Micro Prelim June 2016 – ANSWER KEYS QUESTION 1 If€¦ · Micro Prelim June 2016 – ANSWER KEYS...

Micro Prelim June 2016 ndash ANSWER KEYS

QUESTION 1 2 2

e e e e 1 if e e 10 e e 2 if e e 101 2 1 2 1 2 1 2(a) 15 10

u (e e ) u (e e ) 1 1 2 2 1 2shy

e1 2

e2 2shy

10 if e e 10 10 if e e 10 1 2 1 2

15 10shy2

e(b) The expression e e 1 is maximized at e 75 If e 25 then e 75 10 and thus 1 2

15 1 2 2

e 75 is indeed the optimal choice for Player 1 If e 25 then e 75 would take us into the 1 2 1

range where e1 e2 10 and in that range u1 is strictly decreasing in e1 so that the optimal choice is

75 if e2 25

e1 10 e2 Thus the best reply function of Player 1 is b1(e2 ) 10 e2 if 25 e2 10 Similar 0 if e 10 2

5 if e 5

1

reasoning leads to b ( ) 10 e if 5 e 10e2 1 1 1

0 if e 10 1

(c) See figure below The Nash equilibria are given by the intersection of the two curves that is all the

points (e e ) with e [575] and e 10 e 1 2 1 2 1

e2

10

5

25

Nash equilibria

0 5 e75 10 1

(d) Player 2rsquos choice as a function of Player 1rsquos choice is given by 2rsquos best reply function

5 if e1 5shy

( ) 10 e if 5 e 10b e Thus Player 1 chooses e1 to maximize 2 1 1 1

0 if e 10 1

e 2

e1 5 1 if e1 515

2 The solution is e1 5 and thus the subgame-perfect e

( ) u e b e 10 1 if 5 e 10U e ( ) 1 1 1 1 2 1 15

1shy

e

2

e 1 if e 10115

1

equilibrium is 5 e with actual choices being e e 5b ( ) 2 1 1 2

(e) (e1) A pure strategy for Player 1 specifies his initial effort level as well as his additional effort level as a function of both his initial effort level and Player 2rsquos effort level (e2) A pure strategy for Player 2 specifies her effort level as a function of Player 1rsquos initial effort level

(e3) A pure strategy for Player 1 is ldquomy initial choice is e1 4 and then (1) if my initial choice was 3

and Player 2 chooses 3 then I choose 1 4 while if Player 2 chooses 4 then I choose 1 3 and (2) if

my initial choice was 4 and Player 2 chooses 3 then I choose 1 3 while if Player 2 chooses 4 then I

choose 1 4 A pure strategy for Player 2 is ldquoif Player 1 chooses 3 then I choose 4 and if Player 1

chooses 4 then I choose 3rdquo

e 2

e e 1 1 if e e 101 1 2shy 1 2 1 15(f) The payoff function of Player 1 is 1(e11e2 )

e1 1 2

10 if e e 10 1 2 1 15

We use backward induction At the end of the game as shown in part (b) it is optimal for Player 1 to

target the sum of his efforts to e 75 if e 25 and to target the total effort to 1 1 2

e e 10 if e 25 Hence Player 1rsquos optimal strategy at the end of the game is 1 2 1 2

75 e1 if e2 25 and e1 75

(e e ) 10 e e if e 25 and e e 101 1 2 1 2 2 1 2

0 if e 25 and e 75 or e 25 and e e 10 2 1 2 1 2

In the middle of the game Player 2 realizes that if she chooses e2 25 Player 1rsquos additional effort

will bring the total effort up to at least 10 Thus player 2 should never choose an effort level higher

than 25 If Player 1 chose an effort e1 75 in period 1 then this is exactly what Player 2 should do

But if Player 1 chose a higher effort level then Player 2 need only exert enough effort to bring the total effort up to 10 Thus Player 2rsquos strategy in any subgame perfect equilibrium is described by

25 if e1 75

( ) 10 e if 75 e 10e2 e1shy 1 1

0 if e 10 1

In the initial period Player 1 should never choose an effort higher than 75 since choosing 75 is

enough to guarantee that the total effort will be 10 But any e1 [075] will cause Player 2 to

choose e 25 which Player 1 will follow by choosing 75 e Thus any e [075] can be2shy 1 1 1

chosen initially in a subgame perfect equilibrium and there are many subgame-perfect equilibria

