MICROECONOMIC THEORY IPROBLEM SET 4

MARCIN PĘSKI

Comparative statics of savings decision. An agent chooses consumption in twoperiods, c1 = w−s and c2 = s, where s are the savings. She has discounted separableutility over two periods

u1 (w − s) + βu2 (s) ,

where we assume (only!) that functions ui are weakly increasing.

(1) Consider function

f (s; β) = u1 (w − s) + βu2 (s) .

Show that family f (.; β) is ordered by a single crossing condition. In thelecture notes, we show that this implies that the savings increase with thediscount rate.

In class.

(2) Conclude that savnings increase with discount factor.

In class.

(3) (Much harder) Consider a multi-period version of the consumuer problem

maxc1+...+ct=w

u1 (c1) + βu2 (c2) + ...+ βt−1ut (ct) .

Can you extend the above result, i.e., can you show that the first-periodsavings increase with the discount factor? What if you assume that the utilityfunction is concave? (The truth is that I don’t know whether it can be done,one wat or the other. But I would be very excited if somebody did it, or foundconditions on the utility function that would allow for such an extension.)

I don’t know whether it can be done.1

2 MARCIN PĘSKI

Examples of lattices.

(1) Let X ⊆ R and let ≤ be the relation of "not larger than". Show that X is alattice.

TBA

(2) For any set A, let X be the set of all subsets of A. Show that X is a latticewith partial order induced by the relation of inclusion ⊆: For each x, x′ ⊆ A,

write x ≤ x′ if and only if x ⊆ x′.

The partial order is deifined above.We will sho that any two sets x, y ⊆ A have a unique minimum and maxi-

mum.• Let z = x ∩ y be the intersection (i.e., the common part) of sets x, y. Ofcourse, z ⊆ x and z ⊆ y, which implies that z ≤ x and z ≤ y in the senseof the lattice relation. Moreover, for each z′ such that z′ ≤ x ∈ R andz′ ≤ y, it must be that z′ ⊆ x and z′ ⊆ y, which implies that z′ ⊆ x∩y = z

and z′ ≤ z. Thus, z is equal to the minimum of x and y, z = x ∧ y.• Similarly, we show that z = x ∪ y is the unique “maximum”.

(3) Let Xi ⊆ R for i ≤ n, and let X∗ = X1 × ... × Xn. Define partial order ≤∗:for each x = (x1, ..., xn) , x′ = (x′1, ..., x′n), define x ≤∗ x′ if and only if xi ≤ x′i

for each i ≤ n. Show that (X∗,≤∗) is a lattice.

The partial order is deifined above.We will sho that any two vectors x, y ∈ X∗ have a unique minimum and

maximum.• Let z = (min (x1, y1) , ...,min (xn, yn)) be the vector with coordinatesequal to the smaller of the coordinates of vectors x and y. Of course,z ≤ x and z ≤ y in the sense of the lattice relation. Moreover, for eachz′ such that z′ ≤ x and z′ ≤ y, it must be that z′i ≤ min (xi, yi) whichimplies that z′ ≤ z. Thus, z is equal to the minimum of x and y, z = x∧y.• Similarly, we show that z = (max (x1, y1) , ...,max (xn, yn)) is the unique“maximum”.

ADVANCED MICRO THEORY II 3

(4) Continuation of the previous question. Oshown the other hand, typically,subsets A ⊆ X∗ with partial order ≤∗ are not lattices. Suppose that n = 2,X1 = X2 = {0, 1} . Find all subsets A ⊆ X that are not lattices.

For example, A = {(0, 1) , (1, 0)}.

(a) In general, let (X1,≤1) ., ..., (Xn,≤n) , be a finite collection of lattices.Define product lattice X∗ = X1 × ...×Xn with partial order ≤∗: for eachx = (x1, ..., xn) , x′ = (x′1, ..., x′n), define x ≤∗ x′ if and only if xi ≤i x

′i for

each i ≤ n. Show that (X∗,≤∗) is a lattice.

The proof is similar to (c). The only change is to replace “min (xi, yi)” by“xi ∧i yi” (i.e., the “minimum operation on lattice Xi) and “max (xi, yi)”by “xi ∨i yi”

Graphical representation of the set order. Use a picture to explain set order ineach of the cases of lattices from the excercise Examples of Lattices.

Examples of supermodular functions.

(1) Suppose that (X,≤X) is a lattice and f : X → R is increasing (i.e., for eachx ≤X y, f (x) ≤ f (y)) and supermodular function. Show that for any anyconvex and increasing function v : R→ R, v ◦ f : X → R is supermodular.

Fix any x, y ∈ X and denote

a = f (x ∧ y) ,

b = f (x ∨ y) ,

c = f (x) ,

d = f (y) .

We need to show that

v (c) + v (d) ≤ v (a) + v (b) .

