review 1 and last one

8/10/2019 Review 1 and Last One

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Examiners commentaries 2011


120 Mathematical economics

Important note

This commentary reflects the examination and assessment arrangements for this course in theacademic year 201011. The format and structure of the examination may change in future years,and any such changes will be publicised on the virtual learning environment (VLE).

Specific comments on questions Zone B

SECTION A

Answer ALL questions from this section (8 marks each).

Question 1

Answer all parts of this question.

(a) Define the indirect utility function.

(b) State Roys identity.

(c) Confirm that Roys identity holds for the Cobb-Douglas utility function

u(x1, x2) = x1/21 x

1/22 .

(a) See definition 20, page 38 of the subject guide.

(b) Roys identity is stated as:

xi(p, m) = V(p, m)/piV(p, m)/m

.

(c) We need to solve the utility maximisation problem:

max x1/21 x

1/22 subject to p1x1+p2x2 = m.

First order conditions are:

1

2x1/21 x

1/22 p1 = 0

1

2x1/21 x

1/22 p2 = 0.

Hence:

1

2x1/21 x

1/22

1

2x1/21 x

1/22

= p1

p2

x2x1 =

p1p2

x2 = p1

p2x1.

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(c) Suppose that the utility function is differentiable and that the derivatives of the utilityfunction are strictly positive. Is it possible that the solution to the consumers utilitymaximisation problem does not satisfy the budget constraint as an equality? No. Strictlypositive derivatives imply:

limdx0

u(x+dx, y) u(x, y)

dx >0

hence, when dx >0, we have u(x+dx, y)> u(x, y), which implies nonsatiation:(x+dx, y) (x, y) and in turn the fact that the budget constraint will be satisfied as anequality at the optimum (i.e. utility can be increased if more income is spent onconsumption).

Question 3


(a) Explain why ordinary differential equations have families of solutions rather

than unique solutions.(b) Find the entire family of solutions to the following ordinary differential

equations:

i. y 16

y 16

y= 0

ii. y +y 2y = 0

(a) See page 138 of the subject guide.

(b) i. Roots of the characteristic polynomial are1/3, 1/2. It follows that:

y(x) = Aex/3 +Bex/2

for arbitrary A and B .

ii. Roots of the characteristic polynomial are2, 1. It follows that:

y(x) = Ae2x +Bex

for arbitrary A and B .

Question 4


(a) Define quasiconcavity.(b) Prove that for any quasiconcave functionf(x), the functionf(x) is quasiconvex.

(c) Prove that all concave functions are quasiconcave.

(a) See definition 26, page 69 of the subject guide.

(b) Learning activity 5.1, page 69 of the subject guide.

(c) See theorem 16, page 70 of the subject guide.

Question 5

Let f(x; a) be a continuous function ofx Rn and the scalar a. For anya, considerthe problem of finding the maximum off(x; a) with respect to x. Let x(a) be themaximiser which we assume to be differentiable with respect to a.

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(a) Prove that: ddaf(x(a); a) = a f(x

(a); a).

(b) Provide an economic application of this result.

(a) This is the Envelope Theorem. See page 26 of the subject guide.

(b) Economic applications of the Envelope Theorem are, for example, Shephards lemma, or

Roys identity.

SECTION B

Answer THREE questions from this section (20 marks each).

Question 6

A firm with labour input L and capital input K has a production function,

f(L, K) = L0.25K0.25.

The firm operates competitively in both the output good market and the inputfactor markets. In the firms cost minimisation problem, the firm wishes to find thecombination of inputs (L, K) to minimise its production costC=wL +rK ofproducing an output of at leasty, i.e. with an output constraint off(L, K) y. w isthe wage level and r is the cost of the capital. w and r are both strictly positive.

(a) In the long-run, both the firms capital and labour are freely variable. Using theKuhn-Tucker conditions, find the long-run conditional labour and capitaldemands L(w,r,y) and K(w,r,y) of the cost minimisation problem. What is theresulting minimum cost C(w,r,y) of producing at leasty?

(b) In the short-run the firms choice of capital is constrained byK K. UsingKuhn-Tucker conditions, find the short-run conditional labour and capitaldemands L(w,r,y) and K(w,r,y) of the cost minimisation problem. What is theresulting minimum cost C(w,r,y) of producing at leasty?