QUESTION 2

(a) This is the standard case where Vickreyrsquos theorem applies b1 v1 is a weakly dominant strategy

(b) Recall that p v p No it is not We need to distinguish two cases Case 1 v p in this 1 1 m 1 m

case bidding v1 is not a dominant strategy if Player 2 bids pm then the outcome is (2 v1) and Player

1 would prefer bidding p1 since ndash by benevolence (2 p ) (2 v ) Case 2 v p also in 1 1 1 1 m

this case bidding v1 is not a dominant strategy if Player 2 bids pm then the outcome is (1 v1 pm )

and Player 1 would prefer bidding any pk with k lt m (thus letting Player 2 win) since ndash by

benevolence (2 p ) (2 p ) (1 v ) k 1 m 1 1

(c) Recall that p v p Again we need to distinguish two cases Case 1 v p in this case 1 1 m 1 2

bidding v1 is not a dominant strategy if Player 2 also bids v1 then the outcome is (1v1) and since

(1 v ) (2 p ) (2 p ) Player 1 would prefer bidding p2 inducing the outcome (2 p ) Case 1 1 1 2shy 2

2 v p also in this case bidding v1 is not a dominant strategy if Player 2 bids p4 (recall that m gt1 2

3) then the outcome is (2 1v ) and Player 1 would prefer bidding p3 inducing the outcome (2 3p )

which she prefers to (2 2p ) which in turn is at least as good as (2 1v ) (since 1v 2 p )

(d)The assumption is that both players are selfish and uncaring and B 1 23 v1 45 v2 Since

bidding onersquos own true value is a weakly dominant strategy (35) is a Nash equilibrium however it is not the only Nash equilibrium All of the following are Nash equilibria (15) (25) (35) (45) (14) (24) (34) (13) (23) (51) (52) and (53) Thus a total of 12 equilibria

(e)The assumption is that both players are selfish and benevolent and B 1 23 v1 45 v2 The

Nash equilibria are (13) (14) (15) and (51) A total of 4

(f) The assumption is that both players are selfish and spiteful and B 1 23 v1 45 v2 The Nash

equilibria are (45) (34) (23) A total of 3

(g)In the following figure the first element of the double label refers to Player 1 and the second to Player 2 thus for example UB means that Player 1 is U and Player 2 is B

1 1 3

4 4

BB UB US

2 1 5

6 6

true state

(h1) The extensive form is as follows (note that there is a common prior)

NATURE

15

21 1

21

US

2

2

1 1

BB UB 5

21

1p 1p1p

1p

1p

1p

1p

1p 1p

2p2p 2p

2p 2p2p

2p2p

r s

t u

x y

2p

2

2

(1 p ) (2 p ) (1 p ) (1 p ) (1 p ) (2 p ) (1 p ) (1 p ) (1 p ) (2 p ) (1 p ) (1 p )1 1 1 2 1 1 1 2 1 1 1 2

(h2) It is possible to find a weak sequential equilibrium The reasoning is as follows

At the right-most singleton node of Player 2 Player 2 is of type S and thus prefers (1 p2 ) to (1 p1) so

that p2 is the only sequentially rational choice At the other singleton node of Player 2 Player 2 is

again of type S and prefers (2 p1) to (1 p1) so that the only sequentially rational choice is p2 At both

nodes t and u Player 2 is of type B and thus prefers (1 p1) to (1 p2 ) so that p1 strictly dominates p2 at

information set tu thus p1 is the only sequentially rational choice there At both nodes x and y Player

2 is of type B and prefers (2 p1) to (1 p1) so that p2 strictly dominates p1 at information set xy

thus p2 is the only sequentially rational choice there Hence the game can be reduced to the following

NATURE

15

21 1

21

US

2

1 1

BB UB 5

21

1p 1p1p

2p2p 2p

r s

1(1 )p 1(1 )p

1(2 )p

(1 p2 )

(2 p1)shy (2 p1)

At the singleton Player 1 prefers (1 p1) to (2 p1) and thus p2 is the only sequentially rational choice

At nodes r and s Player 1 is of type U and thus (1) prefers (1 p1) to (1 p2 ) and (2) is indifferent between

