Microeconomics III
Instructor: Prof. Pascal Courty
Problem Set 1
Deadline: January 14, 2009
1
TA session: Michal Lewandowski
Expected Utility Theorem
Let X be a set of alternatives. The objects of choice are lotteries with finite support:
(
)
X
L = P : X → [0, 1] #{x|P (x) > 0} < ∞,
P (x) = 1
x∈X
P
Notice that x∈X P (x) = 1 condition is well defined due to the finite support assumption.
The observable choices are modeled by a binary relation on L, written %⊂ L × L.
Axiom (Weak order). % is complete and transitive
Axiom (Continuity). For every P, Q, R ∈ L,
P Q R ⇒ ∃α, β ∈ (0, 1) : αP + (1 − α)R Q βP + (1 − β)R
Axiom (Independence). For every P, Q, R ∈ L, and every α ∈ (0, 1),
P % Q ⇐⇒ αP + (1 − α)R % αQ + (1 − α)R
Theorem 1 (vNM). %⊂ L × L satisfies Axioms 1-3 if and only if there exists u : X → R
such that, for every P, Q ∈ L
X
X
P % Q ⇐⇒
P (x)u(x) ≥
Q(x)u(x)
x∈X
x∈X
Moreover, in this case u is unique up to a positive linear transformation.
In what follows, you are asked to prove sufficiency of the first part of this
theorem. Necessity is easy to check and is omitted. Second part (uniqueness) is also
omitted.
Please follow steps outlined in this problem set and do not change notation. To save your
time you can use the following lemmas without giving proofs:
Lemma 1. For every P, Q, R ∈ L, and every α ∈ (0, 1),
P Q ⇐⇒ αP + (1 − α)R αQ + (1 − α)R
(1)
P ∼ Q ⇐⇒ αP + (1 − α)R ∼ αQ + (1 − α)R
(2)
Lemma 2. For any P, Q ∈ L and every α ∈ [0, 1],
P % Q ⇒ P % αP + (1 − α)Q % Q
The same holds for strict preference and for indifference.
Prove the following lemmas and intermediary results
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(3)
Problem 1.1
Lemma 3. For every P, Q ∈ L and every α, β ∈ [0, 1],
P Q ⇒ (αP + (1 − α)Q % βP + (1 − β)Q ⇐⇒ α ≥ β)
(4)
Solution:
We assume P Q and want to prove:
αP + (1 − α)Q % βP + (1 − β)Q ⇐⇒ α ≥ β
(5)
Let’s start with (⇐=).
It is easy to verify that in case any of the following occurs: α = 0, α = 1, β = 0, β = 1 or
α = β the claim reduces to triviality or to condition (3) from the previous lemma. Assume
then 1 > α > β > 0. Now define γ = αβ . Note that γ ∈ (0, 1). Now we can write the
following:
βP + (1 − β)Q = γαP + (1 − γα)Q
= γ(αP + (1 − α)Q) − γ(1 − α)Q + (1 − γα)Q
= γ(αP + (1 − α)Q) + (1 − γ)Q
If we denote R = αP + (1 − α)Q then we can write:
βP + (1 − β)Q = γR + (1 − γ)Q
(6)
From assumption that P Q and by lemma 2 we have:
P Q ⇒ P % Q ⇒ αP + (1 − α)Q % Q
Using our definition of R we can write R % Q. Given that let’s apply lemma 2 once more
to obtain:
R % Q ⇒ R % γR + (1 − γ)Q
Substituting equation (6) for the RHS and R = αP + (1 − α)Q for the LHS we get:
αP + (1 − α)Q % βP + (1 − β)Q
This finishes the proof of one direction. Let’s now prove the converse direction, i.e. (=⇒)
of (5)
The contrapositive of the statement to prove is:
¬(α ≥ β) ⇒ ¬(αP + (1 − α)Q % βP + (1 − β)Q)
This is equivalent to:
α < β ⇒ βP + (1 − β)Q αP + (1 − αQ
Notice that this is very similar to the statement which we already proved above. The only
difference is the switched role of α and β and strict preference instead of weak preference.
