Solutions Manual for Advanced Microeconomic Theory I
Jianfei Shen School of Economics, Shanghai University of Finance and Economics
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Contents Part Part 1.
1
Consum Consumer er Theory Theory
Chapter 1. Solution for Exercise I 1. (JR, 1.4) 2. (JR, 1.27) 3. (JR, 1.56) 4. (JR, 1.24) 5. (Homogeneity)
3 3 3 4 5 6
Chapter 2. Solution for Exercise II 1. (JR, 1.36) 2. (JR, 2.10) 3. (MWG, 2.F.7) 4. (JR, 1.62) 5. (JR, 2.9)
9 9 9 10 11 11
Part Part 2.
13
Production Production and Choice under Uncert Uncertaint ainty y
Chapter 3. Solution for Exercise III 1. (JR, 3.36) 2. (JR, 3.39) 3. (The Allais Paradox) 4. (The Simplex) 5. (Independence Axiom)
15 15 16 17 18 19
Chapter 4. Solution for Exercise IV 1. (MWG, 6.C.16) 2. (MWG, 6.C.20) 3. (Betweenness Axiom) 4. (Quadratic v.N-M Utility Function)
21 21 22 23 24
i
Part 1
Consumer Theory
CHAPTER 1
Solution for Exercise I 1. (JR, 1.4) L , and x y, y z. By Definition 1.21, Proof. Suppose that x,y,z ∈ R +
we have
x y ⇔ x y ∧ y x;
(1.1) and
y z ⇔ y z ∧ z y.
(1.2)
The transitivity implies that x z. Suppose that z x. Since y z [by equation (1.2)], the transitivity then implies that y x. But this contradicts (1.1). Thus z x. Hence x z. Suppose x ∼ y, and y ∼ z. By Definition 1.3 2, we have x ∼ y ⇔ x y ∧ y x;
(1.3) and
y ∼ z ⇔ y z ∧ z y.
(1.4) The transitivity implies that
xz
and z x.
Hence x ∼ z.
2. (JR, 1.27) Solution. If x 1 < x2 ,
u(x) = max[ax1 , ax2 ] + min[x1 , x2 ] = ax 2 + x1 ; If x 2 < x1 , u(x) = max[ax1 , ax2 ] + min[x1 , x2 ] = ax 1 + x2 . Case i: a <
p1 p2
< a1 . In this case, x1 = x 2 =
1Jehle & Reny, 2001, P6 2Jehle & Reny, 2001, P7 3
w . p1 + p2
4
1. SOLUTION FOR EXERCISE I
x2
A
C O
B
x1
Figure 2.1. The Indifference Curve
Case ii:
p1 p2
> a1 . In this case, there is a corner solution x1 = 0, x2 =
Case iii:
p1 p2
< a. In this case, there is a corner solution, too x1 =
Case iv:
p1 p2
w . p2
w , x2 = 0. p1
= a1 . In this case, the UMP is
w − p2 x2 w w x1 = , x2 ∈ , . p1 p1 + p2 p2 Case v:
p1 p2
= a. In this case, the UMP is w − p2 x2 w x1 = , x2 ∈ 0, . p1 p1 + p2
3. (JR, 1.56) Solution. Use the monotone transformation of the given utility function 3: n
v(x) = ln u(x) =
bi ln(xi − ai ).
i=1
The UMP can written as n
max
bi ln(xi − ai );
i=1 n
s.t.
pi xi ≤ w.
i=1
3Stone, J.E. (1954): Linear Expenditure Systems and Demand Analysis: An Application to the Pattern of British Demand. Economic Journal 64: 511–27
4. (JR, 1.24)
5
Then form the Lagrangian, we obtain
n
L =
n
bi ln(xi − ai ) + λ w −
i=1
pi xi .
i=1
The first-order condition of the UMP yields ∂ L bi (1.5) = − λpi = 0, ∂x i xi − ai Summarize all of the first-order conditions:
∀i.
n
(1.6)
1 = λ
pi (xi − ai ) ⇒ λ =
i=1
With (1.5) and (1.6), we have bi = So
pi (xi − ai ) n i=1 pi (xi − ai )
xi = a i +
b i (w −
=
1
n i=1 pi (xi −
ai )
.
pi (xi − ai ) . n w − i=1 pi ai
n i=1 pi ai )
pi
,
∀i.