Because f is increasing, we have

a ≤ c, d ≤ b.

4 MARCIN PĘSKI

Because f is supermodular, we have

a+ b ≥ c+ d.

Because v is convex, we have

v (c) ≤ b− cb− a

v (a) + c− ab− a

v (b) ,

v (d) ≤ b− db− a

v (a) + d− ab− a

v (b) .

By summing up both sides of the above inequalities, we get

v (c) + v (d) ≤ 2b− (c+ d)b− a

v (a) + (c+ d)− 2ab− a

v (b)

≤ v (a) + v (b) + (b+ a)− (c+ d)b− a

(v (a)− v (b)) .

Because v is increasing, the latter is not larger than

≤ v (a) + v (b) .

This concludes the proof.

(2) Any convex combination of two supermodular functions is supermodular.

TBA

(3) Suppose that X ⊆ R is an open interval and assume that X is the totallyordered lattice from Exercise 1. Show that any function is supermodular.

TBA

Derivatives of a supermodular function with more than 2 variables. Supposethat twice differentiable function f : R3 → R is supermodular in the partial order onR3: for all x, x′, y, y′, z, z′,

f (x, y, z) + f (x′, y′, z′) ≤f (min (x, x′) ,min (y, y′) ,min (z, z′))

+ f (max (x, x′) ,max (y, y′) ,max (z, z′)) .

Show that all partial derivatives fxy, fxz, fyz are (weakly) positive. Conversely, showthat if these partial derivatives are always weakly positive, then the function f issupermodular.

ADVANCED MICRO THEORY II 5

Instead of solving the excercise directly, we will solve a more general claim (yes,sometimes it is easier!): Fix n ≥ 2. We show two claims:Claim 1: Suppose that a twice differentiable function f : Rn → R is supermodular

in the partial order on Rn, i.e., for all bx = (x1, ..., xn) , x′ = (x′1, ..., x′n) ∈ Rn, wehave

f (x) + f (x′) ≤ f (min (x1, x′1) , ...,min (xn, x

′n)) + f (max (x1, x

′1) , ...,max (xn, x

′n)) .

Then, for all i 6= j, ∂2f∂xi∂xj

≥ 0.Claim 2: The opposite holds as well: If f : Rn → R is a a twice differentiable

function such that for all i 6= j, ∂2f∂xi∂xj

≥ 0, then f is supermodular.Ad Claim 1. Notice first that the claim for n = 2 was proven in the class. We are

goiung to deduce the general case n ≥ 2 from the two dimensional case. Indeed, takeany coordinates i 6= j and fix x∗k ∈ R for all k 6= i, j. Define function

fx∗

ij (xi, xj) = f (x∗1, ..., xi, ..., xj, ...x∗n)

with all but ij coordinates fixed at levels x∗k. Notice that function fx∗ij is superdoular

in two dimensions, which implies that the cross-derivative is positive. In the sametime, the crossderivative of function fx∗

ij is equal to ∂2f∂xi∂xj

.Ad Claim 2. Take any two vectors x, x′. By possibly rearranging the coordinates,

we can assume that x = (a, b) , x′ = (a′b′), where a, a′ ∈ Rm, b, b′ ∈ Rn−m and a ≥ a′

and b ≤ b′. Then,

x ∧ x′ = (a′, b) ,

x ∨ x′ = (a, b′) .

We need to show that if all the partial cross-derivatives are positive, then

f (a, b′)− f (a′, b′) ≥ f (a, b)− f (a′, b)

6 MARCIN PĘSKI

for any a ≥ a′ and b′ ≥ b. Notice that

f (a, b′)− f (a′, b′)− (f (a, b)− f (a′, b))

=1ˆ

0

(a− a′)T

1ˆ

0

D2f (a′ + s (a− a′) , b+ t (b′ − b)) (b′ − b) dt

ds

=1ˆ

0

1ˆ

0

[(a− a′)T

D2f (a′ + s (a− a′) , b+ t (b′ − b)) (b′ − b)]dt

ds.Here, D2f (a′ + s (a− a′) , b+ t (b′ − b)) is a m× (n−m) matrix of partial derivatesevaluated at (a′ + s (a− a′) , b+ t (b′ − b)) for any s, t [0, 1]. Because all the elementsof each matrix D2f are positive, and all the coordinates in vectors a − a′ and b′ − bare positive, the expression in the square brackets is strictly positive. This impliesour claim.

