Problem set 8, solutions

1. Suppose a decision maker with rational % has following (strict) pref-erences over three deterministic outcomes x y z. Define two lotteriesL = αx + (1 − α)y and L′ = αx + (1 − α)z for some α ∈ (0, 1). Supposethat the preferences over lotteries of the decision maker admit the vN-Mrepresentation. Show via two distinct arguments that L L′.

solution: First approach is via the independence axiom (implied by thevN-M representation) which implies that, since we can regards x, y and zas degenerate lotteries, y z ⇔ αy + (1 − α)x αz + (1 − α)x for anyα ∈ (0, 1). Relabelling α to 1− α we have L L′.

The second approach is via the vN-M theorem. Denoting the (Bernoulli)utility of the x, y and z outcomes by ux, uy and uz respectively, L L′ ⇔U(L) = αux + (1− α)uy > U(L′) = αux + (1− α)uz, which is equivalent touy > uz, which holds since y z.

2. Consider the setting in which we discussed the Ellsberg paradox. Thatis, there are two urns: urn one with 50 red and 50 blue balls and urn twowith 100 balls of either red or blue color. A decision maker chooses betweena red bet on urn one or two and separately between a blue bet on urn one ortwo. Suppose she derives utility uw from winning and utility ul < uw fromloosing.

Suppose the decision maker has a set of beliefs C = 0, . . . , 100 aboutthe number of red balls in urn two. A belief p ∈ C assigns probability oneto the number of the red balls in urn two being p. Each p ∈ C also carrieswith it a belief that assigns probability one to the number of the red ballsin urn one being 50.

Finally, suppose the decision maker has preferences over bets that, in-stead of being vN-M, assign ‘expected utility’ to bet bcu on color c ∈ R,Bin urn u ∈ 1, 2 equal to minp∈C Ep[u(bcu)] where Ep is expectation operatorover belief p ∈ C.

1. Show that this model of maximization of minimum expected utility(max-min model for short) generates behavior consistent with the Ells-berg paradox.

2. Which choices over red and blue bets would be consistent with a modelwhere minp∈C Ep[u(bcu)] is replaced by maxp∈C Ep[u(bcu)]. What is thisa model of?

3. Which choices over red and blue bets would be consistent with a modelwhere minp∈C Ep[u(bcu)] is replaced by

∑p∈C

1101Ep[u(bcu)]. What is this

a model of?

1

Problem set 8, solutions

solution: 1. First, let me derive the max-min utility of choosing bet bcuon color c ∈ R,B in urn u ∈ 1, 2. Clearly

minp∈C

Ep[u(bR1 )] = minp∈C

Ep[u(bB1 )] =uw + ul

2

since the believes in C put probability one on the number of red balls in urnone being 50 (that is, Ep[u(bc1)] is independent of p ∈ C for c ∈ R,B). Onthe other hand on urn two

minp∈C

Ep[u(bR2 )] = minp∈C

puw + (100− p)ul100

= ul

minp∈C

Ep[u(bB2 )] = minp∈C

pul + (100− p)uw100

= ul

since betting on, say, red leads to, conditional on belief p ∈ C, expectedutility of puw+(100−p)ul

100 .Max-min model generates behavior consistent with the Ellsberg paradox

since minp∈C Ep[u(bR1 )] > minp∈C Ep[u(bR2 )] as well as minp∈C Ep[u(bB1 )] >minp∈C Ep[u(bB2 )].

Another way to see this is to realize that the max-min model is a modelof ambiguity aversion. Whenever a decision maker has deterministic beliefs(as is the case for urn one), she behaves identically to the vN-M expectedutility maximizer. That is, if Ep[u(bcu)] is independent of p, then the decisionmaker simply calculates expected utility from choosing bcu and chooses theoption with the highest expected utility.

When Ep[u(bcu)] varies with p, that is when the decision maker does notknow precisely the distribution of balls (as is the case for urn two), shecalculates the lowest (most pessimistic) expected utility consistent with herbeliefs p and choose based on those pessimistic beliefs.

solution: 2. With this max-max model

maxp∈C

Ep[u(bR1 )] = maxp∈C

Ep[u(bB1 )] =uw + ul

2

again whereas

maxp∈C

Ep[u(bR2 )] = maxp∈C

puw + (100− p)ul100

= uw

maxp∈C

Ep[u(bB2 )] = maxp∈C

pul + (100− p)uw100

= uw.

