Homework Solutions for 1st Half of Econ 400
Note: With the exception of my “additional question”, these are the solutions provided by
Nicholson and Snyder. I’ve done my best to weed out typos, but use caution and
common sense.
2.1
2.5
U ( x, y)  4 x2  3 y 2 .
a.
U x  8 x, U y  6 y.
b.
Ux = 8, Uy = 12 U x  8, U y  6.
c.
dU  8x dx  6 y dy.
d.
U
dy
4x
 x  .
dx
Uy
3y
e.
4 12  3  22  16.
f.
dy 4 1
2
=
= .
dx 3  2
3
g.
The U = 16 contour line is an ellipse centered at the origin. The slope of
the line at any point is given by dy dx   4 x 3 y .
a.
The height of the ball is given by f (t )  0.5gt 2  40t. The value of t for
which height is maximized is found by taking the first-order condition:
df dt =  gt  40  0, implying t *  40 g .
b.
Substituting for t * ,
2
 40 
 40  800
f (t )  0.5 g    40   
.
g
 g 
 g 
*
Hence,
f (t * )
800
 2 .
g
g
c.
Differentiation of the original function at its optimal value yields
f (t * )
 0.5(t * )2 .
g
Because the optimal value of t depends on g ,
2
 40  800
f (t * )
=  0.5(t * )2  0.5    2 ,
g
g
 g 
as was also shown in part c.
2.6
d.
If g  32, t *  5 4. Maximum height is 800 32  25. If g  32.1,
maximum height is 800 32.1  24.92, a reduction of 0.08. This could
have been predicted from the envelope theorem, since
 800 
 25 
df (t * )  
dg  
 (.01)  0.08.
2 
 32 
 32 
a.
This is the volume of a rectangular solid made from a piece of metal
which is x by 3x with the defined corner squares removed.
b.
The first order condition for maximum volume is given by
V
 3x 2  16 xt  12t 2  0.
t
Applying the quadratic formula to this expression yields
16 x  256 x 2  144 x 2 16 x  10.6 x

 0.225 x.
24
24
The second value given by the quadratic (1.11x) is obviously extraneous.
t
2.7
c.
If t  0.225 x,
V  0.67 x3  .04 x3  .05x3  0.68x3 .
So volume increases without limit.
d.
This would require a solution using the Lagrangian method. The optimal
solution requires solving three non-linear simultaneous equations, a task
not undertaken here. But it seems clear that the solution would involve a
different relationship between t and x than in parts a–c.
a.
Set up the Lagrangian: L  x1  5ln x2   (k  x1  x2 ). The first-order
conditions are
Lx1 = 1   = 0
5
  0
x2
L  k  x1  x2  0.
Lx2 
Hence,   1  5 x2 . With k  10, the optimal solution is x1*  x2*  5.
b.
With k  4, solving the first order conditions yields x1*  1 and x2*  5.
c.
If all variables must be non-negative, it is clear that any positive value
for x1 reduces y. Hence, the optimal solution is x1*  0, x2*  4, and
y*  5ln 4.
d.
If k  20, optimal solution is x1*  15, x2*  5, y*  15  5ln 5. Because x2
provides a diminishing marginal increment to y, whereas x1 does not, all
optimal solutions require that, once x2 reaches 5, any extra amounts be
devoted entirely to x2 . In consumer theory this function can be used to
illustrate how diminishing marginal usefulness can lead to a ceiling in
purchases of certain goods.
2.10
The Cobb-Douglas Function
a.
f1   x1 1 x2  0
f 2   x1 x2 1  0
f11   (  1) x1  2 x1  0
f 22   (   1) x1 x2  2  0
f12  f 21   x1 1 x2 1  0.
Clearly, all the terms in Equation 2.114 are negative.
b.
A contour line is found by setting the function equal to a constant:
y  c  x1 x2 , implying x2  c1  x1  . Hence,
dx2
 0.
dx1
Further,
d 2 x2
> 0,
dx12
implying the countour line is convex.
c.
3.1
Using Equation 2.98, f11 f 22  f122   (1     ) x12 2 x22  2 , which is
negative for     1.
Here we calculate the MRS for each of these functions:
3.2
a.
MRS 
fx 3
 . MRS is constant.
fy 1
b.
MRS 
f x 0.5( y x)0.5
y