Monopoly comparative statics. A monopolist introduces a new product to themarket. Its advertising budget is equal to a ≥ 0. The higher the advertising budgeta, the higher the probability p (a.) that the product is successful and the demand forit is high. In the same time, the monopolist chooses how many units of the productto produce. The expected profits of the monopolist are equal to

p (a) [q (A+ P (q)− c)] + (1− p (a)) [q (P (q)− c)]− a,

where A > 0 is a positive constant that measures ths shift in the demand causedby successful marketing, p (.) is increasing, P (q) is the (inverse) demand function,and c is the marginal cost of production. Witouth making any further assumptions,show that the optimal quantity and the advertising budget are increasing in A anddecreasing in c. UPDATE: I corrected the profit function. I

We will only show that the optimal advertising budegt and quantities are decreasingin c. Let X = {(a, q) : a, q ≥ 0} with the “coordinatewise” partial order. Let w =−cWe need to show that function

f (a, q : w) = p (a) [q (A+ P (q) + w)] + (1− p (a)) [q (P (q) + w)]− a

= q (P (q) + w) + Ap (a) q − a.

ADVANCED MICRO THEORY II 7

is supermodular in (a, q), and that it has increasing differences in (a, q) and w = −c.Then, we can apply the Theorem from the class to conclude that an increase in w (adecrease in c) leads to increase in optimal (a, q).

Supermodularity. For each x, y ∈ X, observe that

f (x ∨ y) + f (x ∧ y)− f (x)− f (y)

=Ap (max (xa, ya)) max (xq, yq) + Ap (min (xa, ya)) min (xq, yq)− Ap (xa)xq − Ap (ya) yq.

We consider four cases:

• If xa ≥ ya and xq ≥ yq, then the above is equal to 0.• If xa ≥ ya and xq < yq, then the above is equal to

=Ap (xa) yq + Ap (ya)xq − Ap (xa)xq − Ap (ya) yq

=A (p (xa)− p (ya)) (yq − xq) ≥ 0.

• All the other two cases are treated in a similar way.

(A short answer would be to notice that function qp (a) is supermodular, functionsthat depend only one one coordinates are supermodular, and the sum of supermodularfunctions is supermodular).

Increasing differences: Take any (q′, a′) ≥ (q, a) and w′ > w. Then,

f (a′, q′;w′)− f (a, q;w′)− f (a′, q′;w) + f (a, q;w)

=q′w′ − qw − q′w + qw = (q′ − q) (w′ − w) ≥ 0.

•

Indifference curves of preferences over AA acts. Assume that the set of theprizes and states are two-element, Z = {a, b} and S = {1, 2}. We interpret AA actsas elements of space X = [0, 1]2, where f = (f1, f2) ∈ X anfd fs is the probability ofprize a in state s ∈ S. Let � be the preference relation over AA acts.

(1) Suppose that preferences are represented by “utility” function over acts u (f1, f2) =min (f1, f2). Do such preferences satisfy the axioms of the expected utility re-presentation? Draw few indifference curves.

8 MARCIN PĘSKI

No, they don’t satisfy Independence. For example, take f = (1, 0), g =(0, 1) and ˙h = 1

2f + 12g =

(12 ,

12

). Then, u (h) > u (f) = u (g).

(2) Suppose that the preferences satisfy the independence axiom. Show that theindifference curves are linear.

TBA

Recovering subjective probability. (This question was written in 2012 duringthe US presidential elections, before Intrade.com closed.)

Intrade.com allows me to buy and sell contracts the provide me a payoff of $10 ifBarack Obama wins US presidential elections in November. Currently (28th August),I can buy one such contract at price $5.65, and I can sell one contract at price $5.63.If I buy the contract, I pay price now and then receive $10 if BO wins the election. IfI sell, I get the price now and then must pay out $10 if BO wins the election. I don’tbuy nor sell the contract, my future consumption is equal to 0, no matter what.)Suppose that I have preferences with a state-independent utility over lotteries overmoney. You observed my preferences over constant (i.e., state-independent) bets anddeduced that my state-independent utility is equal to u (c) =

√c+ 100 for c ≥ −100

(or, more precisely, my state-independen utility is given by u (c) up to positive affinetransformations).Additionally, suppose that I neither buy nor sell any contracts on Intrade.com. Fi-nally, assume that there are no transactions costs.Are all these observations consistent with the expected utility theory? If so, derivethe bounds on my subjective probabilitites of Barack Obama winning presidentialelections. If not, explain.

Let p ve the probability that Obama wins. Because I don’t buy, we have

10 = u (0) ≥ p (u (10− 5.65)) + (1− p)u (−5.65)

≥√

94.35 + p(√

104.35−√

94.35),

or

p ≤ p∗ = 10−√

94.35√104.35−

√94.35

.

ADVANCED MICRO THEORY II 9

Because I don’t sell,

10 = u (0) ≥ p (u (−10 + 5.63)) + (1− p)u (5.63)

≥√

105.63− p(√

105.63−√

95.63),

orp ≥ p∗ =

√105.63− 10√

105.63−√

95.63.

The answer depends on whether p∗ ≤ p∗.