Monotone Comparative Statics
Econ 2100
Lecture 10, October 5
Outline
1
Comparative Statics Without Calculus
2
Supermodularity
3
Single Crossing
4
Topkis’and Milgrom & Shannon’s Theorems
5
Midterm
Fall 2015
Comparative Statics Without Calculus
Remark
Using the implicit function theorem, one can show that if there are
complementarities between choice variable x and parameter q, the optimum
increases in q.
First Order Condition:
By IFT
fx (x; q) = 0.
xq (q) =
Second Order Condition: fxx < 0.
fxq (x; q)
:
fxx (x; q)
Then
xq (q)
0 if and only if fxq (x; q)
0
Here g is the derivative of g with respect to .
Issues with implicit function theorem:
1
IFT needs calculus.
2
The conclusion holds only in a neighborhood of the optimum.
3
The results are dependent on the functional form used for the objective
function.
1
In particular, IFT gives cardinal results that depend on the assumptions on f .
Monotone Comparative Statics
Objectives
With monotone comparative statics, we seek results about “changes” that:
do not need calculus
are not necessarily only local (around the optimum).
are ordinal in the sense of being robust to monotonic transformations.
The objective is to get a similar result without calculus.
The downside is that the results are not as strong.
Main Idea: Complementarities
We want to generalize the notion of complementarities between endogenous
variable and parameters.
With calculus, this is the assumption that fxq (x; q)
0.
We would also like to account for the possibility that the optimum is not
unique so that x (q) is not a function.
What does it mean for a correspondence to be increasing?
Strong Set Order
When can we say that one set is larger?
De…nition
For two sets of real numbers A and B, de…ne the binary relation
A
sB
if
s
as follows:
for any a 2 A and b 2 B
minfa; bg 2 B
A
s
and
maxfa; bg 2 A
B reads “A is greater than or equal to B in the strong set order”.
Generalizes the notion of greater than from numbers to sets of numbers.
According to this de…nition f1; 3g is not greater than or equal to f0; 2g.
This de…nition reduces to the standard de…nition when sets are singletons.
Non-Decreasing Correspondences
De…nition
We say a correspondence g : Rm ! 2R is non-decreasing in q if and only if
q0 > q
implies
g (q 0 )
s
g (q)
This says that q 0 > q implies that for any x 0 2 g (q 0 ) and x 2 g (q):
minfx 0 ; xg 2 g (q) and maxfx 0 ; xg 2 g (q 0 ).
Generalizes the notion of increasing function to correspondences.
Exercise
Prove that if g ( ) is non-decreasing and min g (q) exists for all q, then min g (q) is
non-decreasing.
Exercise
Prove that if g ( ) is non-decreasing and max g (q) exists for all q, then max g (q) is
non-decreasing.
Monotone Comparative Statics: Simplest Case
Set up
Suppose the function f : R2 ! R is the objective function; this is not
necessarily concave or di¤erentiable, and the optimizer could be set valued.
Let
x (q)
arg max f (x; q); subject to q 2
; x 2 S(q)
Notice that for any strictly increasing g ( ), this problem is equivalent to
x (q) arg max g (f (x ; q)); subject to q 2 ; x 2 S (q)
but g ( ) may destroy smoothness or concavity properties of the objective
function.
For now, assume S( ) is independent of q and both x and q are real variables.
Also, assume existence of a solution but not uniqueness.
Supermodularity
De…nition
The function f : R2 ! R is supermodular in (x; q) if
for all x 0 > x
f (x 0 ; q)
f (x; q) is non-decreasing in q:
If f is supermodular in (x; q), then the incremental gain to a higher x is
greater when q is higher.
This is the idea that x and q are “complements”.
Question 1, Problem Set 6.
Show that supermodularity is equivalent to the property that
q0 > q
implies
f (x; q 0 )
f (x; q) is non-decreasing in x:
Di¤erentiable Version of Supermodularity
When f is smooth, supermodularity has a characterization in terms of
derivatives.
Lemma
A twice continuously di¤erentiable function f : R2 ! R is supermodular in (x; q) if
and only if D12 f (x; q) 0 for all (x; q).
The inequality in the de…nition of supermodularity is just the discrete version
of the mixed-partial condition in the lemma.
q0 > q
implies
f (x ; q 0 )
f (x ; q) is non-decreasing in x
Topkis’Monotonicity Theorem
Theorem (Easy Topkis’Monotonicity Theorem)
If f is supermodular in (x; q), then x (q)
arg max f (x; q) is non-decreasing.
Proof.
Question 2, Problem Set 6 (do not use later results).
Supermodularity is su¢ cient to draw comparative statics conclusions in
optimization problems without other assumptons.
Topkis’Theorem follows from the IFT whenever the standard full-rank
condition in the IFT holds.
At a maximum, if D11 f (x ; q) 6= 0, it must be negative (by the second-order
condition), hence the IFT tells you that x (q) is strictly increasing.