(1 p2 ) to (2 p1) so that at information set rs p2 weakly dominates p1 and hence p2 must be the

only sequentially rational choice for Player 1 no matter what her von Neumann-Morgenstern function is

r s because by Bayesian updating she must have beliefs (thus attaching positive probability to node

14

34

r) Thus the following is a weak sequential equilibrium of the original game

Player 1 chooses p2 at her singleton information set and also at her information set rs

Player 2 chooses p2 at information set xy p1 at information set tu and p2 at both

singleton nodesshy

r s t u x y The beliefs are 1 3 1 5 and for any p [01]

4 4 6 6 p 1 p

y VV) 1 0 )~~ ( I 11 T) middot~ middotZ ~~ Mtfdeg6c1 WYl_l ( )

0 d l Y f ~ VJ ~ ( ~ 11 middot ri () ~o ~ -0~ VI

yi)j IV) ~G rv O1fa9 ~() Y ~ deg+f -S (ic l rlL

Jraquo g t ) c t )1 -=3 1-+ 3 2J t ~

J shy

2 C ( ~ Z lt=()v+ -shy 1 T + ~ ~ n() 1 deg1 ) 1-r-v 1 ~1111 ~ -R~

-z ~ s~t g t gt~ ~ - ---= 7-+ -- i ---shyI 3 52 5~_J

Clt10 amp e ltrv ~ ) --1-- -shy-gt 3 ri(J l shyl)nv Y) ~O i ~ 11141 -R~

middot5 pound4 ~~fJ ~tr ri~~ ~ MJ

lt3 (4~dp ~() v~ ~0~~WVl[YlfYj ~ 05 1 = 7 Y) I~ ~ p

~ ~p ~JO~ )i _ (l(S)to~ Y deg3 ~(Hi ~ M) ~ 1[ (9) ~V)

2 1 r ~gt~ ~1~0 --egt~~o ~n~ ( r ) 1

Csmiddotsmiddot11 ( 7 o 1 (s middot s-) (1 J

~ )~~-Vlth3 q ( --I-] bull - --Q 11 ~ -~bO ~0 ~NI(I

-tio i bull f ~ - v __rRo io-Vr([ middot 1 c i toiltaj

I = 1 = zl -)I 7f c 1 ) 7 -middotl-wJ_y 1 6 i middot1deg1fPO i i = d r =gt a ~ Wgt ~ 0- ~I vv TV

1 bull1fo olt1V) (r d ) ~7~ bull iigt4~GJ ( -o middotpound ~ J~ 17 () pound NOJlS30tJ

2 VL_~ 2- 1 ~ ~-+2 [-c-)2l-

4

L ~ ~~Ji)- 1 ~ --_ 4

W0-vJt lltll

dew DgtgtdoJo~~ a Cl r s)

dewW ( o~ ( degi_ (1~ c_~Jl))

-2 1 bull (c~- I~Y- -I~ rr~1~ ~-- - u~t ~ - t

oJ l~m f0 ~ - ~ - (_ ci- ) L uI middotc-o~ ~~(1 ~ook-- ltamp To~aJ lt3IMo ~cul e~ + 3 - ~ - l~J- w-J lta~~

z tS ~-e_M 0(l

rlt~JJtJ ~~u thv1N1

koto l Lugtt~ (l --l-_middotJ) J ~~ 1 oi ~ -L 0~ aI

~~e4 l~ djl- 1 (L-iJ ~ =-_ pound t ~ ct2 ~ d 5 2 ~~

oJ o _-4 -

2middot lti ~~ 1 + ~ r ~ ~ d middot shy

i Ir y - 2 Z- ~ ilt2shyLi

0~~ 0~tll - - shy -- - shy-

(11~ o RQ

-1 ~(

Question 4 - Answer Key

Ali the crocodile lives in the dark water of Putah Creek in the Arboretum just beside the UC Davis campus To the delight of the ducks there Ali is a vegetarian Ali earns income from being employed by UC Davis as a life guard at Putah Creek His task is to rescue students who fall into the water from time to time He is especially busy around prelim exams when more students than usual jump into water out of despair When being rescued by Ali usually their mood lights up quickly in anticipation of a cool selfie with a crocodile that can be shared on Facebook By the time they relapse upon realizing that the water damaged their cell phone beyond repair the authorities have arrived to help (with their cell phones) Anyway letrsquos not digress further While we certainly appreciate Alirsquos work our academic interest is focused on his consumption behavior Being a vegetarian his main diet consists of almonds They have the inconvenient feature of getting stuck between his 72 teeth So he is also a rather heavy consumer of toothpicks Finally there are his frequent gifts he purchases for Mathilda Mathilda She is the duck of the Arboretum who waddles most elegantly with her petite legs by delightfully moving well-shaped hips He is really in love with her at a platonic level of course so that nobody gets physically hurt