4. (JR, 1.24) Solution. A utility function that represents a preference relation is not unique. For any strictly increasing function : f : R → R , v(x) = f (u(x)) is a new
utility function representing the same preferences as u(·)4. a: f (x) = u(x) + [u(x)]2 . ∂f (·) = 1 + 3[u(x)]2 > 0. ∂u(·) So this is a strictly increasing function, and it represents the same . b: f (x) = u(x) + [u(x)]2 . ∂f (·) = 1 + 2u(x). ∂u(·) ∂f (·) ∂u (·)
≤ 0 if u(x) ≤ − 12 . In this case, f (x) does not represents the same ; (·) > 0 if u(x) > − 12 . In this case, f (x) represents the same . • ∂f ∂u (·) c: f (x) = u(x) + ni=1 xi . Not necessarily. f represents the same preferences as u if u(x) is a n monotonic transformation of i=1 xi . Otherwise, they may not represent the same preferences. For example, suppose u(x) = x1 . Then •
u(1, 0, . . . , 0) > u(0, 3, 0, . . . , 0), but f (1, 0, . . . , 0) = 1 + 1 = 2 < 3 = 0 + 3 = f (0, 3, 0, . . . , 0).
4Let x, y ∈ R L . Since u(·) represents , x y iff u (x) ≥ u(y ). Since f (·) is strictly + increasing, u (x) ≥ u (y ) iff v (x) ≥ v (y). Hence x y iff v (x) ≥ v (y ). Therefore v (·) represents .
This completes the proof.
6
1. SOLUTION FOR EXERCISE I
5. (Homogeneity) L L , y ≥ 0, α ≥ 0, and x ∈ R + , if x Proof. a: We want to show that ∀ p ∈ R ++
is the Marshallian demand at ( p, y), then x = αx is the Marshallian demand at ( p, αy). We’ll complete this proof with two steps: Step I: we prove that the consumption bundle x is affordable at ( p, αy).
Note that p · x ≤ y ⇒ p · αy ≤ αy, so x = αx is affordable at ( p, αy) indeed. Step II: we prove that the consumption bundle x is the Marshallian demand function at ( p, αy). L Let x ∈ R + and p · x ≤ αy. Then
x p · ≤ y. α Hence
x u < u(x), (by the Strictly quasiconcave) α since x is the Marshallian demand function at ( p, y). Thus, by the homogeneity of u(·):
u
x 1 = u (x ) < u(x) ⇒ u(x ) ≤ αu(x) = u(αx). α α
Hence αx is the Marshiallian demand function at ( p, αy), and x( p, αy) = αx( p, y). b: By the result of part (a), we have v( p, αy) = u(x( p, αy)) = u(α · x( p, y)) = α · u(x( p, y)) = α · v( p, y). This proves that the indirect utility function v (·) is homogeneous of degree one in y. L c: In this part, we’ll show that ∀ x ∈ R + and α ≥ 0, we have u(αx) = αu(x). We first prove that x( p, αy) = α · x( p, y), where x(·) is the Marshiallian demand function. Since v (·, ·) is homogeneous of degree one in y: v( p, αy) = αv( p, y), we have
∇ p v( p, αy) = α∇ p v( p, y), and
∇y v( p, αy) = α · ∇y v( p, αy) = α · ∇y v( p, y) ⇒ ∇y v( p, αy) = ∇ y v( p, y). Then by Roy’s identity: xi ( p, y) = − we have x( p, αy) = −
∂v( p, y)/∂p i , ∂v( p, y)/∂y
i = 1, . . . , L ,
∇ p v( p, αy) ∇ p v( p, y) = −α · = α · x( p, y). ∇y v( p, αy) ∇y v( p, y)
5. (HOMOGENEITY)
7
Now, again by the homogeneity of v (·, ·), we have u(α · x( p, y)) = u(x( p, αy)) = v( p, αy) = α · v( p, y) = α · u(x( p, y)). This completes the proof of u(α · x( p, y)) = α · u(x( p, y)).