 . Convex; MRS is diminishing.
0.5
f y 0.5( y x)
x
c.
f x 0.5 x 0.5
MRS 

. MRS is diminishing.
fy
1
d.
MRS 
e.
f x ( y ( x  y )  xy ) ( x  y )2 y 2
MRS 

 . Convex; MRS is diminishing.
f y ( x( x  y )  xy ) ( x  y ) 2 x 2
Because all of the first order partials are positive, we must only check the second
order partials.
a.
f11  f 22  f 2  0. Not strictly quasi-concave.
b.
f11 , f 22  0, f12  0. Strictly quasi-concave.
c.
f11  0, f 22  0, f12  0. Strictly quasi-concave.
d.
e.
3.4
fx
2x
x
 .5( x 2  y 2 )0.5 
 2 y  . MRS is increasing.
2
2 0.5
fy
.5( x  y )
y
a.
Even if we only consider cases where x  y, both of the own second order
partials are ambiguous and therefore the function is not necessarily strictly
quasi-concave.
f11 , f 22  0, f12  0. Strictly quasi-concave.
In the range in which the same good is limiting, the indifference curve is
linear. To see this, take the case in which both x1  y1 and x2  y2 . Then
k  U ( x1 , y1 )  x1 and k  U ( x2 , y2 )  x2 , implying
k k
x x y y  x x
U 1 2 , 1 2  1 2 
k
2 
2
2
 2
as well.
In the range in which the limiting goods differ, we can show the
indifference curve is strictly convex. Take the case k  x1  y1 and
k  y2  x2 . Then ( x1  x2 ) 2  k and ( y1  y2 ) 2  k , implying
x x y y 
U  1 2 , 1 2   k.
2 
 2
Hence the indifference curve is convex.
b.
Again, in the range in which the same good is maximum, the indifference
curve can be shown to be linear. Consider a range in which different
goods are maximum, specifically, k  x1  y1 and k  y2  x2 . Then
( x1  x2 ) 2  k and ( y1  y2 ) 2  k , implying
x x y y 
U  1 2 , 1 2   k.
2 
 2
Hence the indifference curve is concave.
c.
Here,
x x y y 
( x1  y1 )  k  ( x2  y2 )  U  1 2 , 1 2  .
2 
 2
Hence he indifference curve is linear.
3.5
a.
Since the four goods are perfect compliments,
U (h, b, m, r )  min(h, 2b, m,0.5r ).
b.
A fully condimented hot dog.
c.
$1.60.
d.
$2.10, an increase of 31%.
e.
Price would increase only to $1.725, an increase of 7.8%.
f.
3.12
Raise prices so that a fully condimented hot dog rises in price to $2.60.
This would be equivalent to a lump-sum reduction in purchasing power.
CES utility
1
a.
b.
U x  x 1   y 
MRS 

  
U y  y 1   x 
, so this function is homothetic.
If   1, MRS    , a constant. If   0,
 y
MRS   ,
 x
This agrees with Problem 3.10.
c.
MRS

   1 y1 x 2 . This is negative if and only if   1.
x

d.
Follows from part a. If x  y, MRS    .
e.
With   .5,


(.9)0.5  .949


.


0.5
MRS (1.1)  (1.1)  1.05


MRS (.9) 
With   1,


(.9)2  .81


.


