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Intermediate Microeconomic Theory

Problems 1. Consider an agent who values consumption in period 0 and 1 according to the following utility function: u(co, el) = ln(co) + 6ln(ci) 6 is a discount factor (8 < 1) which indicates that the agnet prefers to consume today more than he can tomorrow. Suppose that the agent is given a total wealth today of w and that he may save any portion of this money in order to consume tomorrow. If he saves money he is paid interest r. Thus the agents budget constraint is: co + 1 + r—w (a) By renaming variables, notice that this problem is identical to the regular cobb douglas problem. (b) Solve for the agents marshallian demand functions. (c) At what relation of 6 and r will the agent consume the same amount in each period? (d) Suppose that instead of being given a fixed endowment, the agent instead has access to a production technology that can make two products x, y that can be sold at prices ps,py. The technology is limited by inputs so that at most: x2 + y2

2

Solve for the optimal amount of x and y produced that maximize the agents profits. (e) Suppose that x is the amount of production that can be made in period 0 and y is the amount of production made in period 1. Suppose that Ps = 1, py = 1, r = 0 Why can we seperate the production decision from the consumption decision in this problem? Give a value for 6 such that

consumption is held constant in each period and the agent exactly consumes his production.

2. The legrangian multipliers in economics are often called "shadow costs". and build intuition on how to use them to answer economic relevant questions.

In this problem we will try to see where this term comes from

(a) Lets start with a very simple problem: M = max 2x — x2 ST x < K

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1. Solve this problem using as a constraied optimization problem. For a given x, what is the slope of 2x — x2? What is the value of A(K) when K < 1? Suppose that the constraint was increased slightly to K + E with E very small - how much should M change by?

2. Suppose that we wanted to determine how much M increases as K is increased.

One obvious way to solve this problem would be to resolve the problem with the constraint changed, however we can do something else. Use the envelope theorem to show that the change in M when K goes from K0 to K1 is:

f

K1

A(K)

Ko

(b) *Lets take these ideas to a more economic relevant problem.

max a ln(x) + (1 — a) ln(y)

11 (13x, 1. Solve for A(a).

Suppose that we have the following utility function:

x,y

ST

:

PEX

=

2. Find a general formula for how much v (px,py, m) changes as m changes from m* to m** 3. Find a general formula for how much v (px, py, m) changes as Ps changes from px* to px** : 3. Suppose that a safety agency is thinking of establishing a criterion under which an area prone to flooding should be evacuated. The probability of flooding is 1%. There are four possible outcomes: (a) (A) : No evacuation is necessary, and none is performed (B) : An evacuation is performed that is unnecessary (C) : An evacuation is performed that is necessary (D) : No evacuation is performed, and a flood causes a disaster Suppose that the agency is indifferent between the sure outcome B and the lottery of A with probability p and D with probability (1 — p) . Further assume that the agency is indifferent between the sure outcome C and the lottery of B with probability q and D with probability (1 — q) . Suppose also that it prefers A to D and that p E (0,

1), q e (0,1)

(a)Construct a utility function of the expected utility form for the agency.

3 (b)Consider two different policy criteria: • Criterion 1: This criterion will result in an evacuation in 90% of the cases in which flooding will occur and an unnecessary evacuation in 10% of the cases in which no flooding occurs.

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•Criterion 2: This criterion is more conservative. It results in an evacuation in 95% of the cases in which flooding will occcur and an unnecessary evacuation in 5% of the cases in which no flooding occurs.

Derive the probability distributions over the four outcomes under these two creit- eria. Then, by using the utility function in (a), decide which criterion the agency would prefer.

4. Recall that the profit function is given by:

7 ( P)

y

=

m

N

(p)EY ax E

yi(P)p,

(a) Use the envelope theorem to show that if we are given a profit function 71 -(p), we can back out the input demand functions as: thr(p)

Y, (P) = (b) A common notation is to use x's and w's to denote inputs and their prices and to use y's and p' s to denote outputs and output prices.

Consider the single input (x) and single output (y) technology described by:

f 0x<1 Y

1n(x) x > 1

Find the input demand function x(py, wx) and the profit function 7(py, wx). At what ratio of prices py, wx will the firm shut down and produce nothing?