CHAPTER 2
Solution for Exercise II 1. (JR, 1.36) Solution. a. By definition,
e( p, u) min p · x L x∈R+
s.t. u(x) ≥ u The solution to the EMP is known as the Hicksian demand function : h( p, u). Hence u(x0 ) ≥ u 0 , and so e( p, u0 ) ≤ p · x0 , with the equality sign holds when p = p 0 . b. It follows immediately from part a that f ( p) = e( p, u) − p · x0 is maximized at p = p0 : max f ( p) = 0. c. If f ( p) is differentiable at p 0 , (2.1)
∇ p f p0 = 0.
d. If e( p, u) is differentiable in p, then by equation (2.1) we have
∇ p f = ∇ p e − x0 = 0, so h( p, u) = ∇ p e( p, u).
2. (JR, 2.10) Solution. a.
Bundle x x1 x2 42 48 40∗ 33∗ 36 39(∗) 52 48∗ 51 0
p0 Price p1 p2
9
10
2. SOLUTION FOR EXERCISE II
These consumption bundles are plausible in that the WARP is satisfied. In particular, we note p1 x1 = 36 > 33 = p1 x0 , p0 x0 = 42 < 48 = p0 x1 , so x1 R x0 .
(2.2)
p0 x0 = 42 > 40 = p0 x2 , p2 x2 = 50 < 52 = p2 x0 , so x0 R x2 .
(2.3)
p2 x2 = 50 > 48 = p2 x1 , p1 x1 = 36 < 39 = p1 x2 , so x2 R x1 .
(2.4)
b. Now, from (2.2) and (2.3) we have x1 R x0 ,
x0 R x2 ;
but from (2.4), something utterly revolting occurs: x2 R x1 .
So the revealed preference R is intransitivity . 3. (MWG, 2.F.7)
Proof. If the Walrasian demand function x( p, w) satisfies Walras’ law, then
for all p and w: L
(2.5)
p
=1
or
∂x ( p, w) + xk ( p, w) = 0, ∂p k
∀ k = 1, . . . , L ,
p · D p x( p, w) + x( p, w) = 0
(2.6) and
L
(2.7)
p
=1
or
∂x ( p, w) = 1, ∂w
p · Dw x( p, w) = 1.
(2.8)
If the Walrasian demand function x( p, w) is h.o.d. 0, then for all p and w: L
(2.9)
k=1
or (2.10)
∂x ( p, w) ∂ x ( p, w) pk + w = 0, ∂p k ∂w
∀ = 1, . . . , L ,
D p x( p, w) p + Dw x( p, w)w = 0.
5. (JR, 2.9)
11
By the Slutsky Equation: ∂h ( p, u) ∂x ( p, w) ∂ x ( p, w) = + xk ( p, w), ∂p k ∂p k ∂w
(2.11)
∀ , k
or equivalently, D p h( p, u) = D p x( p, w) + Dw x( p, w)x( p, w) ,
(2.12)
where D p h( p, u) is equal to the matrix
··· .. S ( p, w) = . sL1 ( p, w) · · · with sk ( p, w) =
s11 ( p, w) .. .
s1L ( p, w) .. .
sLL( p, w)
∂x ( p, w) ∂ x ( p, w) + xk ( p, w). ∂p k ∂w
So we have p · S ( p, w) = p · D p x( p, w) + p · Dw x( p, w)x( p, w) = p · D p x( p, w) + x( p, w)
(2.13)
= 0 ,
[from (2.12)]
[from (2.8)]
[from (2.6)]
and S ( p, w) p = D p x( p, w) p + Dw x( p, w)x( p, w) p [from (2.12)] = D p x( p, w) p + Dw x( p, w)w
(2.14)
[from Walras’ law]
= 0. [from (2.10)]
4. (JR, 1.62) Solution. Since
p · S ( p, w) = 0, S ( p, w) p = 0 and by symmetric, we have a = − 8,
b = 2,
p = 32.
5. (JR, 2.9) Proof. a.
In the case of two goods, equation (2.13) can be written as
p · S ( p, w) = p1 (2.15)
p2
s11 ( p, w) s12 ( p, w) s21 ( p, w) s22 ( p, w)
= p1 s11 ( p, w) + p2 s21 ( p, w) p1 s12 ( p, w) + p2 s22 ( p, w) = 0 0 .