2
MRS (1.1)  (1.1)  1.21 .


MRS (.9) 
Hence, the MRS changes more dramatically when   1than when
  .5 . The indifference curves are more sharply curved when  is
lower. When    , the indifference curves are L-shaped, implying fixed
proportions.
Additional Problems
Applying the envelope theorem. In class, we solved the following constrained
maximization problem, with P=20:
Max f(x,y) = xy, s.t. P-2x-2y=0
We found the values of x and y that maximized the area of a pen for a fence perimeter of
P=20. From those, you can easily find that the maximum area is f*(x*,y*) = 25. But
what if you wanted to know how the value of f* would change as the fence size (P)
changed? One way of doing this would be to solve the Lagrangian maximization
problem for every value of P that interested you. An easier way of doing it would be to
𝑑𝑓∗
apply the envelope theorem to find 𝑑𝑃 by following the steps below:
a. Find f as a function of x and P by solving the constraint equation for y and
substituting it into f(x,y). This will give you U(x,P).
1
𝑓(𝑥, 𝑃) = 𝑥(𝑃 − 2𝑥)
2
b. Find x*(P), the optimal value of x for all values of P by finding the first order
condition for maximizing f(x,P) and solving for x. (This should be in terms of P and
will be called x*(P).)
𝐹𝑂𝐶:
1
𝑃 − 2𝑥 = 0
2
1
𝑥 ∗ (𝑃) = 4 P
c. Then apply the envelope theorem to find
theorem states that:
𝑑𝑓∗
𝑑𝑃
=
𝑑𝑓∗
𝑑𝑃
as a function of P. Hint: The envelope
𝜕𝑓(𝑥,𝑃)
⃒𝑥=𝑥 ∗ (𝑃) \
𝜕𝑃
𝑑𝑓 ∗ 1 ∗ 𝑃
= 𝑥 =
𝑑𝑃
2
8
d. Now show that the answer you found in part c above will be the same as the answer
found from applying the envelope theorem to the Lagrangian, as shown in class and
in equation (2.75) of the text. (Hint: you will need to use the F.O.C. for maximizing
the Lagrangian to get your answer in terms of P).
𝜕ℒ
𝜕𝑃
4.2
= 𝜆∗ =
𝑃
8
Use a simpler notation for this solution: U ( f , c)  f 2 3c1 3 and I  300.
a.
Setting up the Lagrangian:
L  f 2 3c1 3   (300  20 f  4c).
The first-order conditions are
13
2 c 
L f     20  0
3 f 
1 f
Lc  
3 c
Hence, 5  2c f ,
23

  4  0.

implying 5 f  2c. Substituting into budget constraint
yields f *  10 and c*  25.
b.
With the new constraint, f *  20 and c*  25.
Note: This person always spends 2 3 of income on f and 1 3 on
c. Consumption of California wine does not change when the price of
French wine changes.
c.
In part a,
U ( f , c)  f 2/3c1 3  102/3251 3  13.5.
In part b,
U ( f , c)  202/3251 3  21.5.
This person will need more income to achieve the part-b utility with the
part-a prices. Indirect utility is
23
13
23
13
 2 1
 2 1
21.5      I p f 2 3 pc1 3      I 202 341 3.
 3  3
 3  3
Solving this equation for the required income gives I  482. With such
an income, this person would purchase f  16.1 and c  40.1, and, by
construction, would obtain utility of U  21.5.
4.5
Given U (m)  U ( g , v)  min  g 2, v  .
a.
No matter what the relative prices are (i.e., the slope of the budget
constraint), the maximum utility intersection will always be at the vertex
of an indifference curve where g  2v.
b.
Substituting g  2v into the budget constraint yields 2 pg v  pv v  I , or
I
.
2 pg  pv
Furthermore,
2I
g
.
2 pg  pv
It is easy to show that these two demand functions are homogeneous of
degree zero in pg , pv , and I .
v
c.
Since U  g 2  v, indirect utility is
V ( pg , pv , I ) 
d.
I
.
2 pg + pv
The expenditure function is found by interchanging I ( E ) and V :
E ( pg , pv ,V )  (2 pg  pv )V .
4.7
a.
b.
E ( px , p y ,U )  2 px0.5 p 0.5
y U . With px  1 and p y  4, we have U  2 and
E  8. To raise utility to 3 would require E  12, that is, an income
subsidy of 4.
c.
Now we require E  8  2  px0.5  40.5  3 or px0.5  8 12  2 3. So px  4 9;
that
is, each unit must be subsidized by 5 9. At the subsidized price, this
person chooses to buy x  9. So total subsidy is 5, one dollar greater than
in part c.
d.
and
E ( px , py ,U )  1.84 px0.3 p0.7
y U . With px  1 and p y  4, we have U  2
E  9.71. Raising U to 3 would require extra expenditures of 4.86.
Subsidizing good x alone would require a price of px  0.26, that is, a
subsidy of 0.74 per unit. With this low price, the person would choose
x  11.2, so the total subsidy would be 8.29.
4.8
a.
If U ( x, y)  min( x, y) , utility maximization requires x  y . Substitution
into the budget constraint yields x  I ( px  p y )  y . Hence:
V ( px , p y .I ) 
I
px  p y
E ( px , p y ,V )  ( px  p y )V
If U ( x, y)  x  y , utility maximization requires the purchase of whichever of these
two perfect substitutes has the lower price. So,
If px  py x  0, y  I py . If px  py x  I px , y  0. Given these results:
V ( px , p y .I ) 
I
min( px , p y )
E ( px , p y ,V )  min( px , p y )V
b.
4.10
It is interesting that the discontinuous utility function has continuous indirect utility and
expenditure functions whereas the linear utility function has discontinuous indirect utility
and expenditure functions. Similar dualities occur in many maximization problems.
Cobb-Douglas Utility
a.
The demand functions in this case are
I
x
px
y
(1   ) I
.
py
Substituting these into the utility function gives

