ is general and applies to any point on any isoquant.
s is specific and applies to one specific input bundle on one
particular isoquant.
• The specific input bundle is the cost minimizing input bundle
(l c , k c ).
3
4
5
s only requires input prices for calculation.
σ requires knowledge of MRTS for estimation.
Both σ and s must be non-negative in the two good case
375 / 526
Input substitution III
• In the n good case, the elasticity of substitution is:
sij =
∂(xi /xj ) (wj /wi )
∂ln(xi /xj )
·
=
(wj /wi ) (xi /xj )
∂ln(wj /wi )
∂C
∂C
• Shepherd’s Lemma: ∂w
= xi and ∂w
= xj , rewrite elasticity:
i
j
sij =
where Ci =
∂C
∂wi ,
∂ln(Ci /Cj )
∂ln(wj /wi )
etc.
• Once total cost is known, such as through econometric
estimation, useful information about the substitutability
among inputs can be obtained
376 / 526
Short Run Costs I
• Short-run – the economic actor has limited freedom to vary
some key choices (e.g., inputs may not vary)
• Long-run – the economic actor has more freedom over choice
(e.g., all inputs are variable)
• Assume capital is the fixed input, k̄, in the short-run, but labor
inputs are variable
q = f (l, k̄)
• Short-run total cost
SC = wl + v k̄
• Short-run fixed cost
SFC = v K̄
• Short-run variable cost
SVC = wL
377 / 526
Short Run Costs II
• Short-run average costs
SAC =
SC
q
SMC =
∂SC
∂q
• Short-run marginal cost
• Non-optimality of short-run costs (see graphic)
378 / 526
Capital (K) per period
Short run costs III
10
9
8
7
6
5
vK
4
3
x3
x2
2
1
0
x1
0
L1
L2
L3
6
7
8
9
10
Labor (L) per period
379 / 526
Profit Maximization
380 / 526
Profit and Marginal Revenue
381 / 526
Profit
• A firm’s economic profits are the difference between its
revenues and costs.
π(q) = R(q) − C (q)
• We can write cost as a function of only q because w and v
are given by the market.
• We are going to first focus on the firms choice of q∗.
• Then we will turn to the firms’s choice of an input bundle.
382 / 526
Total Revenue
• Total revenue can be broken down further as:
R(q) = p(q) · q
• p(q), allows for the possibility that the firms choice of q will
affect the price.
383 / 526
The Firms Profit Max Problem
• To maximize profits the firm simultaneously chooses:
• The optimal level of production q.
• The optimal mix of inputs l ∗ and k ∗ .
• It bases these decisions on the market prices:
• The input prices: w and v .
• The price the receive for their output: P.
384 / 526
Profit Maximization
• A necessary conditions for profit max is:
dπ(q)
dR(q) dC (q)
= π 0 (q) =
−
=0
dq
dq
dq
• At the max, the marginal profit of adding q should be zero.
385 / 526
General FOC for Profit Max
dR(q) dC (q)
dπ(q)
= π 0 (q) =
−
=0
dq
dq
dq
• This can be re-written as:
dR(q)
dC (q)
=
dq
dq
Marginal Revenue = Marginal Cost
• To maximize economic profits, the firm should choose output
q ∗ such that MR=MC.
• This is our first order condition for a max.
386 / 526
General SOC for Profit Max
• When we find the optimal q we will need to make sure we
have found a max and not a min.
• Second order condition:
d 2π
dπ 0 (q)
=
<0
dq 2 |q=q∗
dq |q=q∗
• If you are at the profit max level of q and you produce more q
your profits will go down.
387 / 526
Marginal Revenue
• Marginal revenue:
MR =
dR(q)
dq
• The change in total revenue R resulting from a change in
output q
388 / 526
Marginal Revenue: Competitive Market
• Total revenue:
R = p(q) · q
• If the firm’s choice of q has no effect on price then marginal
revenue is the price of the output P.
• Use p(q) for cases where the firm’s choice of q can affect the
output price.
389 / 526
Marginal Revenue: In General
• If the firm’s choice of q may have effect on price then we need
to be more careful.
• Marginal revenue:
dR(q)
dq
d[p(q) · q]
=
dq
dp
= p+q·
(use product rule)
dq
MR(q) =
390 / 526
Elasticity of Output Demand
• We need to define a new elasticity measure.
• The Price Elasticity of (Output Demand)
eq,p =
dq p
dp q
391 / 526
Marginal Revenue: Elasticity Form
• This allows us to express MR in terms of elasticity:
dp
MR(q) = p + q ·
dq
p
dp
=
p+ q·
p
dq
q dp
= p 1+ ·
p dq
1
= p 1+
eq,p
392 / 526
Price and Marginal Revenue
1
MR = p 1 +
eq,p
If...
Elastic
Unit Elastic
Inelastic
eq,p < −1
eq,p = −1
eq,p > −1
Then...
MR > 0
MR = 0
MR < 0
• Elastic: Additional q will not affect price much: MR > 0
• Inelastic: Additional q will affect price a lot: MR < 0
• If eq,p = −∞, back to case where MR = P.
393 / 526
The Marginal Revenue Curve
• Demand Curves can be thought of as Average Revenue
Curves.
• The firm’s Demand Curve has a corresponding Marginal
Revenue curve.
394 / 526
The Marginal Revenue Curve: Graph
10
9
8
7
Price
6
5
4
3
2
1
0
AR(q)=Demand
0
1
2
3
4
5
6
7
8
9
10
Quantity
MR(q)
• As q increases from 0 to 5, Total Revenue (p*q) is increasing.
• For q greater than 5, Total Revenue is decreasing.
395 / 526
Price and Marginal Cost I
• What is the relationship between MC and P?
• Recall at the profit max our FOC is:
MC
MC
p − MC
p
= MR
1
= p 1+
eq,p
1
1
=
=
−eq,p
|eq,p |
396 / 526
Price and Marginal Cost II
p − MC
p
=
1
|eq,p |
• The price firms can charge, even if they have market power, is
still constrained by the elasticity price elasticity of quantity.
• The percentage markup over marginal cost will be higher the
closer eq,p is to −1.
397 / 526
Price-taking firms in the short-run I
• Short-run supply curve.
• Shows how much it will produce at various possible output
prices.
• For a π-maximizing firm that takes P as given, the firm’s
supply curve is:
• The positively sloped segment of the firm’s short-run marginal
cost curve above the SAVC curve.
398 / 526
Market price
Price-taking firms in the short-run II
10
9
SMC
P***
SAC
7
6
P*=MR
5
SAVC
P**
Ps
2
1
0
0
1
2
3
q**
q*
5
q***
6
7
8
9
10
Quantity per period
• For prices below SAVC, the firm’s profit-maximizing decision
is to shut down and produce no output.
399 / 526
Maximizing Profit
400 / 526
The Mathematics of Profit Maximization I
• Maximize the profit function.
• Note: The constraint (total cost) is already a part of the
profit function.
π = TR − TC
π(P, w , v ) = pf (l, k) − vk − wl
• So we can use an unconstrained optimization method.
(FOC 1)
(FOC 2)
∂π(P, w , v )
∂k
∂π(P, w , v )
∂l
= pfk − v = 0
= pfl − w = 0
401 / 526
The Mathematics of Profit Maximization
II
(FOC 1)
(FOC 2)
∂π(P, w , v )
∂k
∂π(P, w , v )
∂l
= pfk − v = 0
= pfl − w = 0
• A system of two equations (FOC1 and FOC2) and two
unknowns (l and k).
• Solve for the unconditional input demand functions:
l(P, w , v ) and k(P, w , v ).
• SOCs: πll = fll < 0, πkk = fkk < 0 and fkk fll − fkl2 > 0
402 / 526
The Mathematics of Profit Maximization
III
• The firm’s profit function tells us the maximum profits they can make
as a function of the prices (P, w, and v).