(c) Prove that the cost functionC(w,r,y) is homogeneous of degree one in inputprices (w, r) in all above cases.

(d) Are the conditional factor demandsL(w,r,y) and K(w,r,y) homogeneous ininput prices (w, r)? If so, of what degree?

(a) First note that Kuhn-Tucker is a sufficient condition for x to be the solution to max g(x)subject to h(x) k, where g(.) and h(.) are both defined on a convex set U Rn, if (i) g isquasi-concave and differentiable, (ii) g not all 0, and (iii) h is quasi-convex anddifferentiable. Here the problem is,

minL,K

wL+rK subject to L0.25K0.25 y maxL,K

wLrK subject to L0.25K0.25 y.

Check: The domain (L, K) 0 is convex, wL rK is linear (and hence is quasi-concave)and differentiable, and L0.25K0.25 is quasi-convex (as L0.25K0.25 is quasi-concave) anddifferentiable. So form the Lagrangian,

L = wL rK+ y+L0.25K0.25

.

The Kuhn-Tucker conditions are:

L

L = w+ 0.25L0.75K0.25 = 0

L

K = r+ 0.25L0.25K0.75 = 0

0y L0.25K0.25

y L0.25K0.25

= 0.

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From the FOCs, we cannot have = 0 when w,r >0. Hence we have >0. Thecomplimentary slackness condition then implies that the output constraint must bind at y .Solving the simultaneous equations yields,

L(w,r,y) =

r

w0.5

y2, K(w,r,y) =

w

r 0.5

y2.

The cost function is then,

C(w,r,y) =wL +rK = 2(wr)0.5y2.

(b) Now we have an additional constraintK K. This constraint is linear, and so isquasi-convex. From the argument above all conditions of the Kuhn-Tucker theorem aresatisfied. The Lagrangian is, this time,

L = wL rK+y+L0.25K0.25

+(KK).

The Kuhn-Tucker conditions are:

L

L = w+ 0.25L0.75

K0.25

= 0L

K = r+ 0.25L0.25K0.75 = 0

, 0

y L0.25K0.25

y L0.25K0.25

= 0

K K

KK

= 0.

The problem is the same as above if= 0. Then the solutions are:

L

(w,r,y) = r

w0.5

y2

, K

(w,r,y) =w

r0.5

y2

C(w,r,y) = 2(wr)0.5

y2

provided that Kis feasible, i.e.

K(w,r,y) K y r

w

0.25K0.5.

Wheny >rw

0.25K0.5 this is not feasible, and we have the binding K= K. Again from the

first FOC cannot equal 0. So the first complimentary slackness implies thatL0.25 K0.25 =y, or

L(w,r,y) = K1y4.

From the FOCs then,

= 4wL0.75 K0.25 = 4wK1y3

= r+ 0.25L0.25K0.75 =w K2y4 r.

is clearly non-negative. is non-negative if

wK2y4 r y r

w

0.25K0.5.

In this case the cost function is

C(w,r,y) =w K1y4 +rK.

Thus in summary:

Ify rw

0.25 K0.5, L =

rw

0.5 y2, K =

wr

0.5 y2 and C= 2(wr)0.5y2.

Ify >rw

0.25K0.5, L = K1y4, K = K andC=w K1y4 +r K.

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(c) For the long-run case

C(tw,tr,y) = 2(twtr)0.5y2 = 2t(wr)0.5y2 =tC(w,r,y).

The short-run case when y rw

0.25K0.5 is the same as in the long-run. When

y > rw0.25 K0.5,C(tw,tr,y) = tw K1y4 +trK= tC(w,r,y).

Therefore the cost functions are homogeneous of degree 1 in all cases.

(d) For the long-run case

L(tw,tr,y) =

tr

tw

0.5y2 =L(w,r,y), K(tw,tr,y) =

tw

tr

0.5y2 =K(w,r,y).

The short-run case when y rw

0.25K0.5 is the same as in the long-run. When

y >rw

0.25K0.5, both factor demands are independent of the factor prices. Hence the

factor demands are homogeneous of degree 0 in all cases.