From equation (2.15) we have
p1 s21 ( p, w) = − s11 ( p, w), p2
12
and
2. SOLUTION FOR EXERCISE II
p2 s12 ( p, w) = − s22 ( p, w). p1 And the equation (2.14) becomes
s ( p, w) s12 ( p, w) S ( p, w) p = 11 s21 ( p, w) s22 ( p, w) (2.16)
= =
From equation (2.16) we have
p1 p2
p1 s11 ( p, w) + p2 s12 ( p, w) p1 s21 ( p, w) + p2 s22 ( p, w) 0 . 0
p1 s12 ( p, w) = − s11 ( p, w), p2 and
p2 s21 ( p, w) = − s22 ( p, w). p1 Thus s12 ( p, w) = s 21 ( p, w).
Part 2
Production and Choice under Uncertainty
CHAPTER 3
Solution for Exercise III 1. (JR, 3.36) Proof. We can get some intuition from the following figure 1.
x2
tx02 x02 O
B A x01
y1
y0 tx01
x1
Figure 1.1. Homothetic production function
Now we give the regular proof. The cost function for all homothetic production function can be written (3.1)
c(w, y) = h(y)φ(w),
where φ(w) is linear homogeneous. 1 Then using the Shephard’s Lemma 2, we have ∂c(w, y) xi (w, y) = = h(y)φi (w), ∂w i ∂c(w, y) xj (w, y) = = h(y)φj (w). ∂w j Hence xi (w, y) φi ( w ) ψ(w), (3.2) = xj (w, y) φj (w) that is, ∂ (xi (w, y)/xj (w, y)) = 0. ∂y 1Any cost function is linear homogeneous in the factor prices. 2JR, Theorem 3.2, p.129 15
16
3. SOLUTION FOR EXERCISE III
2. (JR, 3.39) Proof. a: If the production function is CRS, then n
c(w, y) =
wi xi (w, y)
i=1 n
=
λf i xi (w, y) [ From the F.O.C. ]
i=1 n
= λ
xi (w, y) f i
i=1
= λy =
[ From the Euler Equation ]
∂c(w, y) y ∂y
[ By the interpretation of λ ]
or ∂c(w, y) ∂y = . c(w, y) y
(3.3)
Solve the partial differential equation (3.3), we get (3.4)
c(w, y) = yφ(w),
where φ(w) is a function of factor prices w only. b: If c(w, y) = yφ(w), then from the F.O.C. of the cost minimization problem that for ∀ x(w, y) > 0 (3.5)
wi = λf i ,
and ∂c(w, y) . ∂y Combine equation (3.5) and (3.6), we obtain (3.6)
λ =
wi = φ(w)f i , or xi f i =
wi xi . φ(w)
Hence, for all x i (w, y) n
n
xi f i =
i=1
i=1
=
wi xi φ(w)
1 φ(w)
n
i=1
c(w, y) φ(w) yφ(w) = φ(w) = f (x). =
wi xi
3. (THE ALLAIS PARADOX)
17
It follows from the Euler’s Theorem 3 that (3.7)
f (tx) = tf (x),
that is, the production function is CRS.
3. (The Allais Paradox) Proof. Defining X
= {x1 , x2 , x3 } = {$ 0 ; $ 1, 000, 000;$5, 000, 000},
these four gambles are seen form a parallelogram in the ( p1 , p3 ) triangle, as in the following figure. 1
1
p3
p3 g2
g3
g1
p1 g4
g2
g1
1
(a)
g3
p1 g4 (b)
Figure 3.1. Indifference curves and the Allais Paradox
The Independence Axiom is in fact equivalent to linearity in the probabilities. IA implies that: (1) Indifference curves are straight lines: if, ∀ g, g ∈ G , we have g ∼ g implies αg + (1 − α)g = g ∼ αg + (1 − α)g,
∀ α ∈ (0, 1).
(2) Indifference curves are parallel lines: if, ∀ g, g , g ∈ G , we have g ∼ g , the IA implies that αg + (1 − α)g ∼ αg + (1 − α)g , 3Euler’s Theorem: f (x) is homogeneous of degree r i ff n
rf (x) =
xi f i . i=1
∀ α ∈ (0, 1).