Π(P, w , v ) = max π(k, l)
= max [P · f (l, k) − vk − wl]
= P · f [k(P, w , v ), l(P, v , w )] − vk(P, w , v ) − wl(P, w , v )
• Π is used here instead of π to indicate that the firm is producing the
optimal level of q using the optimal amounts of l and k.
403 / 526
Hotelling’s Lemma (Envelope Results)
• The firm’s supply can be found using Hotelling’s lemma:
∂Π
= q(P, w , v ) ≥ 0
∂p
• Similarly for Demand functions:
∂Π
∂v
∂Π
∂w
= −k(P, w , v ) ≤ 0
= −l(P, w , v ) ≤ 0
404 / 526
Properties of Profit functions.
• Profit functions are homogenous of degree 1 in all prices:
Π(tP, tw , tv ) = t 1 Π(P, w , v )
• Profit functions are non-decreasing in output prices:
∂Π(P, w , v )
>0
∂P
• Profit functions are non-increasing in input prices:
∂Π(P, w , v )
< 0
∂w
∂Π(P, w , v )
< 0
∂v
• Profits functions are convex in output prices:
∂ 2 Π(P, w , v )
>0
∂P 2
405 / 526
To Do
• Producer surplus (p. 386-387)
406 / 526
Input Demand
407 / 526
Marginal Revenue Product of Labor
• Recall one of our first order conditions
∂π
= Pfl − w = 0
∂l
• Pfl : Marginal Revenue Product of Labor.
• Often called the “Value Marginal Product of Labor”
• The extra revenue the firm receives when it uses one more unit
of labor.
408 / 526
Marginal Revenue Product of Capital
• Likewise, recall the other FOC:
∂π
= Pfk − v = 0
∂k
• Pfk : Marginal Revenue Product of Capital.
• Often called the “Value Marginal Product of Capital”
• The extra revenue the firm receives when it uses one more unit
of capital.
409 / 526
Marginal Revenue Product and Input
Prices
• Solve the FOCs for the MRP.
• Pfl = w
• Pfk = v
• In the price-taking case, the marginal revenue product of an
input equals the input’s price.
410 / 526
Marginal Revenue Product and Input
Prices: Graph
10
9
8
7
Amount ($)
6
5
wage
4
3
2
p MPl
1
0
0
1
2
3
L*
4
5
6
7
8
9
10
Labor demand
411 / 526
Anecdote: Worth More than Your Wage?
• In grad school, I walked past a mural with graffiti that said:
“You are worth more than your wage”
• Notice, though – the neoclassical model says that the wage
measures the worker’s value of their marginal product, not
their total product, or even their total worth
412 / 526
Own Price Effects I
• Recall for lc (w , v , q̄)
,v ,q̄)
• ∂lc (w
<0
∂w
• What about: l(w , v , q̄)?
• If there are only two inputs (special case):
,v )
• ∂l(P,w
≤0
∂w
• In a general we have to be concerned about both the income
and substitution effects of a price change.
413 / 526
Own Price Effects II
• lc (w , v , q̄) and l(P, w , v ) are equivalent values when:
q̄ = f (k[P, w , v ], l[P, w , v ])
• Substitute the output supply function for q̄
l(P, w , v ) = lc (v , w , q(P, w , v ))
• Now take the derivative of both sides with respect to
own-input price, w
∂lc
∂lc ∂q
∂l
=
+
·
∂w
∂w
∂q ∂w
414 / 526
Own Price Effects III
∂lc
∂lc ∂q
∂l
=
+
·
∂w
∂w
∂q ∂w
• First term on RHS is substitution effect and non-increasing.
• Second term is output effect.
∂l
• Can we sign second term to determine the overall sign of ∂w
?
415 / 526
Own Price Effects IV
• Take the second term in the expression.
• Then apply Hotelling’s Lemma to the second element:
∂lc ∂q
·
∂q ∂w
∂Π
=
=
=
∂lc ∂p
·
∂q ∂w
∂lc ∂ 2 Π
·
∂q ∂p∂w
∂lc ∂ 2 Π
·
∂q ∂w ∂p
(simplify )
(Young 0 s theorem)
416 / 526
Own Price Effects IV
=
=
=
=
=
∂lc
∂q
∂lc
∂q
∂lc
∂q
∂lc
∂q
∂lc
∂q
·
·
·
·
·
∂Π
∂w
∂p
∂[−l(.)]
(Hotelling 0 s)
∂p
∂[−lc (.)]
[use : l(.) = lc (.)]
∂p
−∂lc ∂q
·
(multiply by 1)
∂p ∂q
−∂lc ∂q
·
(rearrange)
∂q ∂p
417 / 526
Own Price Effects V
∂lc −∂lc ∂q
·
·
∂q
∂q ∂p
∂Π
∂lc −∂lc ∂ ∂p
=
·
·
(Hotelling 0 s)
∂q
∂q
∂p
2
∂lc
∂2Π
= (−) ·
·
∂q
∂p∂p
2 2
∂lc
∂ Π
= (−) ·
·
∂q
∂p 2
=
418 / 526
Own Price Effects VI
∂lc 2 ∂ 2 Π
= (−) ·
·
∂q
∂p 2
= (−) · (+) · (+) ()
∂q
c
• Therefore: ∂l
∂q · ∂w < 0
419 / 526
Input demand V
∂l
∂lc
∂lc ∂q
=
+
·
∂w
∂w
∂q ∂w
∂l
• Negative plus a negative is a negative, and therefore, ∂w
≤0
• When the price of an input increases, two effects cause the
quantity demanded of that input to fall
Substitution effects cause any given output level to be
produced using less of the input, and
2 The increase in costs causes less of the good to be sold,
thereby creating an additional output effect that decreases
demand for the input.
1
420 / 526
Game theory
421 / 526
Basic concepts I
• Every game has three elements: players, strategies and payoffs
Players: each decision-maker in a game is called a player.
Usually label the player i in an n-person game
2 Strategies: each course of action open to a player during the
game is called a strategy. Let S1 denote the set of strategies
open to player 1. Let s1 ∈ S1 be a particular strategy chosen
by player 1 from the set of possibilities. A strategy profile will
refer to a listing of particular strategies chosen by each of a
group of players.
3 Payoffs: The final return to each player at the conclusion of a
game is called the payoff, or payout. Payoffs are measured in
utility or sometimes monetary payoffs (e.g., profits) for firm
games. In a two-person game, u1 (s1 , s2 ) denotes player 1’s
payoff given she chose s1 and her opponent chose s2 .
1
422 / 526
Basic concepts II
• Organization of these three elements is often done using one
of two forms
1
2
Normal form: table form (see next slide)
Extensive form: game tree (see next slides)
423 / 526
Basic concepts III
Row player
Column player
Left
Right
Top
u1 = 0, u2 = 4
u1 = 1, u2 = 0
Bottom
u1 = 3, u2 = 3
u1 = 1, u2 = 1
424 / 526
Basic concepts IV
P1
L
R
P2
L
(a, b)
R
(c, d)
L
(e, f )
R
(g , h)
425 / 526
Example I: Prisoner’s dilemma I
Narrative: Imagine that it’s the 1930s during the Great Depression,
and there’s been a string of bank robberies. All the police know is
that the suspects are a male and female duo of bank robbers,
possibly a couple, but otherwise no identifying witnesses. One day,
a beat cop pulls over a couple for running a stop sign. He notices
that their tail light is out and asks to see what’s in the trunk. In
the trunk are two unlicensed tommy guns (i.e. machine guns). The
officer arrests them both for weapons violations which carries a
minimum of one year in prison. At the station, the district attorney
offers the following deal to each person separately: “we think you
are the notorious bank robbers. If you confess and your partner
doesn’t confess, you’ll get 6 months in prison. If you don’t confess,
but your partner confesses, you’ll get 20 years in prison. If neither
confess, you get one year in prison. If both confess, you get 10
years in prison. What’ll it be?”