Question 7

You find that consumer As Arrow-Pratt coefficient of absolute risk aversionrA isconstant at b >0, where

Coefficient of absolute risk aversionrA= uA(w)

uA(w) (.1)

and uA(w) is the utility function of wealth w for consumer A. Given this, you decideto investigate the family of functionsuA(w) with the property rA= b. You also knowthat the utility of no wealth uA(0) = 0.

(a) By lettingv(w) =uA(w) in (.1), solve the first-order differential equation for thefunction v(w). Hence solve for the family of functionsuA(w) that satisfy rA= b.

In contrast, you find that consumer Bs coefficient of absolute risk aversion islinearly decreasing in wealth, i.e.

uB(w)

uB(w)=rB =

c

w (.2)

where c > 0 is a constant.

(b) Solve for the family of functionsuB(w) that satisfy (.2) for all values ofc >0.Hence show that such function cannot exist with property uB(0) = 0 when c= 1.

(c) What restrictions on the parameters are required for consumers A and B tohave monotonically increasing and strictly concave utility functions? You mayassume that c = 1.

(a) By lettingv (w) = uA(w), (.1) becomes

dv

dw = bv.

Solving this, dv

v = b

dw = ln v= bw+C v= k1e

bw

for some constant k1. Now substituting back uA(w) = v(w),

duAdw

=k1ebw uA=

k1rA

ebw +k2

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for some constant k2. Finally we know thatuA(0) = 0, and hence,

uA(0) = k1b

+k2 = 0 k2 = k1

b .

Thus uA(w) = k1b (1 e

bw).

(b) Similarly by lettingv(w) =uB(w), (.2) becomes

dv

dw =

cv

w.

Solving this, dv

v = c

dw

w ln v= c ln w+D v= k3e

c lnw =k3wc

for some constant k3. Substituting back uB(w) = v(w) then,

du

dw =k3w

c uB =

k31cw

1c +k4, c = 1k3ln w+k4, c= 1

for some constant k4. For c = 1, uB(0) = 0 implies k4 = 0. However, there is no feasiblesolution for c = 1 as uB(0) = for all values ofk3 and k4.

(c) Monotonically increasing utility function meansu(w)> 0; strict concavity requiresu(w)< 0. For A then,

uA(w) = k1ebw uA(w) = bk1e

bw.

Then given that b > 0, we require k1 > 0. For B, forc = 1,

uB =k3wc uB = ck3w

c1.

c >0 then implies that k3 > 0.

Question 8

Dr London has to mark N exam scripts in T days, where 2< T < . If on day t hemarks st scripts, then his disutility on that day is s

2t . Dr London therefore solves the

following dynamic optimisation problem with discount factor, where 0< 1,

min{st}T1

Tt=1

ts2t subject toTt=1

st= N and st 0, t= 1, 2, . . . , T .

(a) Formulate this problem as a dynamic programming problem, and write downthe Bellman equation.

(b) Solve the problem by backward induction to findst, t= 1, . . . , T , in terms ofN, ,t and T. You may guess the form of the Bellman value function, but you mustverify your guess by showing that the Bellman equation is satisfied.

(c) Is st increasing, decreasing or constant over time? In particular, what can yousay about Dr Londons optimal marking schedule when = 1?

(a) Turning the minimisation problem into a maximisation problem by changing the sign of theobjective function, define the Bellman value function as,

V(n, ) = max{st}T1

Tt=

ts2t subject toTt=

st= n.

The Bellman equation is then

V(n, ) = maxs

s2+V(n+1, + 1)

subject to n+1 = n s.

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(b) On the last day Dr London has to mark all remaining scripts, i.e. sT =nT. There istherefore no maximisation on the last day, and his value function is V(nT, T) = n

2T. Next

consider day T 1:

V(nT1, T 1) = maxsT1

s2T1+V(nT, T)

subject to nT =nT1 sT1

= maxsT1s

2

T1+V(nT1 sT1, T)

= maxsT1

s2T1 (nT1 sT1)

2

.

The first order condition is

2sT1+ 2(nT1 sT1) = 0 sT1 =

1 +nT1.

Plugging back in

V(nT1, T 1) =

1 +nT1

2

nT1

1 +nT1

2=

1 +n2T1.

Try another day:

V(nT2, T 2) = maxsT2

s2T2+V(nT2 sT2, T 1)

= max

sT2

s2T2

1 +(nT2 sT2)

2

.