  I   (1   ) I 
  (1 )
V ( px , p y , I )  
  BIpx p y V ,
 
 px   p y 
where B    (1   )1 .
4.11
b.
Interchanging I and V yields E ( px , p y ,V )  B 1 px p1yV .
c.
The exponent  gives the elasticity of expenditures with respect to px .
That is, the more important x is in the utility function, the greater the
proportion that expenditures must be increased to compensate for a
proportional rise in the price of x.
CES Utility
a.
For utility maximization,
U x  x 
MRS 
 
U y  y 
1

px
.
py
Hence,
1

x  px  1  px 
    ,
p 
y  p y 
 y
where

b.
1
.
1 
If   0,
x py

,
y px
implying px x  p y y.
c.
Part a shows
1
px x  px 
  .
p y y  p y 
Hence, for   1, the relative share of income devoted to good x is
positively correlated with its relative price— a sign of low substitutability.
For   1, the relative share of income devoted to good x is negatively
correlated with its relative price— a sign of high substitutability.
d.
The algebra is a bit tricky here, but worth doing once. Let’s solve for
indirect utility:

x  px 
  ,
y  p y 
or

p 
x  y x  .
 py 
 
Substituting into the budget constraint yields
x  px 
 
y  p y 

 p y y,
or
y
Similarly,
Ip y
p1x  p1y
.
Ipx
x  1
.
px  p1y
Hence,

px
 U  x  y  I  1
 px  p1y

Now   1   , so






p y

  I  1
1

 px  p y


 .



1
,
 ( p1x  p1y ) 1 


U  I  
or
1
x
V  I(p
1
1 1
y
p
)
,
1
where V  (U ) . This is the indirect utility function. Clearly, it is
homogeneous of degree zero in income and prices. Inverting the
expression yields the expenditure function:
1
E  I  V ( p1x  p1y )1 .
Clearly, this is homogeneous of degree one in the prices. Note that the
odd form for V here suggests the use of the CES for given in Problem
4.13 in applications involving these functions.
4.13
CES indirect utility and expenditure functions
a.
Given U  ( x  y  )1  . The Lagrangian is
L  ( x  y )1    ( I  xpx  ypy ).
One first-order condition is
1
1 
 
Lx  ( x  y ) (x1 )  px  0,

implying


(x  y )

px
Similarly,
1

1
x 1
.
( x  y )  y 1

.
py
Equating the  yields either
x 1 y 1
(1)

,
px
py
or


1
(2) ( x  y )   0.
Since we assume U  0, equation (2) cannot be a solution. Therefore, the
solution satisfies (1), which upon rearranging gives
1
p 
x  y x  .
 py 
 
Substituting this value of x into the budget constraint yields
 1
p
I  x
p
 y
Solving for y,
y 
1
 1
 ypx  yp y .