426 / 526
Prisoner’s dilemma II
• Discuss the three elements of a game.
• Our job is to organize this material into a tractable game form
that can be solved
• Hint: We need a solution concept before we can solve the
game, though
• Let’s start by organizing the material into the normal form
representation
• Let row player be Bonnie (female), and column player be Clyde
(male)
• Label strategies
• Input utility payoffs – assume players strictly prefer more
freedom to less
427 / 526
Prisoner’s dilemma III
Bonnie
Confess
Clyde
Don’t confess
Confess
u1 = 1, u2 = 1
u1 = 3, u2 = 0
Don’t confess
u1 = 0, u2 = 3
u1 = 2, u2 = 2
428 / 526
Nash equilibrium I
• In the strategic setting of game theory, we will adopt a notion
of equilibrium formalized by John Nash in the 1950s called the
Nash Equilibrium
• Nash equilibrium (NE) involves strategic choices that, once
made, provide no incentives for the players to alter their
behavior further
• A NE is a strategy for each player that is the best choice for
each player given the others’ equilibrium stratgies
• NE can be defined in terms of best responses. In an n-person
game, strategy si is a best response to rivals’ strategies s−i if
player i cannot obtain a strictly higher payoff with any other
possible strategy, si0 ∈ Si given that rivals are playing s−i
429 / 526
Nash equilibrium II
Definition of best response
si is a best response for player i to rivals’ strategies s−i , denoted
si ∈ BRi (s−i ), if ui (si , s−i ) ≥ ui (si0 , s−i ) for all si0 ∈ Si
Definition of Nash equilibrium (NE)
A NE is a strategy profile (s1∗ , s2∗ , . . . , sn∗ ) such that for each player
i = 1, 2, . . . , n, si∗ is a best response to the other players’
∗ . That is s ∗ ∈ BR (s ∗ ).
equilibrium strategies s−i
i −i
i
In a simple 2-person game, (s1∗ , s2∗ ) is a NE if s1∗ and s2∗ are mutual
best responses to each other:
u1 (s1∗ , s2∗ ) ≥ u1 (s1 , s2∗ ) for all s1 ∈ S1 and
u2 (s2∗ , s1∗ ) ≥ u1 (s2 , s1∗ ) for all s2 ∈ S2
430 / 526
Prisoner’s dilemma IV
Bonnie
Clyde
Confess Don’t confess
Confess
1 ,1
3,0
Don’t confess
0,3
2,2
431 / 526
Prisoner’s dilemma V
Bonnie
Clyde
Confess Don’t confess
Confess
1 ,1
3 ,0
Don’t confess
0,3
2,2
432 / 526
Prisoner’s dilemma VI
Bonnie
Clyde
Confess Don’t confess
Confess
1 ,1
3 ,0
Don’t confess
0,3
2,2
433 / 526
Bonnie
Prisoner’s dilemma VII
Confess
Clyde
Don’t confess
Confess
1 ,1
3 ,0
Don’t confess
0, 3
2,2
434 / 526
Bonnie
Prisoner’s dilemma VIII
Confess
Clyde
Don’t confess
Confess
1 ,1
3 ,0
Don’t confess
0, 3
2,2
435 / 526
Prisoner’s dilemma IX
• Confess, confess is a NE in the prisoner’s dilemma game
because confession is a best response to the other player’s
confessing.
• But notice: confessing is the best response to all the other
player’s strategies, confess or don’t confess
• A strategy that is a best response to any strategy the other
players might choose is called a dominant strategy.
• Players do not always have dominant strategies, but when they
do there is a strong reason to believe they will play that way.
Definition of dominant strategies
A dominant strategy is a strategy si∗ for player i that is a best
response to all strategy profiles of other players. That is,
si∗ ∈ BRi (s−i ) for all s−i
436 / 526
Battle of the sexes I
• Prisoner’s dilemma had several unique features
• Unique NE in pure strategy space
• Dominant strategy NE
• Inefficient outcome (i.e., Don’t confess, don’t confess
dominantes utility payoffs)
• Another game will allow us to better understand what is and
is not essential to a game: battle of the sexes
• Multiple equilibria
• Mixed strategy
437 / 526
Battle of the sexes II
Narrative: Scott and Paige are a married couple who have made a
date. After work they will each drive to the movie theater. Scott
prefers Batman to Jane Austen’s Emma adaptation. Paige prefers
Emma to Batman. They each prefer being on a date together than
being alone in either movie. Assume that on the day of the date,
they have forgotten which movie they agreed to, and cannot
coordinate (i.e., they cannot reach one another by phone for
whatever reason).
Discuss the three elements of a game. Set it up in normal form
438 / 526
Battle of the sexes III
Scott (1)
Batman
Emma
Paige (2)
Batman Emma
2,1
0,0
0,0
1,2
439 / 526
Battle of the sexes IV
Paige (2)
Batman Emma
Scott (1)
Batman
2 ,1
0,0
Emma
0,0
1 ,2
440 / 526
Battle of the sexes V
• Compare this with Prisoner’s dilemma
• Two NE in pure strategy space vs. 1
• No dominant strategy
• John Nash, “Equilibrium Points in n-person games”,
Proceedings of the National Academy of Sciences 36 (1950):
48-49
• Existence proof
• All games have at least one NE
• John Harsanyi, “Oddness of the Number of Equilibrium
Points: A New Proof”, International Journal of Game Theory
2 (1973): 235-250.
• Almost all games have an odd number of NE
• PD has one NE, but BoS has two. Where’s the third?
• Exceptions include games in which there are ties between
payoffs. These may have an even or infinite number of NE.
Doesn’t apply with our BoS example though as there are no
ties.
441 / 526
Rock, Paper, Scissors I
• Write down a 3×3 normal form representation of the rock,
paper scissors game
• Assume 1 point if the player wins, −1 points if the player
loses, and 0 if it’s a tie
• How many NE in pure strategy space do you identify?
442 / 526
Rock, Paper, Scissors II
Player 1
Rock
Paper
Scissors
Rock
0,0
1,-1
-1,1
Player 2
Paper Scissors
-1,1
1,-1
0,0
-1,1
1,-1
0,0
443 / 526
Rock, Paper, Scissors III
Rock
Player 1
Player 2
Paper Scissors
Rock
0,0
-1, 1
1 ,-1
Paper
1 ,-1
0,0
-1, 1
Scissors
-1, 1
1 ,-1
0,0
444 / 526
Rock, Paper, Scissors IV
• Contrast with PD and BoS
• No pure strategy NE
• No dominant strategy
• What about Nash and Harsanyi?
• Nash’s proof ensures that a NE exists – just not in pure
strategy space
• Harsanyi’s proof ensures that there will be an odd-number of
NE – so we need only to find the one other NE in the non-pure
strategy space to solve this game
• What else is there besides pure strategy space?
445 / 526
Mixed strategy I
• In poker, sometimes when you are dealt a bad hand, you
should fold, and sometimes you should play the hand
aggressively (i.e., bluff)
• But when? And how often?
• There’s an optimal bluffing behavior in which the player will
mix between folding and playing aggressively which is to
randomize bluffing at some rate that the player has to
calculate
• We now introduce the basic framework and will focus on the
Battle of the Sexes game
• Our goal is to learn how to solve for mixed strategy NE in the
BoS game
446 / 526
Mixed strategy II
• Assume that players’ strategies are probabilities of actions,
rather than the actions themselves
• Ex: Scott plays Batman with probability 12
• Let Scott’s probability of Batman be h and probability of
Emma be 1 − h where 0 ≤ h ≤ 1
• Let Paige’s probability of Batman be w and probability of
Emma be 1 − w where 0 ≤ w ≤ 1
• Calculate Scott’s expected payout
U1 [(h, 1 − h), (w , 1 − w )] = wh2 + (1 − w )(1 − h)1
= 3hw + 1 − w − h
• Calculate Paige’s expected payout
U2 [(w , 1 − w ), (h, 1 − h)] = wh + (1 − w )(1 − h)2
= 3hw + 2 − 2h − 2w
447 / 526
Mixed strategy III
• Assume that Paige is playing 21 , 12 . What should Scott play?