The first order condition is

2sT2+ 2 2

1 +(nT2 sT2) = 0 sT2 =

2

1 ++2nT2,

in which case

V(nT2, T 2) = 2

1 ++2n2T2.

So we conjecture that Dr London will mark son day with the Bellman value functionV(n, ), where,

s = TT

i=0 i

n, V(n, ) = TT

i=0 i

n2. (.3)

To verify this, we show that this solves the Bellman equation for any given date,

V(n, ) = maxs

s2+V(n s, + 1)

= max

s

s2+

T1T1i=0

i(n s)

2

.

Substituting in s in (.3) on the RHS,

TTi=0

in

2

+ TT1i=0

i

n

TTi=0

in

2

=

TTi=0

i

2n2+

TT1i=0

i

T1i=0

iTi=0

i

2n2

=

T

2+T

T1i=0

iTi=0

i2

n2

=

Ti=0

i

T

Ti=0

i2 n2

= TT

i=0 i

n2

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which equals the LHS as given in (.3).

So starting from t = 1, where n1 = N, Dr London marks s1 leaving n2 for day 2 andbeyond, where,

s1 = T1

T1

i=0 iN n2 = N s1 = 1

T1

T

1

i=0 iN =

T2i=0

i

T

1

i=0 iN.

Similarly on day 2,

s2 = T2T2

i=0 i

n2= T2T2

i=0 i

T2i=0

iT1i=0

iN=

T2T1i=0

iN

n3 = n2 s2 =

T2i=0

iT1i=0

i

T2T1i=0

i

N=

T3i=0

iT1i=0

iN

and so on. Hence,

s = TT1

i=0 iN and n =

Ti=0

i

T1i=0 i

N.

(c) For


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(b) The present value Hamiltonian can be written as:

H=et[u(ct) v(t)] +t(act bt d)

wheret is the co-state variable.

(c) The first order condition is

H

ct= 0 etu(ct) +ta= 0.

Hence we obtainu(ct) = ate

t

as requested.

(d) We have that:

t = H

t t= e

tv(t) +tb

t = +H

t t= act bt d.

Also, the transversality condition:limt

t= 0.

Recall thatu(ct) = ate

t.

We differentiate with respect to time, to find:

u(ct)ct = atet ate

t.

Recall:

t = e

t

v

(t) +tbt =

u(ct)

a et.

By substituting, we get:

u(ct)ct = a[etv(t) +tb]e

t atet

= av(t) atbet ate

t

= av(t) at(b+)et

= av(t) +au(ct)

a et(b+)et

= (b+)u(ct) av(t).

Question 10


(a) Consider a consumer with the following indirect utility function:V(p1, p2, m) =

mp1+p2

.

i. Find the consumers demand functionsx1(p1, p2, m) and x2(p1, p2, m).

ii. Find the consumers expenditure functione(p1, p2, u).

iii. From your answer to (ii), what can you say about the type of preferences

that this consumer has? What is his utility functionu(x1, x2)?(b) Consider a consumer with preferences: u(x1, x2) = max{x1, x2}.

i. Find the consumers indirect utility function.

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ii. Find the consumers expenditure function.

(a) i. We use Roys identity to find the consumers demand functions:

x1(p, m) = V/p1V/m

= m

p1+p2

x2(p, m) = V/p2V/m

= m

p1+p2.

ii. We invert the indirect utility to find the expenditure function:

V(p, e(p, u)) = u

e(p, u)

p1+p2= u

e(p, u) = u(p1+p2).

iii. This expenditure function shows us that to getu utils we need to buy u units of good 1and u units of good 2. Hence the two goods are perfect complements and the utility

function is:u(x1, x2) = min{x1, x2}.

(b) i. With utility functionu(x1, x2) = max{x1, x2} the two goods are perfect substitutes.Demands are:

Ifp2 p1, then x1 = mp1

and x2 = 0.

Ifp2 p1, then x1 = 0 and x2 = mp2

.

Hence the indirect utility function is:

V(p, m) = m

min{p1, p2}.

ii. We find the expenditure function by inverting the indirect utility function:

V(p, e(p, u)) = u

e(p, u)

min{p1, p2} = u

e(p, u) = u min{p1, p2}.

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