1
18
3. SOLUTION FOR EXERCISE III
Now consider the conditions in the Allais Paradox . A preference for g1 in the first pair gambles would indicate that the individual’s indifference curves were relatively steep, and hence a preference for g 4 in the second pair. 4 If, on the contrary, g 1 is preferred in the first pair, and g 3 in the second, which implies that indifference curves are parallel but rather fan out , as in figure (b). Now we turn to do the job follows another way. We can rewrite the gambles as g1 = (0.10 ◦ 1, 0.01 ◦ 1, 0.89 ◦ 1); g2 = (0.10 ◦ 5, 0.01 ◦ 0, 0.89 ◦ 1); g3 = (0.10 ◦ 5, 0.01 ◦ 0, 0.89 ◦ 0); g4 = (0.10 ◦ 1, 0.01 ◦ 1, 0.89 ◦ 0). Consider the following three gambles g5 = Get 1 with certainty;
0.10 0.01 ◦ 5, ◦ 0 ; g6 = 0.11 0.11 g7 = Get 0 with certainty . By the Completeness Axiom , we know either g 5 g6 or g 6 g5 . (1) If g5 g6 : by the IA, we have 0.11g5 + 0.89g5 0.11g6 + 0.89g5 , or g1 g2 ; and 0.11g5 + 0.89g7 0.11g6 + 0.89g7 , or g4 g3 . (2) If g6 g5 : we can do the job with the same logic as part (1), and get g2 g1 , and g3 g4 .
4. (The Simplex) Solution. The slope: Keeping the level of v.N.M utility constant
p1 u(x1 ) + p2 u(x2 ) + (1 − p1 − p2 )u(x3 ) = Const., and varying p 1 and p 2 alone, one has, locally, dp 2 dp2 u(x1 ) + u(x2 ) + ( −1 − )u(x3 ) = 0, dp1 dp1 or dp2 u(x3 ) − u(x1 ) = ≥ 0, dp1 u(x2 ) − u(x3 ) 4In the alternative case of relatively flat indifference curves, the gambles g and g would be 2 3
preferred.
5. (INDEPENDENCE AXIOM)
19
since x 2 > x3 > x1 . Direction where the utility increasing: Since upward movements in the triangle increase p 2 at the expense of p 3 (i.e. shift probability from the outcome x 3 up to x 2 ) and leftward movements reduce p 1 to the benefit of p 3 (shift probability from x1 to x3 ), these movements (and more generally, all northwest movements) lead to stochastically dominating gambles and would accordingly be preferred. 1
p2
p1
1
Figure 4.1. Expected utility indifference curves in the simplex diagram
5. (Independence Axiom) Proof. Let us assume without loss of generality that the elements of A have
been indexed so that a1 a2 · · · an . Now, let g k , 0 ≤ k ≤ n, be the gamble that yields outcome k with probability one: gk = (0 ◦ a1 , 0 ◦ a2 , . . . , 1 ◦ ak , . . . , 0 ◦ an ). Then a1 gk an , since all of them can be identified with sure outcomes. Let g = ( p1 ◦ a1 , . . . , pn ◦ an ) be any gamble in G , then n
g =
pk gk .
k=1
If there is only one k, s.th. pk = 1, that is, pj = 0, ∀ j = n, there is nothing to prove. So let N = # {(1, . . . , n) : pk = 0, 0 ≤ k ≤ n } > 1
20
3. SOLUTION FOR EXERCISE III
and suppose the proposition N −1
a1
pk gk an
k=1
is true for
N − 1.
By the definition of a compound gamble , N −1
N
pk gk = (1 − pN )
k=1
k=1
By the induction hypothesis ,
N −1
a1 (1 − pN )
k=1
pk gk + pN g N . 1 − pN
pk gk an . 1 − pN
Hence, by the independence axiom , we have N −1
(1 − pN )a1 + pN g N (1 − pN )
k=1
pk gk + pN g N (1 − pN )an + pN g N . 1 − pN
Applying the axiom once again, we obtain a1 = (1 − pN )a1 + pN a 1 (1 − pN )a1 + pN g N ; (1 − pN )an + pN g N (1 − pN )an + pN a n = an . Hence, by the transitivity, a1 g an , ∀ g ∈
G .