• Calculate Scott’s expected payout
1 1
U1 h, 1 − h),
,
= 3hw + 1 − w − h
2 2
1
1
= 3h( ) + 1 − − h
2
2
1
1
=
h+
2
2
• Now what value of h∗ will maximize expected utility?
• If h∗ = 0: U1 = 12 util
• If h∗ = 1: U1 = 1 util
448 / 526
Mixed strategy IV
• A player will only randomize between two actions if in
equilibrium she is getting the same expected payoff from
either action
• Implication: use an indifference condition to solve solve mixed
strategy equilibria
• Calculate expected payouts under certainty
U2 (Batman, (h, 1 − h)) = 3w + 1 − w − 1 = 2w
U2 (Emma, (h, 1 − h)) = 1 − w
• When is Paige indifferent between playing Batman or Emma
with certainty?
U2 (Batman, (h, 1 − h)) = U1 (Emma, (h, 1 − h))
w
w∗
= 2 − 2w
2
=
3
449 / 526
Mixed strategy V
• Calculate expected payouts under certainty for Scott
U1 (Batman, (w , 1 − w )) = 2w
U1 (Emma, (w , 1 − w )) = 1 − w
• Impose indifference condition and solve for equilibria strategies
U1 (Batman, (w , 1 − w )) = U1 (Emma, (w , 1 − w ))
2w
w
∗
= 1−w
1
=
3
450 / 526
Mixed strategy VI
• We found two pure strategy NE: Batman, Batman and
Emma, Emma
• There’s an odd number of NE – the third was in mixed
strategy space
• The mixed strategy NE was w ∗ = 23 ,1 − w ∗ = 13 and h∗ = 13 ,
1 − h∗ =
2
3
451 / 526
Continuum of actions I
• Now we will look at situations where the players’ actions can
be drawn from a continuum of possibilities, such as when
firms set prices or quantity
• Two shepherds are sharing a “commons” area for grazing
sheep. The market price for sheep is based on the total
number of sheep brought to market
v (q1 , q2 ) = 120 − Q
where Q = q1 + q2 and qi is shepherd i’s output
• This implies that the value of grazing a given number of sheep
is lower the more sheep are around competing for grass
• We cannot use a matrix to represent the normal form of this
game of continuous actions; the normal form is simply a
listing of the herder’s payoff functions:
u1 (q1 , q2 ) = q1 v = q1 (120 − q1 − q2 )
u2 (q2 , q1 ) = q2 (120 − q1 − q2 )
452 / 526
Continuum of actions II
• Solve shepherd 1’s revenue-max problem; find BR function
maxq1 U1 = q1 (120 − q1 − q2 ) = 120q1 − q12 − q1 q2 = 0
1
BR1 (q2 ) = 60 − q2
2
• Solve shepherd 2’s revenue-max problem; find BR function
maxq2 U2 = q2 (120 − q1 − q2 )
1
BR2 (q1 ) = 60 − q1
2
∗
∗
• The NE is given by the pair (q1 , q2 ) that satisfies both best
response functions. Use substitution
1
1
1
q1 = 60 + (60 − q2 ) = 30 + q1
2
2
4
q1∗ = 40
• Value = 120 − q1 − q2 = 120 − 80 = 40
453 / 526
Continuum of actions III
• Why is this such a tragedy? To see let’s consider the amount
of grazing that would maximize the revenue function
max
Q U(Q)
= Q(120 − Q)
= 120Q − Q 2
= 120 − 2Q = 0
Q
∗
= 60
• Contrast this to what happens under competition:
Q = q1 + q2
= 40 + 40
= 80
• Value = 120 − Q = 120 − 60 = 60
454 / 526
Sequential games I
• Previous games assumed players made decisions
simultaneously either implicitly or explicitly
• Now we relax this assumption and introduce the idea of a
sequential decision-making
• To motivate the game, we will return to the Battle of the
Sexes set up
• Paige will make her decision; Scott will observe her decision
then make his decision
• Contingent strategy: Scott’s decisions are conditional upon, or
contingent, on what Paige does
455 / 526
Sequential games II
• Paige has only two strategies (no contingent strategies):
Batman or Emma
• Scott has four contingent strategies:
1 Always go to Batman: (Batman | Batman, Batman | Emma)
2 Follow Paige: (Batman | Batman, Emma | Emma)
3 Do the opposite of Paige: (Emma | Batman, Batman | Emma)
4 Always go to Emma: (Emma | Batman, Emma | Emma)
• First use the normal form to solve for the NE
• Then we will draw the game out using the extensive form
456 / 526
Sequential games III
Scott (1)
Batman
Emma
Always Batman
2,1
0,0
Paige (2)
Follow Paige Opposite
2,1
0,0
1,2
0,0
Always Emma
0,0
1,2
• The hard part was incorporating the conditional strategies into normal form
• Everything else is the same: find best responses column by column, row by row
to find NE
457 / 526
Sequential games IV
Always Batman
Scott (1)
Paige (2)
Follow Paige Opposite
Always Emma
Batman
2 ,1
2 ,1
0 ,0
0,0
Emma
0,0
1,2
0 ,0
1 ,2
458 / 526
Sequential games V
Always Batman
Scott (1)
Paige (2)
Follow Paige Opposite
Always Emma
Batman
2 ,1
2 ,1
0 ,0
0,0
Emma
0,0
1, 2
0 ,0
1 ,2
459 / 526
Sequential games VI
Always Batman
Scott (1)
Paige (2)
Follow Paige Opposite
Always Emma
Batman
2 ,1
2 ,1
0 ,0
0,0
Emma
0,0
1, 2
0 ,0
1 ,2
• Three NE (oddness of NE)
• But two of these are not like the other
• Always Batman and Always Emma
• Paige really would play “Always Emma” if Scott went to Batman?? Is that
credible?
• We aren’t done. Let’s break this down using the sequential method
460 / 526
Sequential games VII
Scott
Batman
Emma
Paige
Batman
Emma
Batman
Emma
(1, 2)
(0, 0)
(0, 0)
(2, 1)
461 / 526
Sequential games VIII
Scott
Batman
Emma
Paige
Batman
Emma
Batman
Emma
(1, 2)
(0, 0)
(0, 0)
(2, 1)
462 / 526
Sequential games IX
• The first tree is the sequential form. Note the lack of a
dashed circle on player 2.
• The second tree is the extensive form representation of the
simultaneous game. The dashed circle represents ignorance
and is called an information set.
• We introduce a new NE concept called “subgame-perfect
equilibrium”
Definition of subgame-perfect equilibrium
A subgame-perfect equilibrium is a strategy profile (s1∗ , s2∗ , . . . , sn∗ )
that is a NE on every proper subgame
• How do we identify the subgame-perfect NE? We use
backwards induction.
463 / 526
Sequential games X
Scott
Batman
Emma
Paige
Batman
Emma
Batman
Emma
(1, 2)
(0, 0)
(0, 0)
(2, 1)
464 / 526
Sequential games XI
• When we break the game into three subgames, we find that
only (Emma, Follow Paige) survives
• What happened to (Emma, Always Emma) and (Batman,
Always Batman)?