5. *Suppose that we are given the production function f (xi, x2) = min(xi, OC2)" (a) Draw an isoquant for the case where a = 1, and f (xi, x2) = 10. What are the possible combinations of (x1, x2) that will be chosen by a cost minimizing firm? (b) Find the input demand functions xl (p, w1, w2, ce), x2 I,(pW2, , w1, w2, a), the supply funcC(w tion y(p, w1, w2, a) and profit funciton 71 (p,wl, w2, a). For what values y)of a=is xi + ,w2x2 a , ST : min wimin(x x2)" — 7F < 00? What is this requirement called? (c) Just like in utility maximization, we can start with a cost minimization problem instead of a profit maximization problem.

1. Start by solving a cost minimization problem: 2. From the minimization problem you will have xi are conditional on y. We call these "conditional factor demand functions".

, w2, a, y), x2(wi, w2, a, y), c(w1, w2). Note that these x's you get here

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3. Given your function c(wi, w2, a, y), no 71 (p,

w solve: y

r, co2) = max py — c(c4.7 , 2, a, y) Notice that this 71 is identical to the one w you found in part (i). Just likewin the utility case, we can solve a (typically simpler) cost minimization problem and then use our results to solve for profit maximization as a function of a single variable.

4. Calculate the supply function y (p, wr, w2, a). Plug this into the conditional factor demand functions to get the same results as part (ii).

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Solutions 1.

(a) if we were to replace 6 with a, ±1E,, with pl, and co, c1 with xo, x1 our problem is identical to:

max u(xo, xi) = ln(x0) + 81n(xl) X0 ,X1

:st

xo +pixi = u.)

(b) Taking the FOC we have: 1

= A co =

6 c1 ci

SO:

1 1+r

= (1+ r)

oco

Substitution into the budget constraint yields:

4 (8'r) = lw+ = 6 (11 ++ 60 xi(sr) (c) when 6(1 + r) = 1, then the marshallian demands are equal. In macro the usual condition is phrased: "when the discount rate is equal to the interest rate" (d) To maximize profits, the agent solves: IllaXpxX +pyy St

:

X2 +

y2 = 2

Note that we do not have to worry about corner solutions because the production possiblity set is a circle and thus when the production set hits the corners, the slopes are zero and infinity respectively. Thus, there is no price vector athat is not tangent to the PPC before hitting the corners. Solving the FOC we get: Px A2x py + 2y Thus:

0

0

Px

Plugging these into the production constraint yields:

Py

Y (1 13

= =

+ 1)

2

y2 = 2

Py

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P2

(e) px = py = 1, thus: x(Px,Py) = Y(Px,Py) = 1 We must find the discount rate such that production equals consumption. This will be met if: co=c1=1 From part c we know that this must be 6 = 1.

2.

(a) Setting up the legrangian we have: OK

K: 2x — x2 + A(K — x)

al 2 — 2x — A = 0 ax (a : (K — x) > 0,A > 0, (K — x)A = 0 = max((2 — 2K), 0). Notice s that the slope of 2x — 2 — 2x. In places where the constraint

The FOC of this is:

A(K)

x2 =

binds x = K and thus the legrangian multiplier is exactly equal to the slope of the function.

In economics, A is often called the "shadow cost." It corresponds to the amount that the agent would be willing to pay to change the constraint slightly. Notice that if the constraint K is binding and is moved slightly from K to K + E M would change by: lim M(K + e) — M(K) E—>o As E goes to zero and we multiply the top and the bottom of this by a little bit we get: lim M(K + E) — M(K) = = EA( K) M(K) M ( KE W + E) — lim E e->0

When the constraint is not binding: The second term in this is simply the slope. Thus lim M(K e) — M(K) = 0 dM E—>o But this is identical to \E since A = 0. Thus: lim M(K + E) — M(K) = EA(K) c—>o

In the entire domain.