We write xa xt xg ge 0 for the amounts of his consumption of almonds toothpicks and gifts respectively We assume for simplicity that these goods are infinitesimally divisible Let x = (xa xt xg) We assume that his utility function is given by u(xa xt xg) = α θ 1minusαminusθx x x with α θ isin (0 1) and α + θ lt 1 His income or wealth is denoted by a t g

w gt 0 Finally we denote by pa pt pg gt 0 the prices of almonds toothpicks and gifts respectively and p = (pa pt pg)

a Use the Kuhn-Tucker approach to derive step-by-step the Walrasian demand funcshytion x(p w) Verify also second-order conditions

Our optimization problem is given by maxxisinR3 u(x) st p middot x le w We set up the Lagrange function

+

α θ 1minusαminusθL(xa xt xg λ) = xa xt xg minus λ(paxa + ptxt + pgxg minus w) (1)

The first-order conditions are

partL(xa xt xg λ) partxa

= αxαminus1 a x θ

t x 1minusαminusθ g minus λpa equiv 0 (2)

partL(xa xt xg λ) partxt

= θxα a x θminus1

t x 1minusαminusθ g minus λpt equiv 0 (3)

partL(xa xt xg λ) partxg

= (1 minus α minus θ)x α a x θ

t x minusαminusθ g minus λpg equiv 0 (4)

partL(xa xt xg λ) = minuspaxa minus ptxt minus pgxg + w equiv 0 (5)

partλ

1

The second-order conditions are

part2L(xa xt xg λ) partx2

a

= (α minus 1)αxαminus2 a x θ

t x 1minusαminusθ g lt 0 (6)


t = (θ minus 1)θxα

a x θminus2 t x 1minusαminusθ

g lt 0 (7)


= minus(α + θ)(1 minus α minus θ)x α a x θ

t x minusαminusθminus1 g lt 0 (8)

g

part2L(xa xt xg λ) = 0 (9)

partλ2

The second-order conditions show that first-order conditions are both necessary and sufficient

Next we divide equation (2) by (3) as well as (2) by (4) to obtain Marginal Rates of Substitutions

θ 1minusαminusθαxαminus1x x λpaa t gminus = minus (10)1minusαminusθθxα

a x θt minus1 xg λpt

θ 1minusαminusθαxαminus1x x λpaa t gminus θ = minus (11)

α minusαminusθ(1 minus α minus θ)xa xt xg λpg

which simplify to

αxt paminus = minus (12)θxa pt

αxg paminus = minus (13)(1 minus α minus θ)xa pg

Rearranging yields

α paxa = (14)

θ ptxt α paxa

= (15)(1 minus α minus θ) pgxg

These equations show already that expenditure ratios are equal to ratios of powers in the Cobb-Douglas utility function Solving for xt and xg as a function of xa we obtain

θpaxa xt = (16)

αpt (1 minus α minus θ)paxa

xg = (17)αpg

2

Substituting these two equations into (5) we get

θpaxa (1 minus α minus θ)paxa paxa + pt + pg = w (18)

αpt αpg

and simplifying yields

α θpaxa (1 minus α minus θ)paxa paxa + + = w (19)

α α α paxa

= w (20)α

We solve for xa(p w)

w xa(p w) = α (21)

pa

Plugging this solution into (16) and (17) allows us to solve for

w xt(p w) = θ (22)

pt w

xg(p w) = (1 minus α minus θ) (23) pg

We note that the demand for a good does not depend on the price of other goods which is a convenient but often unrealistic feature of Cobb-Douglas utility functions

b Verify that the demand function is homogenous of degree zero and satisfies Walrasrsquo Law

For homogeneity of degree zero we need to show xi(λw λp) = xi(w p) for all λ gt 0 i isin a t g But this is obvious eg xa(λw λp) = α λw = α w = xa(p w)λpa pa