In the max-max model the decision maker is ambiguity seeking or, alterna-tively, optimistic about the composition of urn two (notice that this opti-mism depends on which bet is being chosen, if a red bet, then the decisionmaker acts based on beliefs p = 100 and if a blue bet is chosen, then thedecision maker acts based on beliefs p = 0).

2

Problem set 8, solutions

solution: 3. In this max-average model again∑p∈C

1101Ep[u(bR1 )] =

∑p∈C

1101Ep[u(bB1 )] =

uw + ul2

since |C| = 101. On the other hand∑p∈C

1101Ep[u(bR2 )] =

∑p∈C

1101

puw + (100− p)ul100

=puw + (10100− p)ul

101 · 100∑p∈C

1101Ep[u(bB2 )] =

∑p∈C

1101

pul + (100− p)uw100

=puw + (10100− p)ul

101 · 100.

where p =∑100

i=0 i = 5050. Hence∑

p∈C1

101Ep[u(bR2 )] =∑

p∈C1

101Ep[u(bB2 )] =uw+ul

2 so that the decision maker behaves as if her beliefs about urn two wereidentical to the beliefs about urn one. Namely, she is indifferent between ared bet on urn one and two as well as being indifferent between a blue beton urn one and two.

3. Given a lottery L = pi, xini=1 that gives xi with probability pi (so that∑ni=1 pi = 1), describe how to model L as a random variable. Be explicit

about the underlying probability space.

solution: Given L with finite set of n outcomes, we can define the set ofstates Ω = 1, . . . , n. Using Ω we can define F as, say, the power set ofΩ. We know that for finite sets, such a rich σ-algebra causes no problems.Given (Ω,F) we can define probability measure P : F → [0, 1] via P[ω] = pωfor ∀ω ∈ Ω (notice that we have just started building link to L). Since∑

ω∈Ω pω = 1, P will be a well defined probability measure (we can extendit to all of F by taking unions of states). Finally, we construct a randomvariable l : Ω→ X, where X = ∪ni=1xi is the set of all outcomes of L, suchthat l(ω) = xω for ∀ω ∈ Ω. Given this construction, for any state ω ∈ Ω, ωdetermines the realization of l, l(ω) = xω. Since ω occurs with probabilityP[ω] = pω, xω is the realization of l with probability pω.

4. Consider CARA utility u(x) = x1−a−11−a with a ∈ [0, 1] defined on x > 0.

Show that lima→1 u(x) = log x for any x > 0.

solution: From L’Hopital’s rule, lima→1x1−a−1

1−a = lima→1−x1−a−a log (x) =

log (x).

5. Show that rA(x) and rR(x) derived for u(x) ∈ C2 are invariant topositive linear transformation of u.

3

Problem set 8, solutions

solution: Take u and its positive linear transformation v = cu+ k, where

k ∈ R and c > 0. Then v′ = cu′ and v′′ = cu′′. Hence rA,u(x) = −u′′(x)u′(x) =

− cu′′(x)cu′(x) = −v′′(x)

v′(x) = rA,v(x). Identical arguments proves rR,u(x) = rR,v(x).

6. Consider a knowledge function defined by Ki(E) = ω ∈ Ω|Pi(ω) ⊆ Efor some (finite) set of states Ω and information partition Pi inducing Pi(ω)for ∀ω ∈ Ω. Let F be the power set of Ω.

1. Show that E ⊆ F ⇒ Ki(E) ⊆ Ki(F ) for ∀E ∈ F .

2. Show that Ki(E) ∩Ki(F ) = Ki(E ∩ F ) for ∀E,F ∈ F .

3. Show that Ki(E) = Ki(Ki(E)) for ∀E ∈ F .

4. Give an example of Pi for i ∈ 1, 2 on Ω = 1, 2, 3, 4, 5 such that forany ω ∈ Ω no event is common knowledge in ω.

5. Give an example of Pi for i ∈ 1, 2 on Ω = 1, 2, 3, 4, 5 such that forany ω ∈ Ω, ω is common knowledge in ω.

solution: 1. Take E and F such that E ⊆ F . We need to show thatω ∈ Ki(E)⇒ ω ∈ Ki(F ). Take arbitrary ω ∈ Ki(E). Then, by definition ofKi, Pi(ω) ⊆ E and hence Pi(ω) ⊆ F . Thus ω ∈ Ki(F ).