Example
Pro…t Maximiization Without Calculus
A monopolist chooses output q to solve max p(q)q
c(q; ).
p( ) is the demand (price) function
c( ) is the cost function
costs depend on the existing technology, described by some parameter
.
Let q ( ) be the monopolist’s optimal quantity.
Suppose c(q; ) is supermodular in (q; ); then the entire objective function
is also supermodular in (q; ).
this follows because the …rst term of the objective does not depend on .
Notice that supermodularity says that for all q 0 > q,
nondecreasing in .
c(q 0 ; ) + c(q; ) is
in other words, the marginal cost is decreasing in .
Conclusion: by Topkis’theorem q is nondecreasing as long as the marginal
cost of production decreases in the technological progress parameter .
A Useful Trick
Sometimes one can “invent” an objective function in order to apply the
theorem.
Compare the Optima of Di¤erent Optimization Problems
Suppose one wishes to compare the solutions to two di¤erent maximization
problems,
max g (x)
and
max h(x)
x 2S
x 2S
Apply the theorem to an arti…cial function, f
(
g (x) if = 0
f (x; ) =
h(x) if = 1
If h(x) g (x) is nondecreasing, f is supermodular and therefore the solution
to the second problem is greater than the solution to the …rst.
Single-Crossing
In constrained maximization problems we have x 2 S(q), and supermodularity
is not enough to establish Topkis’theorem.
De…nition
The function f : R2 ! R satis…es the single-crossing condition in (x; q) if for all
x 0 > x, q 0 > q
f (x 0 ; q) f (x; q) 0
implies
f (x 0 ; q 0 ) f (x; q 0 ) 0
and
f (x 0 ; q)
f (x; q) > 0
implies
f (x 0 ; q 0 )
f (x; q 0 ) > 0:
As a function of the second argument, the marginal return can cross 0 at most
once; whenever it crosses 0, as the second argument continues to increase, the
marginal return is going to remain positive.
Theorem
If f satis…es single crossing in (x; q), then x (q) = arg maxx 2S (q) f (x; q) is
nondecreasing. Moreover, if x (q) is nondecreasing in q for all constraint choice
sets S, then f satis…es single-crossing in (x; q).
Monotone Comparative Statics
n-dimensional choice variable and m-dimensional parameter vector
Before we had a function of a real number and a parameter.
Next, we generalize to functions of vectors where the parameter is also a
vector.
De…nitions
Suppose x; y 2 Rn .
The join of x and y is de…ned by
x _ y = (maxfx1 ; y1 g; maxfx2 ; y2 g; : : : ; maxfxn ; yn g):
The meet of x and y is de…ned by
x ^ y = (minfx1 ; y1 g; minfx2 ; y2 g; : : : ; minfxn ; yn g):
Draw a picture.
A Simple Result
Lemma
(i) :
(ii)
x _y =x ,x
:
x ^y =x ,x
y , and
y
Proof.
Let x; y 2 Rn . Note that
x _ y = x , (maxfx1 ; y1 g; : : : ; maxfxn ; yn g) = (x1 ; : : : ; xn )
, maxfxi ; yi g = xi 8i 2 f1; : : : ; ng
, xi
Similarly,
,x
yi 8i 2 f1; : : : ; ng
y:
x ^ y = x , (minfx1 ; y1 g; : : : ; maxfxn ; yn g) = (x1 ; : : : ; xn )
, minfxi ; yi g = xi 8i 2 f1; : : : ; ng
, xi
,x
yi 8i 2 f1; : : : ; ng
y
Strong Set Order
De…nition (Strong set order in Rn )
The binary relation
A
s
on subsets of Rn is de…ned as follows: for A; B
sB
if
Rn ,
for any a 2 A and b 2 B
a^b 2B
and a _ b 2 A
This is the same as the previous de…nition.
We can use this to talk about non-decreasing Rn -valued correspondences.
Quasi-Supermodularity
De…nition
Let f : Rn Rm ! R. Then f (x; q) is quasi-supermodular in x if, for all x; y 2 Rn
and q 2 Rm :
1
f (x; q)
2
f (x ^ y; q)
)
f (x _ y; q)
f (y; q);
f (x; q) > f (x ^ y; q)
)
f (x _ y; q) > f (y; q):
This is a generalization of the mixed second partial derivatives typically used
to make statements about complementarities or substitutability.
Quasi-supermodularity is an ordinal property, and for di¤erentiable functions
there is a su¢ cient condition for quasi-supermodularity.
Exercise
1
2
Prove that if f is quasi-supermodular in x, then h f is quasi-supermodular in
x for any strictly increasing h : f (Rn Rm ) ! R.
Suppose f (x; q) is twice di¤erentiable in x and
with i 6= j. Then f is quasi-supermodular in x.