Ip1y ( 1)
*
px ( 1)  p y ( 1)
,
implying
x* 
Ip1x ( 1)
.
px ( 1)  p y ( 1)
Further,
V  ( x   y  )1 
1



 
 
Ip1y ( 1)
Ip1x ( 1)
   ( 1)
 
 
 p y ( 1)   px ( 1)  p y ( 1)  
 px


1
 pxr  p yr 
I r
 ( p  p r ) 
y
 x

 I(p  p )
r
x
y
r
1

 I ( pxr  p yr ) 1 r ,
where
r

 1
.
b.
Scale all variables by t and the function is unchanged.
c.
The partial derivative of V with respect to I is positive as the prices are
positive.
d.
Again, the partial derivatives of V with respect to the prices are both
negative. For example,
 (1 r )
V
  Ipxr 1 ( pxr  p yr ) r  0.
px
e.
Simply reversing the positions of V and I in the indirect utility function
yields
E  V ( pxr  p yr )1 r .
f.
Multiplying prices by any factor t multiplies expenditures by t.
g.
For example,
1 r
E
 Vpxr 1 ( pxr  p yr ) r  0.
px
h.
Differentiating the expression from part g,
1 2 r
1 r
2 E
2 r 2
r 2
r
r

(1

r
)
Vp
K

(
r

1)
Vp
K
,
x
x
2
px
where K  pxr  p yr . Division of this expression by Vpxr 2 k
1 r
r
yields
(r  1)(1  k )  0, where k  p K  1.
r
x
5.1
a.
Utility = Quantity of water
b.
If
and
 .75x  2 y.
p x < 3 p y 8, then x* = I px and y*  0. If p x < 3 p y 8, then x* = 0
y*  I p y .
c.
d.
Increases in
I shift demand for x outward. Reductions in p y do not affect demand for
x until p y < 8 p x 3 . Then the demand for x falls to zero.
e.
The compensated demand curve for good
x is a vertical line so long as
px  3 p y 8. If the person buys only x, holding utility constant requires that
x  U 0.75 no matter what the price of x is.
5.2
a.
To avoid confusing goods’ names with prices, let b stand for peanut
butter. Utility maximization requires b  2 j. The budget constraint is
Substitution gives
.05b .1j  3.
b  30 and j  15.
*
*
p j  $0.15, substitution now yields j*  12 and b*  24.
b.
If
c.
To continue buying j  15, b  30. David would need to buy 3 more ounces of
jelly and 6 more ounces of peanut butter. This would require an increase in income of
*
*
3(.15)  6(.05)  .75.
d.
e.
Because David uses only peanut butter and jelly to make sandwiches (in fixed
proportions), and because bread is free, it is just as though he buys the good
“sandwiches,” where ps  2pb  p j .
ps  .20 and qs  15.
In part b, ps  .20 and qs  12.
In general, q s = 3 p s , so the demand curve for sandwiches is a hyperbola.
In part a,
f.
There is no substitution effect due to the fixed proportion. A change in
results in an income effect only.
5.4
a.
Since
x
0.7 I
0.3I
,
and y 
py
px
we have
U = .3.3 .7.7 I px.3 p y.7 = BIpx.3 p y.7 ,
where
b.
B  .30.3.70.7. The expenditure function is then E = B 1Up.3x p.7y .
The compensated demand function is x
c
 E px  0.3B1Upx0.7 p0.7
y .
price
c.
It is easiest to show the Slutsky Equation in elasticities by just reading exponents from
the various demand functions: ex , px  1, ex , I  1, exc , p  .7, sx  0.3. Hence
x
ex, px  exc , p  sx ex, I implies 1  0.7  (0.3)(1).
x
5.5
a.
The Lagrange method yields
p
y
 x,
x  1 py
or
py y  px x  px . Substitution into the budget constraint yields
x=
I  px
I + px
and y =
.
2 px
2py
Hence, changes in
b.
p y do not affect x, but changes in px do affect y.
The indirect utility function is
V=
(I + p x )2
,
4 p x py
which yields an expenditure function of
E  V 4 px p y  px .
c.
Clearly the compensated demand function for
x depends on p y whereas the
uncompensated function did not. By Shepherd’s Lemma:
xc 
5.7
a.
E
 V 0.5 px0.5 p 0.5
y  1.
px
Because of the fixed proportions between
h and c, we know that the
h  I ( ph  pc ). Hence,
p ( p  pc )
 ph
h ph
I