solution: 2. We need to show that ω ∈ Ki(E) ∩Ki(F )⇒ ω ∈ Ki(E ∩ F )as well as ω ∈ Ki(E ∩ F ) ⇒ ω ∈ Ki(E) ∩ Ki(F ). Take arbitrary ω ∈Ki(E) ∩ Ki(F ). Since ω ∈ Ki(E) and ω ∈ Ki(F ), by definition of Ki,Pi(ω) ∈ E and Pi(ω) ∈ F . Hence Pi(ω) ∈ E ∩ F and thus ω ∈ Ki(E ∩ F ).To prove the reverse implication take arbitrary ω ∈ Ki(E∩F ). By definitionof Ki we have Pi(ω) ∈ E ∩ F and thus Pi(ω) ∈ E and Pi(ω) ∈ F and henceω ∈ Ki(E) and ω ∈ Ki(F ).

solution: 3. Since Ki(Ki(E)) ⊆ Ki(E) it suffices to show that ω ∈Ki(E) ⇒ ω ∈ Ki(Ki(E)). Take arbitrary ω ∈ Ki(E). By definition of Ki,Pi(ω) ∈ E. Note that for any ω′ ∈ Pi(ω), ω′ ∈ Ki(E). Hence Pi(ω) ⊆ Ki(E)and hence ω ∈ Ki(Ki(E)).

solution: 4. Such an information structure does not exist. If it did itwould be equivalent to P1 ∧P2 being empty, which is not possible (union ofelements of meet is Ω by definition). Another way to see this is to note thatKi(Ω) = Ω for any information partition Pi, for ∀i ∈ 1, 2. Hence, as anexample, K1(K2(Ω)) = Ω. That is, Ω is always common knowledge at anyω ∈ Ω. The information partition close to being almost empty is such P1

4

Problem set 8, solutions

and P2 such that P1 ∧ P2 contains unique element, Ω. An example of thisinformation structure is, for the Ω given,

P1 = 1, 2, 3, 4, 5P2 = 1, 3, 2, 4, 5.

It is easy to see that P1 ∧ P2 = Ω.

solution: 5. One trivial example is, with the second equality easy to check,P1 = P2 = P1 ∧ P2 = 1, 2, 3, 4, 5.

7. Suppose a consumer lives for two periods. In period 1 she is born withcapital k ≥ 0. Given her capital a chance determines her income zk wherez is a r.v. with log-normal distribution lnN (0, κ).1 Given her income, theconsumer splits her income between period-1 consumption c and period-2capital k′. For simplicity assume that the period-1 draw of z is z = 1. Inperiod 2 the consumer’s income is zk′ (where z is another draw from thelog-normal) and since there are no further periods, period-2 consumption c′

is equal to zk′.Suppose the consumer has Epstein-Zin preferences. That is, her period-1

objective function aggregates period-1 consumption c and period-2 certainty

equivalence y into V (c, y) =[(1− β)c1−ρ + βy1−ρ] 1

1−ρ . The consumer has

CARA attitude towards risk in period-2 and hence y =[E[(c′)1−α]

] 11−α .

Assume α ∈ (0, 1) and ρ ∈ (0, 1).

1. Show that EX [xα] = exp (αµ+ α2σ2

2 ) for α > 0.[Hint: Not that the

Gaussian integral∫R exp (−ax2 + bx+ c)dx =

√πa exp

(b2

4a + c).]

2. Noting that c′ = zk′, derive y =[E[(c′)1−α]

] 11−α .

3. Show that period-1 objective of the consumer is to maximize, by choos-

ing c, V(c, (k − c) exp (1−α)κ

2

).

4. Solve the consumer’s period-1 optimization problem (ignore secondorder conditions) and write optimal c as a function of k. Write impliedoptimal k′ and hence implied optimal E[c′].

5. Show that the elasticity of intertemporal substitution of E[c′]c with

respect to β1−β is 1

ρ .