• Neither was “subgame perfect” – Scott wouldn’t play Always
Emma in a world where Paige played Batman, nor Always
Batman in a world where Paige played Emma
• But “Follow Scott” was subgame perfect
• What was the “problem” with these other two NE we just
eliminated? They were based on non-credible threats
• Examples of credible threats: disabling the steering column in
a game of chicken; burning ships when invading; placing
attorneys on retainer
• Scott had no reason to respect the threat; backwards
induction ruled them out; subgame perfection is a NE solution
concept that eliminates these somewhat implausible NE
465 / 526
Incomplete information I
• Everything so far has assumed that players have complete
information
• This means they are aware of the preferences of themselves,
their opponents, and their rivals’ own knowledge
• But in some situations, it may not be realistic to assume that
players are completely informed
• Examples: Teams may conceal injuries; Firms may be unable
to ascertain the quality of a job applicant
• Harsanyi introduced the concept of “player types”
• At some initial node (“chance node”), a player’s type is drawn
from some distribution
• She may be injured or not injured for instance
• We introduce a new kind of NE concept called Bayesian NE
466 / 526
Incomplete information II
Definition of Bayesian NE
In a two player, simultaneous move game in which player 1 has
private information, a Bayesian NE is a strategy profile (s1∗ (t), s2∗ )
such that s1∗ (t) is a best response to s2∗ for each type t ∈ T of
player 1
U1 (s1∗ (t), s2∗ , t) ≥ U1 (s10 , s2∗ , t) for all s10 ∈ S1
and such that s2∗ is a best response to s1∗ (t) given player 2’s beliefs
Pr (tk ) about player 1’s types:
X
X
Pr (tk )U2 (s2∗ , s1∗ (tk ), tk ) ≥
Pr (tk )U2 (s20 , s1∗ (tk ), tk )
tk ∈T
tk ∈T
for all s20 ∈ S2
467 / 526
Incomplete Information II
Player 1
U
D
Player 2
L
R
t,2 0,0
2,0 2,4
• t = 6 with probability 12 ; t = 0 with probability 21
468 / 526
Incomplete information III
Player 1
U
D
Player 2
L
R
t,2 0,0
2,0 2,4
• t = 6 with probability 12 ; t = 0 with probability 21
• If t = 0, player 1’s best response is D if L; D if R
• If t = 6, player 1’s best response is U if L; D if R
• Only 2 candidates for an equilibrium in pure strategies
• 1 plays (U|t = 6, D|t = 0) and 2 plays L
• 1 plays (D|t = 6, D|t = 0) and 2 plays R
• Player 2’s expected payout from L if first strategy is played
1
1
(2) + (0) = 1
2
2
• Player 2’s expected payout from R if first strategy is played
1
1
(0) + (4) = 2
2
2
• Player 2’s expected payout if second strategy is played is 4.
• So player 2 plays R. Player 1’s BR to R is to play D
469 / 526
Incomplete information IV
• If the probability that player 1 is of type t = 6 is high enough,
can the first candidate be a Bayesian NE? If so, compute the
threshold probability
470 / 526
Bayes Rule I
• Next we talk explicitly about sequential games with
incomplete information
• Example: you’re playing poker, and the person in front of you
raises. What’s your best response?
• It depends on what you learned from that raise and what cards
you’re holding.
• Before we can do this, we’re going to talk about a classic
piece of statistics known as Bayes Rule
• Rev. Thomas Bayes, 19th century statistician and pastor
• Let A and B be two events. We want to calculate the
conditional probability, Pr (A|B)
471 / 526
Bayes Rule II
Pr (A and B)
Def’n conditional prob
Pr (B)
Pr (A and B) = Pr (A|B)Pr (B)
Pr (B and A)
Pr (B|A) =
Pr (A)
Pr (B and A) = Pr (B|A)Pr (A)
Pr (A|B) =
Pr (A and B) = Pr (B and A)
Pr (B|A)Pr (A)
Pr (A|B) =
Naive Bayes Rule
Pr (B)
472 / 526
Bayes Rule III
Pr (B) = Pr (B and A) + Pr (B and ∼ A)
= Pr (B|A)Pr (A) + Pr (B| ∼ A)Pr (∼ A)
Pr (B|A)Pr (A)
Pr (A|B) =
Pr (B)
Pr (B|A)Pr (A)
Pr (A|B) =
Pr (B|A)Pr (A) + Pr (B| ∼ A)Pr (∼ A)
473 / 526
Bayes Rule IV
Narrative: There are three doors labeled 1, 2 and 3. Behind one of
the doors is a million dollars; behind the other two are goats. The
game show host, Monty Hall, asks you to select a door. Assume
you select door 1. After you’ve selected your door, he opens door 2
and shows you a goat behind it. Monty Hall then asks you “would
you like to change your pick from door 1 to door 2?” What’s the
probability that the money is behind door 3 vs. door 1?
474 / 526
Bayes Rule V
• Let D1 =“money is behind door 1”, D2 =“money is behind door 2, and
D3 =“money is behind door 3
• Let O =“Door 2 was opened”
• We want to know the probability that the money is behind door 3 given he
opened door 2. Calculate Pr (D3 |O) using Bayes Rule
Pr (D3 |O) =
Pr (O|D3 )Pr (D3 )
Pr (O|D3 )Pr (D3 ) + Pr (O|D1 )Pr (D1 ) + Pr (O|D2 )Pr (D2 )
• “Prior beliefs”: Pr (D1 ) = Pr (D2 ) = Pr (D3 ) = 31
• “Posterior beliefs”: Pr (D3 |O)
• Signals: Pr (O|Di ). Given the money is behind door i, what’s the probability
Monty Hall opens door 2? Hint: Monty Hall would never open door 2 if it had
the money behind it
475 / 526
Bayes Rule VI
Pr (D3 |O) =
1
2
3
Pr (O|D3 )Pr (D3 )
Pr (O|D3 )Pr (D3 ) + Pr (O|D1 )Pr (D1 ) + Pr (O|D2 )Pr (D2 )
Pr (O|D1 ) = 0.5. Monty Hall can pick either door 2 or 3.
Pr (O|D2 ) = 0.0. Monty Hall would never open door 2 if it had the money
behind it
Pr (O|D3 ) = 1.0. Monty Hall has no choice but to open door 2 if the money is
behind door 3
Pr (D3 |O) =
=
1·
2
3
1
3
1 · 13
+ 12 · 13 + 0 ·
1
3
• We know the money isn’t behind door 2
• By the law of total probability, Pr (D1 |O) = 13 (Calculate this value to convince
yourself)
• Definitely switch to door 3!
476 / 526
Job Market Signaling I
• Player 1 is a worker.
• There are two types t of workers: high (H) and low types (L).
• Nature assigns Player 1’s type at probability Pr (H), and
Pr (H) + Pr (L) = 1.0.
• Player 1 observes her own type
• Player 1 can choose to invest in a signal, E , (e.g., education)
which is costly (ct )
• After deciding whether to invest in education, player 1 sends a
job application to Player 2
• Player 2 is a firm.