 h h

.
2
ph h ( ph  pc )
I
( ph  pc )
demand for ham is
eh, ph
Similar algebra shows
 pc
.
( ph  pc )
So ph  pc , eh, ph  eh, pc  0.5.
eh, pc 
b.
With fixed proportions, there are no substitution effects. Here the
compensated price elasticities are zero, so the Slutsky equation shows that
ex, px  0  sx  0.5.
ph  2 pc , part a shows that eh, ph  2 3 and eh, pc  1 3.
c.
With
d.
If this person consumes only ham and cheese sandwiches, the price elasticity of demand for those
must be 1 . Price elasticity for the components reflects the proportional effect of a change in the
price of the component on the price the whole sandwich. In part a, for example, a 10% increase in
5.12
the price of ham will increase the price of a sandwich by 5% and that will cause quantity
demanded to fall by 5%.
Quasi-linear utility (revisited)
a.
First, we need to find the demand functions for both the goods. This is done by
straightforward application of the Lagrange method to give:
x
p
I  px
and y  x .
py
px
The income effect for x is
x px  I

.
I
px2
The income elasticity for x is
x I
I
ex , I  . 
.
I x I  px
The income effect for y is
y
 y  0.
I
x
The income elasticity for y is
ey , I 
b.
y I
.  0.
I y
Now, we need to find the compensated demand functions for both the goods by first
finding the indirect utility function:
I  px
 ln px  ln p y .
px
V
So the expenditure function is
E  px (V  ln px  ln p y  1),
implying
xc 
E
 V  ln px  ln p y
px
yc 
px
.
py
and
The substitution effect for x is
x c
1

.
px
p xx
The compensated own price elasticity for x is
exc , p 
x
x c px
1
. c 
.
px x
ln px  ln p y  V
The substitution effect for y is
p
y c
  x2 .
p y
py
The compensated own price elasticity for y is
eyc , p 
y
c.
y c p y
.
 1.
p y y c
For the Slutsky equation, the own price effects of the demand functions are also needed:
p
y
x
I
  x2 .
  2 and
p y
py
px
px
Now, put all the terms into the Slutsky equation. For x,
(1)
x
I
 2
px
px
(2)
x c
x
1 p I
I
x   x 2  2
px
I
px
px
px
and
imply
x x c
x

x .
px px
I
For y,
(3)
p
y
  x2
p y
py
(4)
p
y c
y
 y   x2  0
p y
I
py
and
imply
y y c
y

y .
p y p y
I
Thus, the Slutsky equation holds for both goods.
For the elasticity version of the equation, we need the own price elasticity of
each good:
ex , px 
I
and ey , py  1.
px  1
Put all this into the elasticity version of the Slutsky equation. For x,
ex , px 
I
px  I
and
exc , p  sx ex , I =
x

ex, px  exc , p
x
px
1
px  I
I
px  I
 sx ex, I . For y,

imply
p  I
1

 1  x 
ln px  ln p y  V 
I  I  px
ey , py  1
and
eyc , p  s y ey , I  1  0  1
y
imply
ey , py  eyc , p  s y ey , I . This confirms the elasticity version of the Slutsky
y
equation.
.1
a.
As for all Cobb-Douglas applications, first-order conditions show that
pmm  ps s  0.5I . Hence, s  0.5I ps and s pm  0.
b.
own
Because indifference curves are rectangular hyperboles (ms = constant)
substitution and cross- substitution effects are of the same proportional
size, but in opposite directions. Because indifference curves are
homothetic, income elasticities are 1.0 for both goods, so income effects
are also of same proportionate size. Hence, substitution and income
effects of changes in pm on s are precisely balanced.
c.
We have the two conditions
s
s
s
0