6. Explain the meaning of a claim that the Epstein-Zin preference breakthe link between the elasticity of intertemporal substitution and therelative risk aversion.

1 Recall that if a r.v. X ∼ lnN (µ, σ2), then ln(X) ∼ N (µ, σ2). Also, EX [x] =

exp (µ+ 12σ2) and fX(x) = 1

xσ√2π

exp(− (ln (x)−µ)2

2σ2

)for ∀x ∈ R>0.

5

Problem set 8, solutions

solution: 1. With the pdf given

E[xα] =

∫R>0

xα1

xσ√

2πexp

(−(ln (x)− µ)2

2σ2

)dx.

Using change of variables x = exp (z), xα = exp (αz), dz exp z = dx thisbecomes

E[xα] =

∫R

exp (αz)1

exp (z)σ√

2πexp

(−(z − µ)2

2σ2

)exp (z)dz

=

∫R

exp (αz)1

σ√

2πexp

(−(z − µ)2

2σ2

)dz

=

∫R

1

σ√

2πexp

(−(z − µ)2

2σ2+ αz

)dz

=1

σ√

2π

√π1

2σ2

[( µσ2 + α)2

4 12σ2

− µ2

2σ2

]

=

[αµ+

α2σ2

2

].

solution: 2. We know that c′ = zk′ where z is r.v. with lnN (0, κ). Hence

E[(c′)1−α]1

1−α can be computed using the expression in the previous part.

E[(c′)1−α] = E[z1−α](k′)1−α = exp(

(1−α)2κ2

)(k′)1−α. From this it follows

easily that E[(c′)1−α]1

1−α = exp(

(1−α)κ2

)k′.

solution: 3. We know from the previous part that the certainty equivalent

of the period-2 consumption, as a function of k′, is exp(

(1−α)κ2

)k′. Since the

period-1 income of the consumer is k (recall the normalization of period-1z to unity), we have budget constraint k = c + k′. Thus the certainty

equivalent, assuming the budget constraint holds, is exp(

(1−α)κ2

)(k − c).

Period-1 consumption is c and hence the consumer maximizes, by choosing

c, V(c, exp

((1−α)κ

2

)(k − c)

).

solution: 4. Using the functional form of V and the expression from theprevious part, the optimization problem in period 1 is

maxc∈[0,k]

[(1− β)c1−ρ + β(k − c)1−ρ exp

((1−α)(1−ρ)κ

2

)] 11−ρ

with foc after slight rewriting (and noting that the objective function to the1

1−ρ−11

1−ρpower cannot equal zero)

(1− β)(1− ρ)c−ρ − β(1− ρ)(k − c)−ρ exp(

(1−α)(1−ρ)κ2

)= 0.

6

Problem set 8, solutions

Further algebra shows

c = (k − c)[

β1−β exp

((1−α)(1−ρ)κ

2

)]− 1ρ.

Denoting the term in the square brackets by m this gives c = k m1+m . From

k′ = k − c this gives k′ = k1+m . Since c′ = zk′, E[c′] = E[z] k

1+m =

exp (κ2 ) k1+m .

solution: 5. From the previous part E[c′]c = exp (κ2 ) 1

m . Elasticity of E[c′]c

with respect to β1−β is

∂E[c′]c

∂ β1−β

β1−βE[c′]c

. Note that ∂m−1

∂ β1−β

= 1ρm

1ρ−1

− 1ρ exp

((1−α)(1−ρ)κ

2

).

Hence

∂ E[c′]c

∂ β1−β

β1−βE[c′]c

= exp (κ2 )1ρm

1ρ−1

− 1ρ exp

((1−α)(1−ρ)κ

2

)β

1−βm

exp (κ2 )

= 1ρm−1+ρ exp

((1−α)(1−ρ)κ

2

)β

1−βm

= 1ρm−1+ρm−ρm = 1

ρ .

solution: 6. Epstein-Zin preferences permit the consumer’s attitude to-wards risk to be governed by a different coefficient than the attitude towardsintertemporal changes in consumption. We have just seen that the elasticityof intertemporal substitution is 1

ρ . Because the agent has CARA attitudetowards randomness in the second period (see the expression for certaintyequivalence) her coefficient of relative risk aversion is α.

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