(analogously for toothpicks and gifts)

To verify Walrasrsquo Law we need to verify

paxa(p w) + ptxt(p w) + pgxg(p w) = w

Substituting expressions for the demand functions into this equation yields

paxa(p w) + ptxt(p w) + pgxg(p w) = w (24) w w w

paα + ptθ + pg(1 minus α minus θ) = w (25) pa pt pg

αw + θw + (1 minus α minus θ)w = w (26)

w = w (27)

3

c To be honest we do not really know whether Ali has the Cobb-Douglas utility function stated above Would Ali want to differently substitute a marginal amount of almonds for some toothpicks when having a differentiable utility function differshyent from the one above and optimally demanding positive amounts of all goods Explain

As long as Ali consumes all goods in positive amounts he must be at an interior solution Thus no matter what differentiable utility function he has his marginal rates of substitution must equal to the price ratios So no he wouldnrsquot want to differently substitute a marginal amount of almonds for some toothpicks

d You would expect that the more almonds Ali eats the more they get stuck in his teeth and the more toothpicks he purchases In light of such considerations does it make sense to assume Ali has the utility function above

If the price of almonds goes up then it follows from xa(p w) that his demand for almonds goes down Yet it follows from xt(p w) that his demand for toothpicks stays constant in pa So either he has also some other use for toothpicks or the Cobb Douglas utility function doesnrsquot make much sense given the story

e Suppose the university would slightly raise Alirsquos income (Assuming at most small changes of income is a very realistic assumption at UC Davis) Can we learn from the Lagrange approach by how much his utility would change

This question can be answered by observing that the Lagrange multiplier λ represhysents the marginal value of relaxing the budget constraint We can use for instance equation (2) plug in the demands for almonds toothpicks and gifts and solve for λ and simplify ie αminus1 θ 1minusαminusθ

w w w α α θ (1 minus α minus θ) = λpa (28)

pa pt pg α α

θ θ

1 minus α minus θ 1minusαminusθ

= λ (29) pa pt pg

f Derive Alirsquos indirect utility function (denote it by v(p w)) Simplify

We have

v(p w) = u(x(p w)) (30)

= (xa(p w))α(xt(p w))

θ(xg(p w))1minusαminusθ (31) α θ 1minusαminusθ

w w w = α θ (1 minus α minus θ) (32)

pa pt pg α θ 1minusαminusθα θ 1 minus α minus θ

= w (33) pa pt pg

4

g Can we also answer e also using Alirsquos indirect utility function Briefly explain partv(pw)Yes We note that = λ

partw

h Verify that Ali satisfies Royrsquos identity with respect to almonds

Royrsquos identity with respect to almonds reads

partv(pw)

xa(p w) = minus partpa (34)partv(pw)

partw

We compute the partial derivative of the indirect utility function with respect to the price of almonds α θ 1minusαminusθ

partv(p w) α α θ 1 minus α minus θ = minus w (35)

partpa pa pa pt pg

Now α θ 1minusαminusθ partv(pw) minus α α θ 1minusαminusθ w

pa pa pt pg w minus partpa = minus α θ = α = xa(p w) (36)partv(pw) 1minusαminusθ

α θ 1minusαminusθ papartw

pa pt pg

i When Professor Schipper interviews Ali about how exactly he arrives at his optimal consumption bundle Ali expresses ignorance about maximizing utility subject to his budget constraint Instead he seems to minimize his expenditure on consumpshytion such that he reaches a certain level of utility A smart undergraduate student walks by and claims that this is clear evidence against the assumption of utility maximization in economics Since Professor Schipper hates Cobb-Douglas utility functions and boring calculations he sends the student to you so that you can show him how expenditure minimization works Again use the Kuhn-Tucker approach to derive the Hicksian demand function

The expenditure minimization problem is minxisinR3 p middot x subject to u(x) ge u where +

u gt u(0) First we set up the Lagrange function α θ 1minusαminusθL(xa xt xg λ) = paxa + ptxt + pgxg minus λ xa xt xg minus u (37)