This is where we use the strict preference version of equation (3). This finishes the proof.
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Problem 1.2
Step 1. Assume that there are worst and best outcomes in X denoted x∗ and x∗ , respectively.
For every P ∈ L there exists a unique αP ∈ [0, 1] such that:
P ∼ αP x∗ + (1 − αP )x∗
where x∗ and x∗ are treated as degenerate lotteries.
Solution:
It is easy to verify that condition (4) from lemma 3 holds for strict inequality and strict
preference. Suppose we have three lotteries such that P Q R. By continuity we know
that there exist α, β ∈ (0, 1) such that the following holds:
αP + (1 − α)R Q βP + (1 − β)R
. From lemma 3 and transitivity we know that it must be that α > β. Again by continuity,
transitivity and lemma 3 we know that there exists α1 , β1 ∈ (0, 1) such that α1 < α, β1 > β
and:
α1 P + (1 − α1 )R Q β1 P + (1 − β1 )R
If we continue in this way we can construct a decreasing sequence (αi )i=1,...,n and an increasing sequence (βi )i=1,...,n such that:
P αi P +(1−αi )R αi+1 P +(1−αi+1 )R Q βi+1 P +(1−βi+1 )R βi P +(1−βi )R R
It is easy to see now that the limits of two constructed sequences must be the same:
lim αn = αQ = lim βn
n→∞
n→∞
and αQ P + (1 − αQ )R ∼ Q. If we assume now that x∗ Q x∗ , the result follows
immediately (just replace P with x∗ , R with x∗ and Q with P ). If x∗ ∼ Q x∗ , then
αQ = 1, if x∗ Q ∼ x∗ , then αQ = 0 and finally if x∗ ∼ x∗ , then ∼= L × L and a constant
u does the job. Uniqueness of αQ follows from lemma 3.
Notice that this proof also shows that the sets:
{α ∈ [0, 1]|αP + (1 − α)R Q}
{α ∈ [0, 1]|Q αP + (1 − α)R}
are open or alternatively that the sets:
{α ∈ [0, 1]|αP + (1 − α)R % Q}
{α ∈ [0, 1]|Q % αP + (1 − α)R}
are closed which are alternative definitions of continuity.
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Problem 1.3
Step 2. Consider a lottery P ≡ (x1 , p1 ; x2 , p2 ; ...; xn , pn ), where P (xi ) ≡ pi > 0. Index
∗
αP such
Pnthat P ∼ αP x + (1 − αP )x∗ is linear in probabilities pi i.e. it can e written as
αP = i=1 pi αxi , for some scalars αxi .
Hint: Eliminate step by step each outcome of the lottery replacing it with the mixture of
x∗ and x∗ .
Solution:
Construct a sequence of lotteries Pi , i = 1, ..., n such that:
• P0 = P
• Pi ∼ Pi−1
• supp(Pi ) = {x∗ , x∗ , xi+1 , ..., xn }
Observe that
P
= p1 x1 + (1 − p1 ) x2 ,
p2
pn
; ...; xn ,
1 − p1
1 − p1
We can now replace x1 by αx1 x∗ + (1 − αx1 )x∗ , where x1 ∼ αx1 x∗ + (1 − αx1 )x∗ . We obtain:
P1 = p1 (αx1 x∗ + (1 − αx1 )x∗ ) + p2 x2 + ... + pn xn
We can write is as:
P1
p1 (1 − αx1 )
p3
pn
p 1 α x1
; x∗ ,
; x3 ,
; ...; xn ,
= p2 x2 + (1 − p2 ) x ,
1 − p2
1 − p2
1 − p2
1 − p1
∗
Again we can replace x2 by αx2 x∗ + (1 − αx2 )x∗ , where x2 ∼ αx2 x∗ + (1 − αx2 )x∗ . If we
continue this way, at the end we will end up with the following lottery:
!
n
n
X
X
∗
pi αxi , x∗ ,
pi (1 − αxi )
Pn = x ,
i=1
i=1
But notice that
Pn
i=1 pi (1
− αxi ) =
Pn =
Pn
x∗ ,
i=1 pi
n
X
−
Pn
pi αxi , x∗ , 1 −
i=1
This shows that αP =
Pn
i=1 pi αxi
i=1 pi αxi )
n
X
=1−
!