• Player 2 decides whether to offer a job J or not NJ to Player 1
at wage w
• Player 2 observes whether Player 1 is educated
• Player 2 does not observe Player 1’s type
• If Player 2 hires a high type, he earns π minus w
• If Player 1 hires a low type, he earns 0 minus w
477 / 526
Job market signaling III
478 / 526
Job market signaling IV
Definition of subgame perfect bayesian NE
A subgame perfect Bayesian NE consists of a strategy profile and a
set of beliefs such that
• at each information set, the strategy of the player moving
there maximizes his or her expected payoff, where the
expectation is taken with respect to his or her beliefs, and
• at each information set, where possible, the beliefs of the
player moving there are formed using Bayes’ rule (based on
prior beliefs and other players strategies/signals)
479 / 526
Job market signaling V
• Expected payout from hiring educated worker
EU2 (J|E ) = Pr (H|E )(π − w ) + (1 − Pr (H|E ))(−w )
= Pr (H|E )π − w
• Expected payout from not hiring educated worker
EU2 (NJ|E ) = 0
• Offer job if and only if EU2 (J|E ) > 0
Pr (H|E )π − w
> 0
w
Pr (H|E ) >
π
• Calculate right-hand-side using Bayes Rule
480 / 526
Job market signaling VI
Pr (H|E ) =
Pr (E |H)Pr (H)
Pr (E |H)Pr (H) + Pr (E |L)Pr (L)
• Substitute into previous equation
Pr (H|E ) >
Pr (E |H)Pr (H)
Pr (E |H)Pr (H) + Pr (E |L)Pr (L)
>
w
π
w
π
• We need to know something about player strategies: Pr (E |H)
and Pr (E |L)
481 / 526
Job market signaling VII
1
Separating equilibrium: High types always signal and low
types never signal; firms only hire educated workers
• Pr (E |H) = 1.0, Pr (E |L) = 0.0
• Pr (J|E ) = 1.0, Pr (J|NE ) = 0.0
Pr (E |H)Pr (H)
Pr (E |H)Pr (H) + Pr (E |L)Pr (L)
Pr (H)
Pr (H)
w
π
w
>
π
w
1 >
π
π > w
>
• Requires that π > w
482 / 526
Job market signaling VIII
• For a separating equilibrium, we also need for high types to
always signal and low types to never signal
• This will occur under the following conditions
• Educated high types earn w − cH . Non-educated high types
earn 0. So education is an equilibrium if w − cH > 0, or
w > cH
• Educated low types earn w − cL . Non-educated low types earn
0. So no education is an equilibrium if w − cL < 0 or w < cL
• Separating equilibrium requires that cH < w < cL
• Question: Why does the worker sometimes obtain an
education even though it does not raise her skill level? Would
the separating equilibrium exist if a low type could obtain an
education more easily than a high type?
483 / 526
Job market signaling IX
2
Pooling equilibrium: High and low types always signal; firms
only hire educated workers
• Pr (E |H) = Pr (E |L) = 1.0
Pr (H|E ) >
Pr (E |H)Pr (H)
Pr (E |H)Pr (H) + Pr (E |L)Pr (L)
Pr (H)
Pr (H) + Pr (L)
>
>
Pr (H) >
w
π
w
π
w
π
w
π
• The probability of a high type has to be sufficiently high to
support a pooling equilibrium
484 / 526
Job market signaling X
• What about hiring non-educated workers?
Pr (H|NE )(π − w ) + (1 − Pr (H|NE ))(−w ) > 0
Pr (H|NE )π − w
> 0
w
Pr (H|NE ) >
π
• Summarize: Pr (H|NE ) < w
π < Pr (H). The likelihood a
non-educated worker is a high type must also be sufficiently
low in the population.
485 / 526
Job market signaling XI
3
Hybrid equilibrium: High types always signal, low types signal
with probability e, firms hire educated workers with probability
j, and never hire non-educated workers
• For low types to mix their strategy between education and
non-education, they must get the same payout either way
(indifference)
jw − cL = 0
cL
j∗ =
w
486 / 526
Job market signaling XII
• For firms to mix, they must get the same payout from hiring
as they get from not hiring (0)
Pr (H|E )π − w
= 0
w
Pr (H|E ) =
π
Pr (E |H)Pr (H)
w
=
Pr (E |H)Pr (H) + (1 − Pr (H))e
π
Pr (H)
w
=
Pr (H) + (1 − Pr (H))e
π
(π − w )Pr (H)
e∗ =
w [1 − Pr (H)]
487 / 526
Allocation of Time
488 / 526
Gary Becker
• Gary Becker (Columbia, University of Chicago) won the Nobel
Prize in Economics in 1992 “for having extended the domain
of microeconomic analysis to a wide range of human
behaviour and interaction, including nonmarket behaviour”.
• His contributions to economic science include theories of crime
and punishment, human capital, the family, fertility, marriage
and divorce, the allocation of time within households, and
much more
• His theory of the allocation of time within households was
developed at some point while working with Jacob Mincer in
the famous labor economics seminar at Columbia, where
Becker spent many years early in his career
• Becker, Gary S. (1965). ”A Theory of the Allocation of Time”.
Economic Journal
• In this model, each household acts as if it were an individual –
i.e., the model ignores bargaining between spouses and
children and other allocation problems within the household
489 / 526
Allocation of time
• This individual maximizes her utility that is a function of
various activities she likes to do
• Sleeping, eating, watching TV, playing soccer, volunteering at
her church, going to parties, working
• Becker calls these activities “commodities” but I think it’s
more intuitive to think of them as activities
• Amartya Sen, “things we like to do or states we like to be”
(e.g., eating is something we like to do, being healthy is a
state we like to be)
• The model will have a few moving parts, which we will
simplify into the familiar utility maximization model with two
parts: utility and a resource constraint
490 / 526
Allocation of time
• Suppose there are two activities, Z1 and Z2 , that this person
likes. To be concrete, let Z1 be playing basketball outside and
Z2 be watching TV shows inside
U = U(Z1 , Z2 )
where U1 > 0; U2 > 0; U11 < 0; U22 < 0
• These conditions ensure that utility is increasing and concave
function of the two activities (i.e., positive but diminishing
marginal utility)
491 / 526
Allocation of time
• Each activity (basketball and watching TV) is carried out, or
“produced”, by combining market purchased basketball
equipment and time spent playing basketball (and vice versa
for TV)
• Imagine one of the Z goods was getting an education
• Then the activity of education is achieved by combining goods,
like textbooks, with study time
• The activity production functions are:
Z1 = Z1 (X1 , t1 )
Z2 = Z2 (X2 , t2 )
• The goods budget constraint is:
P1 X1 + P2 X2 = wL + V
where P1 and P2 are the prices of consumption goods, w is
the market wage, L is time spent working, V is non-labor
income (e.g., parent allowance)
492 / 526
Allocation of time
• The second constraint is a time constraint:
t1 + t2 + L = T
where T is her time endowment (i.e., 24 hours minus time spent
sleeping)
• Combine the first and second constraint:
P1 X1 + P2 X2 + wt1 + wt2 = wT + V
P1 X1 + wt1 + P2 X2 + wt2 = wT + V
• The first two terms on LHS is total expenditure on activity Z1 (e.g.,
playing basketball). Note expenditure has two components:
the money cost of the goods used for this activity (e.g., the ball and
shoes used for basketball)
2 the opportunity cost of the time used for this activity (lost earnings due
to time spent playing basketball)
1
• The right hand side is “full income”, or the maximum possible income
for this person – this is what she would earn if she allocated her entire
time endowment to work
493 / 526
Example 1: Perfect complements
• Becker’s simplest model adds an additional assumption:
goods and time will be perfect substitutes in the production of
each activity
• You always need one ball, one pair of shoes, a court and an
hour to play basketball.
• You always need one TV, and an hour of time to watch TV
• Activity production functions can now be written in the
following form:
Xi
= bi Zi
ti
= ci Zi
where bi and ci are constant parameters that represent the
fixed input intensity of each activity. If b1 is 2, then each unit
of basketball activity, Z1 is produced with 2 units of the good
X1 (i.e., shoes and ball)
494 / 526
Example 1: Perfect complements
• This is a simplifying assumption that allows us to write the full
income budget constraint into something simpler:
(P1 b1 + wc1 )Z1 + (P2 b2 + wc2 )Z2 = wT + V
• Note the budget constraint has now been expressed as a function of
the two activities, rather than the four inputs (X1 , X2 , t1 , t2 )
• The coefficient on Zi (the activities) tells us the “unit cost” of that
activity
• In order to produce one unit of basketball, she has to spend P1 b1 + wc1
where P1 is the price of one good (e.g., basketball and shoes) and b1 is
how many goods she needs to produce one unit of basketball (e.g., 2
goods).