@2 f
@xi xj
> 0 for all i; j = 1; : : : ; n
Single-Crossing Property
De…nition
Let f : Rn Rm ! R. Then f (x; q) satis…es the single-crossing property if, for all
x; y 2 Rn and q; r 2 Rm such that x y and q r:
1
f (x; r)
2
f (y; r)
)
f (x; q)
f (y; q);
f (x; r) > f (y; r)
)
f (x; q) > f (y; q):
The “marginal return” f (x; ) f (y; ) as a function of the second argument
can cross 0 at most once.
The single-crossing property is an ordinal property, and for di¤erentiable
functions there is a su¢ cient condition for single-crossing.
Exercise
1
Prove that if f satis…es the single-crossing property, then h f satis…es the
single-crossing property for any strictly increasing h : f (Rn Rm ) ! R.
2
2
@ f
Suppose f (x; q) is twice di¤erentiable and @x
> 0 for all i = 1; : : : ; n and
i xj
j = 1; : : : ; m. Then f satis…es the single-crossing property.
Hint: Use induction on dimensions.
Increasing Di¤erences and Supermodularity
De…nition
Let f : Rn Rm ! R. We say that f (x; q) satis…es the increasing di¤erences if, for
all x; x0 2 Rn and q; q0 2 Rm such that x0 x and q0 q:
f (x0 ; q0 )
f (x; q0 )
f (x0 ; q)
f (x; q)
Strict increasing di¤erences has a strict inequality.
Similar to single crossing, but stronger. The “marginal return” f (x0 ; ) f (x; )
is increasing in the second argument.
@2 f
If f (x; q) is twice di¤erentiable and @x
> 0 for all i and j, then f has
i xj
increasing di¤erences in (x; q).
De…nition
f (x; q) is supermodular in x if, for all x; y 2 Rn and q 2 Rm :
f (x _ y; q) + f (x ^ y; q)
f (x; q) + f (y; q);
Similar to quasi-supermodularity, but stronger.
@2 f
If f (x; q) is twice di¤erentiable in x and @x
> 0 for all i; j = 1; : : : ; n with
i xj
i 6= j, then f is supermodular in x.
Unlike quasi-supermodularity and single-crossing, increasing di¤erences and
Monotone Comparative Statics
Theorem (easy Milgrom and Shannon)
Let f : Rn Rm ! R. De…ne x (q) = arg maxx 2Rn f (x; q). Suppose jx (q)j = 1
for all q and f (x; q) is quasi-supermodular in x and satis…es the single-crossing
property. Then
q r ) x (q) x (r ):
This is ‘easy’since it assume the optimum is unique (the proof does not use
the ‘strict’part of the de…nitions of quasi-supermodularity and single-crossing).
Proof.
Suppose q
)
r . Then:
f (x (r ); r )
f (x (q) _ x (r ); r )
) f (x (q) _ x (r ); q)
)
)
f (x (q) ^ x (r ); r ) by de…nition of x (q)
f (x (q); r )
by quasi-supermodularity in x
f (x (q); q)
x (q) _ x (r ) = x (q)
x (q)
x (r )
by Single Crossing
since jx (q)j equals 1
by the earlier Lemma
This result can be extended to constrained optimization problems (Question 3,
Problem Set 6).
Problem Set 6 (Incomplete)
Due Monday, 13 October, at the beginning of class
1
Assume the function f : R2 ! R is supermodular in (x; q) and show that
supermodularity is equivalent to the property that
q0 > q
2
3
implies
f (x; q 0 )
f (x; q) is non-decreasing in x:
Without using Milgrom and Shannon’s theorem, prove that if f : R2 ! R is
supermodular in (x; q), then x (q) arg max f (x; q) is non-decreasing.
Consider the following extension of the monotone comparative statics results allowing
for constrained optimization. Suppose f : Rn Rm ! R is quasi-supermodular in x and
satis…es the single crossing property. Suppose ' : Rm ! Rn is a correspondence such
that '(q 0 ) s '(q) for every q 0 > q. Set
x (q) = arg max f (x; q)
x 2'(q)
4
and assume jx (q)j = 1 for each q. Prove that x (q 0 ) x (q) for every q 0 > q.
Suppose a …rm purchases two inputs x1 and x2 at unit costs w1 and w2 to produce a
single product with production function f (x1 ; x2 ) to sell at a unit market price of p. Its
pro…t is therefore:
(x1 ; x2 ; p; w1 ; w2 ) = pf (x1 ; x2 )
2
w1 x1
w2 x2 :
Suppose f is C 2 and @x@ 1 fx2 > 0, i.e. x1 and x2 are complements. Assume that
x (p; w1 ; w2 ) = arg max (x1 ; x2 ; p; w1 ; w2 ) is unique. Use monotone comparative
statics to prove that if p 0 p, w10 w1 , and w20 w2 , then
x (p 0 ; w10 ; w20 ) x (p; w1 ; w2 ). (Hint: you will have to do some change of variables.)