We derive first-order conditions

partL(xa xt xg λ) α α θ 1minusαminusθ = pa minus λ xa xt xg equiv 0 (38)partxa xa

partL(xa xt xg λ) θ α θ 1minusαminusθ = pt minus λ x x x equiv 0 (39)a t gpartxt xt

partL(xa xt xg λ) 1 minus α minus θ α θ 1minusαminusθ = pg minus λ x x x equiv 0 (40)a t gpartxg xg

partL(xa xt xg λ) α θ 1minusαminusθ = minusxa xt xg + u equiv 0 (41)partλ

5

Taking the ratio of (38) and (39) as well as (38) and (40) respectively we obtain

pa αxt = (42)

pt θxa pa αxg

= (43) pg (1 minus α minus θ)xa

We solve for xt and xg respectively as a function of xa to obtain (16) and (17) respectively Substituting into the utility function and factoring out any variable and parameter pertaining to almonds

α θ 1minusαminusθαpaxa θpaxa (1 minus α minus θ)paxa

u(xa xt xg) = (44)αpa αpt αpg

α θ 1minusαminusθ paxa α θ 1 minus α minus θ

= (45)α pa pt pg

Let u = u(xa xt xg) and solve for xa as a function of u and p This is the Hicksian demand function for almonds So may want to denote it like in class by ha(p u) ie

α u(p u) = (46)ha α θ 1minusαminusθpa α θ 1minusαminusθ

pa pt pg

Plugging this into (16) and (17) respectively we solve for the Hicksian demands of toothpicks and gifts respectively

θ uht(p u) = (47)

α θ 1minusαminusθpt α θ 1minusαminusθ pa pt pg

1 minus α minus θ uhg(p u) = (48)

α θ 1minusαminusθpg α θ 1minusαminusθ pa pt pg

j Derive the expenditure function Show that the expenditure function is homogeshyneous of degree 1 in prices strictly increasing in u as well as nondecreasing and concave in the price of each good

α u θ ue(p u) = pa + ptα θ 1minusαminusθ α θ 1minusαminusθpa α θ 1minusαminusθ pt α θ 1minusαminusθ

pa pt pg pa pt pg

1 minus α minus θ u+pg (49)


u= (50)

α θ 1minusαminusθ α θ 1minusαminusθ pa pt pg

6

It is easy to see that it is homogenous of degree 1 in prices strictly increasing in u and nondecreasing in prices For concavity in prices differentiate the expenditure function wrt pa

parte(p u) α u= (51)

α θ 1minusαminusθpartpa pa α θ 1minusαminusθ pa pt pg

which is clearly positive since u gt u(0) (analogous wrt to pt and pg)

k Professor Schipper cannot compute the Hicksian demand using Kuhn-Tucker withshyout a coffee Unfortunately Ali has no coffee to offer Yet he could offer Professor Schipper his expenditure function Is there a way to quickly calculate the Hicksian demand from the expenditure function without coffee

Thatrsquos pretty obvious from the last line of j The right hand site is the Hicksian demand for almonds This is essentially an implication of Shephardrsquos lemma

Verify the (own price) Slutsky equation for the example of almonds

We need to verify

partha(p u) partxa(p w) partxa(p w) = + xa(p w) (52)

partpa partpa partw

for u = v(p w)

Focus first on the left-hand side Taking the partial derivative of Hicksian demand for almonds wrt to pa yields

partha(p u) partpa

= minusα p2 a

α pa

α θ pt

u θ

1minusαminusθ pg

1minusαminusθ + α2

p2 a

α pa

α θ pt

u θ

1minusαminusθ pg

1minusαminusθ(53)

Since u = v(p w) we substitute (33) and simplify in order to obtain

partha(p u) partpa

= minusα w p2 a

+ α2 w p2 a

(54)

Now turn to the right-hand side

partxa(p w) partpa

= minusα w p2 a

(55)

partxa(p w) partw

= α pa

(56)

Thus

partxa(p w) partw

xa(p w) = α2w p2 a

(57)

Adding (55) and (57) finishes the verification

7

m Which term in the (own price) Slutsky equation refers to the substitution effect As mentioned previously we donrsquot really know whether Ali has a Cobb-Douglas utility function Suppose Ali has a continuous utility function representing locally nonsatiated preferences and that his Hicksian demand function is indeed a function (rather than a correspondence) Would a change in prices still have qualitatively the same effect on Hicksian demand as when he has the above Cobb-Douglas utility function If yes provide a short proof If not argue why not