Pn
i=1 pi αxi ).
So
pi αxi )
i=1
is indeed linear in probabilities.
In principle the proof is finished since we can define u(xi ) = αxi and U (P ) = αP
and thenPfrom the above step it follows that the utility from lottery P may be written as
U (P ) = ni=1 pi u(xi ). For the sake of completeness, below I give the last step, which you
do not have to prove.
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Step 3 (Final). %⊂ L × L satisfies Axioms 1-3. Then there exists u : X → R such that for
any lotteries P, Q ∈ L:
X
X
P % Q ⇐⇒
P (x)u(x) ≥
Q(x)u(x)
x∈X
x∈X
Consider any two lotteries: P ≡ (x1 , p1 ; x2 , p2 ; ...; xn , pn ) and Q ≡ (y1 , q1 ; y2 , q2 ; ...; ym , qm ),
such that x∗ xi , yj x∗ , i = 1, ..., n j = 1, ..., m. We know that P ∼ Pn and Q ∼ Qm ,
where Pn and Qm are constructed as in the last step:
!
n
n
X
X
Pn = x∗ ,
pi αxi , x∗ , 1 −
p i α xi )
Qm =
∗
i=1
m
X
x ,
qi αyi , x∗ , 1 −
i=1
m
X
i=1
!
q i α yi )
i=1
Define αx = u(x). By construction P % Q is equivalent to:
!
!
n
n
m
m
X
X
X
X
∗
∗
x ,
pi u(xi ), x∗ ,
pi (1 − u(xi )) % x ,
qi u(yi ), x∗ , 1 −
qi u(yi ))
i=1
i=1
i=1
i=1
By lemma 3 since x∗ x∗ the above is in turn equivalent to:
n
X
i=1
2
pi u(xi ) ≥
m
X
qi u(yi )
i=1
Risk attitudes
Prove the following claims and propositions:
Problem 2.1
Proposition 2.1. All CARA utility function with U (0) = 0 and U 0 (0) = 1 are of the
following form: Ua (x) = a1 − a1 e−ax , a > 0. Show that risk neutrality occurs for a → 0+
Solution:
00
(x)
We have to solve the differential equation: − UU 0 (x)
= a > 0, where a > 0 is a constant
representing absolute risk aversion. We have two initial conditions which makes it a Cauchy
problem. Integrate the equation once from 0 to x to obtain log U 0 (x) − log U 0 (0) = −ax + 0
1-5
or by using the fact that U 0 (0) = 1, log U 0 (x) = −ax. Rewrite as U 0 (x) = e−ax . Integrate
it once more from 0 to x to obtain U (x) − U (0) = − a1 e−ax + a1 or by using another initial
condition U (0) = 0, U (x) = a1 − a1 e−ax
Now we calculate the limit as a goes to zero. Note that we can use L’Hospital rule:
lim
a→0+
1 − e−ax
xe−ax
= lim
=x
a
1
a→0+
Problem 2.2
Proposition 2.2. Given a twice continuously differentiable utility function U , the following
holds:
4
1
ARA(W ) = lim
p(W, h) −
(7)
2
h→0+ h
00
(W )
where ARA(W ) = − UU 0 (W
) and p(W, h) is a probability premium defined implicitly by:
p(W, h)U (W + h) + (1 − p(W, h))U (W − h) = U (W )
(8)
Moreover, for h = λW , we have:
4
RRA(W ) = lim
+
λ→0 λ
1
p(W, λW ) −
2
(9)
Hint: Use second order Taylor approximation of U .