CHAPTER 4
Solution for Exercise IV 1. (MWG, 6.C.16) Solution. The maximum amount the person is willing to buy the
gamble: (4.1)
0.5u(w − Rb − y) + 0.5u(w − Rb + x) = u(w),
where R b is the maximal buying price. u(·)
A
u(w)
B
Rb
Rb
w
w−y
w + x
w
Figure 1.1. The maximum amount the person is willing to buy the gamble
The minimum amount the person is willing to sell the gamble: (4.2)
0.5u(w − y) + 0.5u(w + x) = u(w + Rs ),
where R s is the minimal selling price. In general, these two prices are different. However, if u(·) is CARA, then they are the same. In face, equation (4.1) and (4.2) can be restated as (4.3)
CE w−R = w,
(4.4)
CE w = w + Rs ,
b
21
22
4. SOLUTION FOR EXERCISE IV
u(·) C
E [u(·)] u(w)
D
Rs
w w + Rs
w−y
w + x
w
Figure 1.2. The minimum amount the person is willing to sell the gamble
where C E w−R and C E w are certainty equivalence for (4.1) and (4.2), respectively. The CARA implies that1 b
(w − Rb ) − CE w−R = w − CE w , b
thus Rb = Rs .
2. (MWG, 6.C.20) Proof.
u(CE ) = 0.5u(x + ε) + 0.5u(x − ε), where C E is the Certainty Equivalent . Hence u (CE )
(4.5)
(4.6)
u (CE )
∂C E = 0.5u (x + ε) − 0.5u (x − ε), ∂ε
∂C E ∂ε
2
+ u (CE )
∂ 2 CE = 0.5u (x + ε) + 0.5u (x − ε). ∂ε 2
Thus 0.5u (x + ε) + 0.5u (x − ε) − u (CE ) ∂ 2 CE lim = lim ε↓0 ∂ε 2 ε↓0 u (CE ) u (x) = u (x) = − R a (x), 1See MWG (1995) Section 6.C.
∂CE 2 ∂ε
3. (BETWEENNESS AXIOM)
23
since lim ε↓0
∂C E 0.5u (x + ε) − 0.5u (x − ε) = lim ε↓0 ∂ε u (CE ) = 0,
and lim u(CE ) = u(x). ε↓0
3. (Betweenness Axiom) Solution. The Betweenness Axiom 2 only requires that indifference sets be convex , i.e., if an individual is indifferent between two lotteries, then any probability
mixture of these two is equally good: if g ∼ g , then λg + (1 − λ)g ∼ g,
∀ λ ∈ [0, 1].
Essentially, the betweennss axiom is a substantially weaker version of the controversial independence axiom. 3 The axioms 1-4, 6, and the betweenness axiom means that the indifference curves are straight lines can be established in the same way as in Chapter 3, exercise (3). Note that we do not use the independence axiom in that exercise, in fact, betweenness axiom is suffices. Note also that these straight indifference curves need not be parallel, because the betweenness axiom imposes restrictions only on straight indifference curves and nothing on the relative positions of different indifference curves. 1
p2
p1
1
Figure 3.1. Betweenness axiom means the indifference curves are straight lines, but need not be parallel.
2See Dekel, E. (1986) for further discussion. 3See MWG Exercise 6.B.1A .
24
4. SOLUTION FOR EXERCISE IV
4. (Quadratic v.N-M Utility Function) a. Solution. The restrictions are
u (w) > 0,
u (w) < 0,
so b > − 4cw; ¯ c < 0.
b. Solution. R a (w) = −
so
4c , b + 4cw
∂ R a (w) 16c2 = ∂w (b + 4cw)2 > 0. This means that the quadric utility functions are unsatisfactory . Not only do they imply that utility reaches a maximum, they also entail that the absolute degree of risk aversion is increasing in wealth, approaching infinity as utility approaches its maximum. Consequently, one is led to the absurd result that the willingness to gamble for a bet of fixed size should decrease as wealth is increased. c. Proof. Before going through the proof, it is worthwhile to consider the intuition of the representation. The expected utility hypothesis suggests that preferences toward gambles can be represented by the expected value of a v.N-M utility function E u(w) , where w is a random variable that represents the income from an uncertain gamble. Expected utility in general depends on the form of the function u(·) and on the distribution of w. Suppose the distribution of w can be completely characterized by a vector of parameters α. In particular, let w be distributed on the real line with a P.D.F. f (w, α). Then4
E
u(w) =
u(w)f (w, α) dw.