• Similarly, w is the price of one unit of time, and c1 is how much time
she needs to produce one unit of basketball (e.g., one hour)
• Because these input intensities are fixed, her maximization problem is
straightforward: maximize utility by choosing optimal levels of Z1 and
Z2
495 / 526
Example 1: Perfect complements
• Use the Lagrangean:
maxZ1 ,Z2 = U(Z1 , Z2 ) + λ(wT + V − π1 Z1 − π2 Z2 )
where πi = Pi bi + wci , which are the combined time and
goods cost of each activity (i.e., the full price of activity Zi )
• First order conditions:
U1 − λπ1 = 0
U2 − λπ2 = 0
π1 Z1 + π2 Z2 = wT + V
• Combine the first two FOCs like we always do:
U1
π1
P1 b1 + wc1
=
=
U2
π2
P2 b2 + wc2
• The third FOC tells us the optimal lies on the budget
constraint. Taking these conditions together, we see the
optimum lies at the tangency between the indifference curve
and the budget constraint
496 / 526
Example 1: Perfect complements
• Notice that the slope of the budget constraint (i.e., the
relative price of the activities) can be influenced by changes in
seven different parameters (P1 , b1 , c1 , P2 , b2 , c2 , and w )
• Three of these parameters are prices (P1 , P2 , w )
• Four are input input intensities of the activities (b1 , b2 , c1 , c2 )
• To see this more clearly, let’s examine how each of these
parameters influences the relative price R =
π1
π2
∂R
∂R
∂R
> 0;
> 0;
>0
∂P1
∂b1
∂c1
∂R
∂R
∂R
< 0;
< 0;
<0
∂P2
∂b2
∂c2
497 / 526
Example 1: Perfect complements
• The following events would make us substitute away from
basketball (i.e., activity Z1 ) towards watching TV (i.e.,
activity Z2 )
Increase in the price of goods used for basketball
Decrease in the price of goods used for watching TV
Increase in the goods and/or time intensity used for playing
basketball
4 Decrease in the goods and/or time intensity used for watching
TV
1
2
3
• Examples:
• Gym membership price increases, she substitutes away from
basketball and towards watching TV
• If the price of both basketballs and DVDs fell by the same
proportion, the relative price of watching a DVD could
decrease if the activity of watching movies is more goods
intensive than the activity of playing basketball. In other
words, the impact of a price change of a good is greater on
activities that are more goods intensive
498 / 526
Example 1: Perfect complements
• The time and goods intensities of activities may change with
technological improvements in household consumption activity
• A microwave reduces the time intensity of cooking and eating,
as does “fast food”
• Pipe-borne hot water reduces the time intensity of showering
• Netflix and DVRs will reduce the time intensity of watching
movies
• The impact of an increase in wages on the relative price of the
activities is tricker to establish, though, because the wage is
the implicit price of time used for both activities
• However, it’s reasonable to suppose that the impact of a wage
increase would be to increase the relative price of the more
time intensive activity
499 / 526
Example 1: Perfect complements
∂R
∂w
=
=
c1 (P2 b2 + wc2 ) − c2 (P1 b1 + wc1 )
(P2 b2 + wc2 )2
P2 b2 c1 − P1 b1 c2
(P2 b2 + wc2 )2
• The denominator is always positive. The numerator is positive only if:
P2 b2 c1 − P1 b1 c2 > 0
P1 b1
c1
>
c2
P2 b2
• In other words, an increase in the wage will increase the relative price
of basketball if basketball is sufficiently more time intensive than
watching TV. Therefore she substitutes away from basketball and
towards watching TV
• The lower the relative price of goods used in basketball, and the lower
the relative goods intensity of basketball, this substitution will happen
at a lower level of c1 (i.e., the inequality condition will be satisfied at a
lower level of cc12 )
500 / 526
Demand functions
• If we specify the utility function, we can get specific forms for
the demand functions. Let utility be Cobb-Douglas
U(Z1 , Z2 ) = Z1α Z2β
• Recall the first order conditions from earlier:
U1 − λπ1 = 0
U2 − λπ2 = 0
wT + V − π1 Z1 − π2 Z2 = 0
• Solve for Z1∗ and Z2 ∗
Z1 =
Z2 =
α wT + V
α+β
π1
α wT + V
α+β
π2
501 / 526
Demand functions
• Because of the assumption of fixed time intensities, we can
easily obtain the demand functions for goods and time. Recall
their form
• Substituting
Zi∗
Xi
= bi Zi
ti
= ci Zi
and πi into these equations, we get
X1 =
X2 =
t1 =
t2 =
α b1 (wT + V )
α + β P1 b1 + wc1
β b2 (wT + V )
α + β P2 b2 + wc2
α c1 (wT + V )
α + β P1 b1 + wc1
β c2 (wT + V )
α + β P2 b2 + wc2
502 / 526
Comparative statics
• What happens to your demand for the goods used to play
basketball (watch TV) when your preference for basketball (α)
increases?
∂X1
>0
∂α
∂X2
<0
∂α
• Proposition 1: When the consumer’s preference for the first
activity increases, then her demand for the goods used to
produce that activity (e.g., basketballs, shoes) increases and
her demand for the goods used to produce the alternative
activity (e.g., DVDs, TVs, apps) decreases
503 / 526
Comparative statics
• What happens to the optimal time spent on basketball (TV
watching) when your preference for basketball (α) increases?
∂t1
>0
∂α
∂t2
<0
∂α
• Proposition 2: When preference for the first activity increases,
then the optimal time spent on it increases and the optimal
time spent on the alternative activity falls
504 / 526
Comparative statics
• What happens to your demand for the goods used to produce
basketball (watch TV) if basketball becomes more good
intensive (i.e., basketball, shoes, and a uniform)?
∂X1
>0
∂b1
∂X2
=0
∂b1
• Proposition 3: When the first activity becomes more good
intensive (i.e., more stuff needed to play basketball), your
demand for those goods increases. There is no effect on the
demand for the alternative activity’s goods, though.
505 / 526
Comparative statics
• What happens to the optimal time spent playing basketball,
though, if it becomes more goods intensive (i.e., uniforms in
addition to basketballs and shoes)?
∂t1
<0
∂b1
∂t2
=0
∂b1
• Proposition 4: When the first activity becomes more good
intensive (i.e., more stuff needed to play basketball), the
optimal time spent playing basketball decreases. There is no
effect on the optimal time spent on the alternative activity,
though.
506 / 526
Comparative statics
• What happens to the demand for the goods used to produce
basketball (watching TV) if basketball becomes more time
intensive (i.e., longer playing times), t1 ?
∂X1
>0
∂t1
∂X2
=0
∂t1
• Proposition 5: When the first activity becomes more time
intensive, then the optimal time spent on that activity
increases, but there is no change in the optimal time spent on
the alternative activity.
507 / 526
Matching
508 / 526
Introduction
• In economics, we are interested in how societies allocate
resources
• Most allocation problems are solved by price system
• There is a price that responds to supply and demand that
clears the market
• Some allocation problems do not allow for the use of the price
system
• Organ donors, public school places to children, marriages
• How do we solve for the allocation problem in this kind of
market?
• The 2012 Nobel prize was awarded jointly to Alvin Roth and
Lloyd Shapley “for the theory of stable allocations and the
practice of market design”.
509 / 526
Introduction
• Assume individuals are of several observable types and
two-sided market
• Buyer-seller, worker-firm, worker-task, male-female marriages
• We will use the label “male” and “female” to refer to the
general case, but this is just metaphorical
• We are interested in answering the question about the optimal
way men and women should match
510 / 526
Pairwise matching with non transferable
utility
• Finite number of men and women
• Men propose to women, there is no trade or side payments
after the match
• Payoffs are equivalent to utils (ordinality) and there is non
transferable utility (NTU), or there is no money
• Who matches with who?