Solution:
Rewriting equation (8) by using second order Taylor expansion of U around W , we
obtain for small h:
1
p(W, h)[U (W ) + U 0 (W )h + U 00 (W )h2 ]
2
1
+(1 − p(W, h))[U (W ) − U 0 (W )h + U 00 (W )h2 ] ≈ U (W )
2
And after simplifying:
1 00
U (W )h + U 0 (W )(2p(W, h) − 1) ≈ 0
2
Or
ARA(W ) = lim
h→0+
4
h
1
p(W, h) −
2
as was to be shown. If we set h = λW , equation (9) immediately follows.
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Problem 2.3
A gamble is a real-valued random variable with finite support. We can denote it by x ≡
(x1 , p1 ; ...; xn , pn ), where xi denotes monetary values and pi the corresponding probabilities,
i = 1, ..., n. In what follows we will implicitly assume that the arguments of utility function
belong to the domain. For instance if the domain of U is (0, +∞), then W + x ∈ (0, +∞).
Assume that we are in the world of Expected Utility, i.e. there exists a vNM utility function
U that represents the decision maker’s preferences.
Definition 1. A strategy whether to accept a gamble x or not is called wealth-invariant iff
the following holds:
EU (W1 + x) ≥ U (W1 ) ⇐⇒ EU (W2 + x) ≥ U (W2 ),
∀W1 , W2
(10)
Prove the following proposition:
Proposition 2.3. If strategy is wealth-invariant then utility is CARA.
Remark: It is easy to check that the converse direction is also true.
Solution:
Given utility function U , consider two lotteries xi ≡ (h, p(Wi , h); −h, 1−p(Wi , h)), where
Wi , h > 0, i ∈ {1, 2}, such that:
EU (Wi + xi ) = U (Wi )
(11)
Contrary to what we are to show, assume that ARA(W1 ) > ARA(W2 ). By proposition
2.2 equation (7), for h sufficiently small we know that p(W1 , h) > p(W2 , h). Let q be
between the two probability premiums: p(W1 , h) > q > p(W2 , h). I define another lottery
y ≡ (h, q; −h, 1 − q). By definition of a probability premium utility function U rejects
lottery y at wealth W1 and accepts it at wealth W2 , which contradicts wealth invariance.
Problem 2.4
Definition 2. A strategy whether to accept a gamble x or not is called homogeneous (or
scale-invariant) iff the following holds:
EU (W + x) ≥ U (W ) ⇐⇒ EU (λW + λx) ≥ U (λW ),
Prove the following proposition:
Proposition 2.4. If strategy is homogeneous then utility is CRRA.
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∀λ > 0
(12)
Remark: It is easy to check that the converse direction is also true.
Solution:
We will be using proposition 2.2, equation (9). In order to avoid notational problem I
will rewrite this equation using symbol θ instead of λ, since λ is used in the definition of a
homogeneous strategy (see equation (12)). So we will use the following equation:
4
1
RRA(W ) = lim
p(W, θW ) −
(13)
2
θ→0+ θ
where p(W, θW )U (W + θW ) + (1 − p(W, θW ))U (W − θW ) = U (W ).
Given utility function U , consider lottery x ≡ (θW, p(W, θW ); −θW, 1 − p(W, θW )), where
W, θ > 0, such that:
EU (W + x) = U (W )
(14)
Contrary to what we are to show, assume that RRA(λW ) > RRA(W ), for some λ > 0 and
λ 6= 1. Define x0 ≡ (θW, p(λW, λθW ); −θW, 1 − p(λW, λθW )), such that:
EU (λW + λx0 ) = U (λW )
(15)
For θ sufficiently small, by equation (13), we know that: p(λW, λθW ) > p(W, θW ). Therefore, we can find q such that: p(λW, λθW ) > q > p(W, θW ). Define y ≡ (θW, q; −θW, (1 −
q)).
By equation (15), since p(λW, λθW ) > q:
EU (λW + λy) < U (λW )
By equation (14), and since q > p(W, θW ), we have:
EU (W + y) > U (W )
which is a contradiction.
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