The integral on the right-hand side of this equation is a function of this integral be represented by u(α), then u(α) = E u(w) is a valid representation of preferences.
.5 If we let
α
4From this subsection through the end of the chapter, we focus on continuous monetary
variable for convenience. 5It is not a function of w since w is just the variable of integration.
4. (QUADRATIC V.N-M UTILITY FUNCTION)
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Many problems in the economics of uncertainty are related to the trade-off between the average level of income and its degree of riskiness. Since the mean is a summary measure of average and the variance is a summary measure of risk, it will be particularly convenient to represent preferences by a function of the mean and variance of the income distribution. Unfortunately , this is not always possible, because in general the mean and variance do not completely determine the distribution of a random variable. There are many income streams that have the same mean and variance but different probability distributions. The expected utility associated with these income streams are different. Although u(α) is a valid representation of preferences, the vector α generally contains more than two parameters. Thus a utility function that depends only on mean and variance can be at best be viewed as an approximation to expected utility. There are some special cases, however, when a function involving only the mean and variance of the income distribution can be used to represent preferences. The quadratic utility function in this exercise is a such example. u(w) = a + bE w + 2cE w2
E
d.
= a + bE w + 2c
E
w
2
+ 2cVar w .
6
Solution. We prove this proposition by an indifference curve in the meanvariance plane.7 To establish this result, consider two gambles g1 , g2 ∈ G such
that g1 ∼ g2 . Then, the individual must be indifferent between g 1 , g 2 , and a compound gamble gq = (q ◦ g1 , (1 − q ) ◦ g2 ),
q ∈ [0, 1]
where q denotes the probability of obtaining g1 , and consequently (1 − q ) is the probability of obtaining g 2 . gq ∼ g1 ∼ g 2 . Letting µ i and σ i2 denote the mean and variance, respectively, of the distributions corresponding to the gambles (i = 1, 2, and q ). 6Markowitz (1959) demonstrated that if the ordering of alternatives is to satisfy the v.N-M
axioms of rational behavior, only a quadratic utility function is consistent with an ordinal expected utility function that depends solely on the mean and variance of the return. Consequently, even if the return for each alternative has a normal distribution, the mean-variance framework cannot be used to rank alternatives consistently with the v.N-M axioms unless a quadratic v.N-M utility function is specified. 7Our proof gives here follows Baron (1977).
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4. SOLUTION FOR EXERCISE IV
µ V (µ, σ 2 ) = β µ2
g2
µq
gq g1
µ1
σ12
σq2
σ22
σ2
Figure 4.1. An indifference curve in the mean-variance plane.
Note that distribution functions preserve the linear structure of gambles (as do P.D.F.’s),8 so µq = (4.7)
w d qF 1 (w) + (1 − q )F 2 (w) w dF 1 (w) + (1 − q )
= q
= qµ 1 + (1 − q )µ2 . 2
σq = (4.8)
=
w dF 2 (w)
2
w − qµ 1 − (1 − q )µ2 d qF 1 (w) + (1 − q )F 2 (w) 2
q (w − µ1 ) + (1 − q )(w − µ2 ) d qF 1 (w) + (1 − q )F 2 (w)
= qσ 12 + (1 − q )σ22 + q (1 − q )(µ1 − µ2 )2 . Solving for q from (4.7) yields (4.9)
q =
µq − µ2 . µ1 − µ2
Substituting (4.9) into (4.8) yields µq − or
µ1 − µ2 2 µ1 − µ22 + σ12 − σ22
µ2q + σq2 =
µ2 µ1 (µ1 − µ2 ) + µ2 σ12 − µ1 σ22 , µ21 − µ22 + σ12 − σ22
µq − α µ2q + σq2 = k,
(4.10) where
µ1 − µ2 , µ22 + σ12 − σ22 µ2 µ1 (µ1 − µ2 ) + µ2 σ12 − µ1 σ22 k = . µ21 − µ22 + σ12 − σ22 α =
8See MWG (1995) p.183.
µ21 −