511 / 526
Pairwise matching with non transferable
utility
• Example: Hip-hop and classic rock artists. A hip-hop artist
(male) has to match with a classic rock artist (female) to
collaborate on a song. The utilities for each man and woman
are the following
Snoop Dogg
50 Cent
Jay-Z
Pink Floyd
2,23
4,16
6,9
Bob Dylan
4,24
8,18
12,12
Hendrix
6, 25
12,20
18,15
512 / 526
Pairwise matching with non transferable
utility
• Every rapper asks Hendrix, since he is most preferred for
everyone. Hendrix accepts Snoop and dumps 50 Cent and
Jay-Z. In the next round, 50 Cent and Jay-Z ask Dylan. Dylan
accepts 50 Cent and dumps Jay-Z. Finally, Jay-Z and Pink
Floyd match. Thus the male optimal stable marriages are:
(Jay-Z, Pink Floyd); (50 Cent, Dylan); (Snoop, Hendrix)
Snoop Dogg
50 Cent
Jay-Z
Pink Floyd
2,23
4,16
6,9
Bob Dylan
4,24
8,18
12,12
Hendrix
6, 25
12,20
18,15
513 / 526
Deferred acceptance algorithm
• David Gale and Lloyd Shapley, (1962) “College Admissions and
Stability of Marriage”, American Mathematical Monthly
• The deferred acceptance algorithm involves a number of rounds of
proposals to determine who matches with who and solve the stable
marriage problem. It assumes no costs of waiting, searching dissolving
a match. The algorithm involves the following stages:
Each unengaged male proposes to the most-preferred woman to whom
he has not yet proposed assume he is better than remaining single
2 Each woman, regardless if she is engaged, considers all her suitors and
accepts to match with the one she prefers the most (if better than
single). All the others are declined. A temporal match would arise in
this stage that we call provisional engagement ( to all men she says
“no” and to one “maybe”).
3 Repeat 1 and 2
1
• If there is a finite number of men and women, the algorithm finishes
in finite time.
• If there is an equal number of men and women, and matching is
preferred to non-matching, then everybody ends up matched when
the algorithm finishes.
514 / 526
Definitions
Matching
S
A matching is a one-to-one correspondence from the set M onto
itself or order two (that is, µ2 (x) = x) such that if µ(m) 6= m then
µ(m) ∈ W and if µ(w ) 6= w ten µ(w ) ∈ M. We refer to µ(x) as
the mate of x.
Rationality
The matching µ is individually rational if each agent is
acceptable to his or her mate. That is, a matching is individually
rational if it is not blocked by any (individual) agent.
Stability
A matching µ is stable if it is not blocked by any individual or any
pair of agents.
515 / 526
Stability
P(m1 ) = w2 , w1 , w3 ; P(w1 ) = m1 , m3 , m2
P(m2 ) = w1 , w3 , w2 ; P(w2 ) = m3 , m1 , m2
P(m3 ) = w1 , w2 , w3 ; P(w3 ) = m1 , m3 , m2
Consider this matching:
µ = (m1 , w1 ); (m2 , w2 ); (m3 , w3 )
This is not stable because (m1 , w2 ) is a blocking pair. However the
matching:
µ = (m1 , w2 ); (m2 , w3 ); (m3 , w1 )
is stable because there does not exist a blocking pair.
516 / 526
Theorems
Theorem 1: The marriages produced by the algorithm are
stable
Proof: Let Alice and Bob be married, but not to each other. After
the algorithm, it is impossible that both prefer each other over
their current partners. If Bob prefers Alice to his current partner,
he must have proposed to Alice before he proposed to his current
partner. If Alice accepted his proposal, yet is not married to him at
the end, she must have dumped him for someone she likes more,
and so doesn’t like Bob more than her current partner. If Alice
rejected his proposal, she was already with someone she liked more
than Bob.
517 / 526
Theorems
Theorem 2: With an equal number of men and women,
everyone gets married.
Proof: by contradiction, suppose that at the end there is a man
unmatched. Then given that the number of men and women is the
same, there is also a woman unmatched. But, there cannot be a
man and a women both unengaged, as he must have proposed to
her at some point and, being unengaged, she would have to have
said yes.
518 / 526
Example of the deferred acceptance
algorithm
• Find the matching µ using the deferred acceptance algorithm.
Let males do the proposing.
P(m1 ) = w1 , w2 , w3 , w4 ; P(w1 ) = m2 , m3 , m1 , m4 , m5
P(m2 ) = w4 , w2 , w3 , w1 ; P(w2 ) = m3 , m1 , m2 , m4 , m5
P(m3 ) = w4 , w3 , w1 , w2 ; P(w3 ) = m5 , m4 , m1 , m2 , m3
P(m4 ) = w1 , w4 , w3 , w2 ; P(w4 ) = m1 , m4 , m5 , m2 , m3
P(m5 ) = w1 , w2 , w4
519 / 526
Example 2
• An example in which all men prefer the same woman, and all
women prefer the same man
P(m1 ) = w1 , w2 , w3 ; P(w1 ) = m1 , m2 , m3
P(m2 ) = w1 , w2 , w3 ; P(w2 ) = m1 , m3 , m2
P(m3 ) = w1 , w3 , w2 ; P(w3 ) = m1 , m2 , m3
Exercise: Find the stable matching using the deferred acceptance
algorithm when males propose and compare it to the matching you
find when females propose. Call this µM and µW , respectively
520 / 526
Example 2
µM
= (m1 , w1 ); (m2 , w2 ); (m3 , w3 )
µW
= (m1 , w1 ); (m2 , w3 ); (m3 , w2 )
• Notice that the deferred acceptance algorithm does not pin
down all the stable matches – it depends on who does the
proposing.
• Notice that all the men like the matching µM matching at
least as well as the matching µW .
• Man m1 is indifferent between the two matchings (he ends up
with w1 regardless of who does the proposing), and the other
men prefer µM .
• Similarly, the women are not in any disagreement that µW is
the optimal stable matching
521 / 526
M-optimality; W-optimality
Definition of M- (W-) optimality
For a given marriage market (M, W , P), a stable matching µ is
M-optimal if every man likes it at least as well as any other stable
matching; that is, if for every other stable matching µ0 , µ ≥M µ0 .
Similarly, a stable matching v is W-optimal if every woman likes it
at least as well as any other stable matching, that is, if for every
other stable matching v 0 , v ≥W v 0 .
522 / 526
M-optimality; W-optimality
• Each individual agent compares alternative matchings in
terms of his or her preferences for his or her own mates at
those matchings.
• So in examining the set of stable matchings, an agent is
involved in comparing those mates who he or she might have
at some other stable matching.
• In a marriage market in which all men and women have strict
preferences, each man and woman who has any achievable
mates has (only) one favorite among these.
523 / 526
M-optimality; W-optimality
Gale and Shapley
When all men and women have strict preferences, there always
exists an M-optimal stable matching, and a W-optimal stable
matching. Furthermore, the matching µM produced by the
deferred acceptance algorithm with men proposing is the
M-optimal stable matching. The W-optimal stable matching is the
matching µW produced by the algorithm when the women propose
524 / 526
M-optimality; W-optimality
• Proof by induction. Assume that up to a given step in the
procedure no man has yet been rejected by a woman who is
achievable for him. At this step, suppose woman w rejects
man m. If she rejects m as unacceptable, then she is
unachievable for him, and we are done. If she rejects m in
favor of man m0 , whom she keeps engaged, then she prefers
m0 to m. We must show that w is not achievable for m.
• We know m0 prefers w to any woman except for those who
have previously rejected him, and hence are unachievable for
him. Consider a hypothetical matching µ that matches m to
w and everyone else to an achievable mate. Then m0 prefers
w to his mate at µ. So the matching µ is unstable, since it is
blocked by m0 and w who each prefer the other to their mate
at µ.
• Therefore there is no stable matching that matches m and w ,
and so they are unachievable for each other, which completes
the proof.
525 / 526
M-optimality; W-optimality
Knuth
When all agents have strict preferences, the M-optimal stable
matching is the worst stable matching for the women; that is, it
matches each woman with her worst stable matching for the
women; that is, it matches each woman with her least preferred
achievable mate. Similarly, the W-optimal stable matching
matches each man with his least preferred achievable mate.
See proof on page 33 of Roth and Sotomayor.
526 / 526