m
pm pm U
I
0
m m
m

s
.
ps ps U
I
But
s
pm
So
m
d.
6.7

U
m
.
ps U
s
m
s
.
I
I
From part a,
 0.5 
 0.5 
 0.5 
s
m
m  m
.
  m
  s
s
I
I
 ps 
 pm m s 
 pm 
Assume xi  ai I and x j  a j I . Hence,
x j
xi
= ai a j I = xi
.
I
I
So income effects (in addition to substitution effects) are symmetric.
xj
7.2
See graph.
This would be limited by the individual's resources. Since unfair bets are
continually being accepted, he or she could run out of wealth.
7.3
a.
Strategy One
Outcome
12 Eggs
Probability
.5
0 Eggs
.5
Expected Value = .5 (12)  .5( 0)  6.
Strategy Two
Outcome
Probability
12 Eggs
.25
6 Eggs
.5
0 Eggs
.25
Expected Value = .25 (12)  .5( 6)  .25(0)  3  3  6.
b.
7.4
a.
The insurance company has a 50% chance of paying out $10,000. Its cost
is thus $5,000. The consumer has a certain wealth of $15,000 with fair
insurance compared to a 50-50 chance of wealth of $10,000 or $20,000
without insurance.
b.
7.7
a.
Cost of the policy is .5  5,000   2,500. Hence, wealth is 17,500 with no
illness and 12,500 with the illness.
The farmer will plant corn since
U  wheat   .5 ln  28,000   .5 ln 10,000   9.7251.
U  corn   .5 ln 19,000   .5 ln 15,000   9.7340.
b.
With half in each, YNR  23,500 and YR  12,500.
U  .5 ln  23,500  .5 ln 12,500  9.7491.
The farmer should plant a mixed crop. Diversification yields an increased
variance relative to corn only, but takes advantage of wheat’s high yield.
c.
Let   percent in wheat.
U  .5 ln  28, 000  19, 000(1   ) 
 .5 ln 10, 000  15, 000(1   ) 
 .5 ln 19, 000  9, 000   .5 ln(15, 000  5, 000 ).
Taking the first-order condition,
dU
4,500
2,500
=

= 0.
d 19, 000 + 9, 000 15, 000  5, 000
Rearranging,
45(150  50 )  25 190  90  ,
implying   .444. Plugging  into the utility function yields
U  .5 ln  22,996   .5 ln 12,780   9.7494.
This is a slight improvement over the 50-50 mix.
d.
If the farmer plants only wheat, YNR  24,000 and YR  14,000.
U  .5 ln  24,000  .5 ln 14,000   9.8163.
Availability of this insurance will cause the farmer to forego
diversification.
2
7.8
a.
E (v 2 )   p( xi ) f ( xi )  0.5(1) 2  0.5(1) 2  1.
i 1
b.
E(h2 )  0.5(k )2  0.5(k )2  k 2 .
c.
If U (W )  ln(W ), then for W  0,
U (W ) 1 W 2 1
r (W )  

 .
U (W ) 1 W W
d.
p  0.5E (h2 )r (W )  0.5k 2
1
k2

.
W 2W
Calculate p when W=10
k
p
0.5
0.0125
1
0.05
2
0.2
Calculate p when W=100
k
p
0.5
0.00125
1
0.005
2
0.02
Risk premium is higher when the level of initial wealth is lower. The greater the size of
risk faced ( larger the k ), higher will be the risk premium. Because k enters as a
quadratic, increasing k and W in the same proportion will increase p.
7.11 Prospect theory
Scenario 1
Gamble
Expected wealth
A
1,000  1 2  (1,000  0)  1,500
B
1,000  500  1,500
Scenario 2
Gamble
Expected wealth
C
2,000  1 2  (1,000  0)  1,500
D
2,000  500  1,500
a.
Scenarios A-D provide the same expected wealth—$1,500—so a risk
neutral Stan should be indifferent among them.
b.
The expected wealth levels are the same. Thus a risk-averse Stan should
select the safe option B in Scenario 1 and D in Scenario 2.
c.
It is natural to suppose subjects are risk averse, so more should choose D
in Scenario 2.
d.
(1)
Pete should make the same choices as the majority of experimental
subjects. Scenario 1 involves gains, so Pete behaves as predicted
by expected utility theory there. Scenario 2 involves losses. The
certainty of a small loss may be worse than a smaller chance of a
large loss, so C may be preferred.
(2)
Utility
Scenario 2
•
Scenario 1
•
1,000 2,000
Wealth
The utility curve has to shift because of the kink at the anchor
point. Prospect Pete’s curve changes from convex to concave at
the anchor point; Standard Stan’s is linear or concave everywhere,
so doesn’t have to shift.