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(b) If we think about Reimann Integration, we will recognize the above difference as exactly one of our rectangles underneath a curve. Thus, to find the rate of change over an area, we are simply going to add up a bunch of rectangles as their widths go to zero. A bit more formally: = lim E N->D9 i=1

M ( KO - M ( K0)

[M ( K0 + ( K1

Notice that as N gets bugger the difference between identical to the above: lim

E

[m(K

-v—>oc a=1

+

+1

-

+ (K1

1 Ko)

) gets smaller and smaller. Eventually we get something that looks

and i

M(K0

)

+1

)

m (K

i )1

+

=f

K1

A(K)dK

K0

We could get the same thing using the envelope thm, Recall that the envelope thm says that we can ignore the marginal effects of K on the optimal x to find the change of M with a change in K. Thus: A

dM(x(K), K) dK

— M(K0) In order to find the change over a larger area we dK = f A (K) M(KO) =f simply add up all the marginal changes: dK K) dM(x(K),

(c) Setting up the FOC: K : a ln(x) ±

(1 — a) ln(y) + A(m — pxx + PO))

Thus:

O K

• x

Y(Px,Py, in) x(Px,Py,m) A(a,px,m)

a

= Apx

1-a

a y1—a

Ox

and:

(d) 1. A(a) = f7,'„

— x

OK Ox

= Apy px py (1 — a)m Py

am Px a °1,x11 Ps

1

m

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2. e:; =)(a). Thus:

m** 1 ) = f -mdm = ln(m**) — ln(m*) m* You could check this by plugging everything into the indirect utiltiy function and noting m*.as a seperate term.

14* that=the function ln(m) 3. v (pr , py, m) v (74, py, m) = — f ÷„„x* dm — findirect ;i dm =utility ln(px* ) — a has ln (pr

Also from the text: u(C) = qU (b) (1 — q)U (D) = pq u(A) 1 Thus a utility function of the expected utitliy form that represents these prefer- ences has: u(B)

pq

u(C)

u(D)

=

U(L)

iE{A,B,C,D)

0

E

Aiu(t),

E

Ai = 1

ie (11,B,C,D)

(b) To judge a criterion we look at the probability of the four cases: Criterion 1: Pr(No Evacuation Nec&None Performed) Pr(A) = * 9 9 No Flood EvacNoFlood Pr(No Evacuation Nec&Performed) Pr(Evacuation Pr(B) = .99 * .1 = .099 Nec&Performed) Pr(Evacuation Nec&None Performed) Pr(C) = .01 * .9 = .009

= .891

Pr(C) = .01 * .1 = .001

Thus the total utility in this case is:

.891U(A) + .099U(B) + .009U(C) + .001U(D) = .99(.9 + .1p) + .01 (.9pq) Criterion 2: (.99 * .95)U(A) + (.99*.05)U(B) + (.01 * .95)U(C) = .99(.95 + .05p) + .01(.95pq) Subtracting Criterion 2 from Criterion 1 yields: .99 (.05 — .05p) + .01(.05pq)

4

Since IDE (0,1) criterion 2 is strictly preferred.

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4. (a) This is a simple result of the envelope theorem: yCY

From the FOC: joi So:

7r(P) =

P +

uYi

Oyi

p3

=0

ayi E P Y(P)

Y(P) + A

yEY

ayi

= Y(P)

Pi n 0

pi j#i

(b) Ignoring shut down for a minute, the firm that is forced to produce would maxi- mize: pin(x) — wx p

The FOC is:

p

— w —> xx=— w Plugging these into the maximand we have: 71-(p, w) = p ln(1w1) —p = p [ln(-w11) — 11 This is only positive when

> 1. Thus: x

5.

_

ln( ) > 1 0w otherwise p [1n(li) — 1] ln(1w1) > 1

7r(P, 0 (a) We see that the isoquant will always have a kink at xi — x2. w) produce (x1 ratio of = x2), or use the same factor one and two

otherwise Thus a profit maximizing firm will either choose to not

(b) plugging in xi = x2 MaXpXai — (Wi

W2)X1

FOC: xl

Thus:

pax

X2

(

= 1 = (Wi + w2) ((w 1 ± W2))

1

0-1

Pa (w1 + w) (

)

pa (wi )1_ w2) pa w (i(w p ( 74, w1, W27 a) pa The second derivative is negative if a < 1. CRTS or(71).1.

IRIS and xi Otherwise the production function is w2) = x2 = oc y(p,

w2,

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(c) 1. See ii 2.

ST

min wix

We don't need to take the FOC, simply note that for a given Y: xi(w), w2, c(w), w2,

3. Now solving:

+ w2x2 x", =xz =y

= x2(wi, w2, ,

(wi w2)y

Y) =

a

maxpy — (wi + w2)Y`' We get:

1 —Y a

a

p =w11 + W2

ap

y = ( W1 + w2)

ifoz

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