Yes it has the same effect The substitution effect is always negative This is known as the Compensated Law of Demand

Note that we did not assume differentiability We show that for any p p gt 0 the Hickisan demand function satisfies

(p minus p ) middot (h(p u) minus h(p u)) le 0 (58)

Since h(p u) minimizes expenditures at prices p over any other consumption bundle yielding a utility of at least u Thus we must have

p middot h(p u) le p middot h(p u) (59)


This is equivalent respectively to

p middot (h(p u) minus h(p u)) le 0 (61)

0 le p middot (h(p u) minus h(p u)) (62)

Adding these to inequalities and bringing all terms to the left-hand side yields (58)

n Because of the drought the price of almonds changes from p0 a to p1

a UC Davis is committed to keep Ali as well off as before the price change The newly hired Senior Vice Provost for Crocodile Welfare turns to Professor Schipper for advice on the exact amount to be deducted from the budget of the university and paid to Ali as compensation for the price change Unfortunately Professor Schipper is so immersed in exciting new research that he is extremely slow in answering his email Luckily the Senior Vice Provost for Crocodile Welfare spends most of his time in the hammocks on the quad where he meets you studying for the prelims Help him calculate the amount

Essentially the problem asks for the Compensating Variation It is a measure of change of welfare of a consumer due to a price change It measures in units of wealth the amount a social planner must pay to Ali after the price change to compensate him so as to make him just as well off as before the price change It is negative if and only if the transfer goes from the social planner to Ali

0 0 1 1 0 0 1 1Let p = (pa pt pg) p = (pa pt pg) u = v(p w) and u = v(p w) The Compensating Variation is defined by

0 1 1 1CV (p p w) = e(p u 1) minus e(p u 0) (63)

8

Since Ali spends his entire wealth in the expenditure minimizing consumption bunshydle ie e(p1 u1) = w we have

0 1 1CV (p p w) = w minus e(p u 0) (64)

We compute

0ue(p 1 u 0) = (65)

α θ 1minusαminusθ α θ 1minusαminusθ 1 pt pgpa

α θ 1minusαminusθ α θ 1minusαminusθ w 0p pt pga

(66)= α θ 1minusαminusθ

α θ 1minusαminusθ 1p pt pga

α1pa (67)= w p0 a

Thus

α1p0 1 aCV (p p w) = w minus w (68) p0 a

α1p= w 1 minus a (69)

p0 a

Thus the Compensating Variation is negative if and only if pa 1 gt p0

a

9

(e) (e1) A pure strategy for Player 1 specifies his initial effort level as well as his additional effort level as a function of both his initial effort level and Player 2rsquos effort level (e2) A pure strategy for Player 2 specifies her effort level as a function of Player 1rsquos initial effort level

(e3) A pure strategy for Player 1 is ldquomy initial choice is e1 4 and then (1) if my initial choice was 3

and Player 2 chooses 3 then I choose 1 4 while if Player 2 chooses 4 then I choose 1 3 and (2) if

my initial choice was 4 and Player 2 chooses 3 then I choose 1 3 while if Player 2 chooses 4 then I

choose 1 4 A pure strategy for Player 2 is ldquoif Player 1 chooses 3 then I choose 4 and if Player 1

chooses 4 then I choose 3rdquo

e 2

e e 1 1 if e e 101 1 2shy 1 2 1 15(f) The payoff function of Player 1 is 1(e11e2 )

e1 1 2

10 if e e 10 1 2 1 15

We use backward induction At the end of the game as shown in part (b) it is optimal for Player 1 to

target the sum of his efforts to e 75 if e 25 and to target the total effort to 1 1 2

e e 10 if e 25 Hence Player 1rsquos optimal strategy at the end of the game is 1 2 1 2

75 e1 if e2 25 and e1 75

(e e ) 10 e e if e 25 and e e 101 1 2 1 2 2 1 2

0 if e 25 and e 75 or e 25 and e e 10 2 1 2 1 2

In the middle of the game Player 2 realizes that if she chooses e2 25 Player 1rsquos additional effort

will bring the total effort up to at least 10 Thus player 2 should never choose an effort level higher

than 25 If Player 1 chose an effort e1 75 in period 1 then this is exactly what Player 2 should do

But if Player 1 chose a higher effort level then Player 2 need only exert enough effort to bring the total effort up to 10 Thus Player 2rsquos strategy in any subgame perfect equilibrium is described by

25 if e1 75

( ) 10 e if 75 e 10e2 e1shy 1 1

0 if e 10 1

In the initial period Player 1 should never choose an effort higher than 75 since choosing 75 is

enough to guarantee that the total effort will be 10 But any e1 [075] will cause Player 2 to

choose e 25 which Player 1 will follow by choosing 75 e Thus any e [075] can be2shy 1 1 1

chosen initially in a subgame perfect equilibrium and there are many subgame-perfect equilibria

QUESTION 2

(a) This is the standard case where Vickreyrsquos theorem applies b1 v1 is a weakly dominant strategy

(b) Recall that p v p No it is not We need to distinguish two cases Case 1 v p in this 1 1 m 1 m

case bidding v1 is not a dominant strategy if Player 2 bids pm then the outcome is (2 v1) and Player

1 would prefer bidding p1 since ndash by benevolence (2 p ) (2 v ) Case 2 v p also in 1 1 1 1 m

this case bidding v1 is not a dominant strategy if Player 2 bids pm then the outcome is (1 v1 pm )

and Player 1 would prefer bidding any pk with k lt m (thus letting Player 2 win) since ndash by

benevolence (2 p ) (2 p ) (1 v ) k 1 m 1 1

(c) Recall that p v p Again we need to distinguish two cases Case 1 v p in this case 1 1 m 1 2

bidding v1 is not a dominant strategy if Player 2 also bids v1 then the outcome is (1v1) and since

(1 v ) (2 p ) (2 p ) Player 1 would prefer bidding p2 inducing the outcome (2 p ) Case 1 1 1 2shy 2











1 1 3

4 4

BB UB US

2 1 5

6 6

true state


NATURE

15

21 1

21

US

2

2

1 1

BB UB 5

21

1p 1p1p

1p

1p

1p

1p

1p 1p

2p2p 2p

2p 2p2p

2p2p

r s

t u

x y

2p

2

2

(1 p ) (2 p ) (1 p ) (1 p ) (1 p ) (2 p ) (1 p ) (1 p ) (1 p ) (2 p ) (1 p ) (1 p )1 1 1 2 1 1 1 2 1 1 1 2









NATURE

15

21 1

21

US

2

1 1

BB UB 5

21

1p 1p1p

2p2p 2p

r s

1(1 )p 1(1 )p

1(2 )p

(1 p2 )

(2 p1)shy (2 p1)






14

34




singleton nodesshy


4 4 6 6 p 1 p




Jraquo g t ) c t )1 -=3 1-+ 3 2J t ~

J shy


-z ~ s~t g t gt~ ~ - ---= 7-+ -- i ---shyI 3 52 5~_J





2 1 r ~gt~ ~1~0 --egt~~o ~n~ ( r ) 1






2 VL_~ 2- 1 ~ ~-+2 [-c-)2l-

4

L ~ ~~Ji)- 1 ~ --_ 4

W0-vJt lltll



-2 1 bull (c~- I~Y- -I~ rr~1~ ~-- - u~t ~ - t


z tS ~-e_M 0(l

rlt~JJtJ ~~u thv1N1



oJ o _-4 -



0~~ 0~tll - - shy -- - shy-

(11~ o RQ

-1 ~(







+













partλ

1



a






g lt 0 (7)




g


partλ2







which simplify to



Rearranging yields

α paxa = (14)

θ ptxt α paxa



θpaxa xt = (16)


xg = (17)αpg

2



αpt αpg



α α α paxa

= w (20)α


w xa(p w) = α (21)

pa


w xt(p w) = θ (22)

pt w












w = w (27)

3








pa pt pg α α

θ θ


= λ (29) pa pt pg


We have

v(p w) = u(x(p w)) (30)





= w (33) pa pt pg

4


partw



partv(pw)


partw



partpa pa pa pt pg




pa pt pg









5


pa αxt = (42)

pt θxa pa αxg






= (45)α pa pt pg



pa pt pg


θ uht(p u) = (47)






pa pt pg pa pt pg



u= (50)


6








We need to verify


partpa partpa partw

for u = v(p w)


partha(p u) partpa

= minusα p2 a

α pa

α θ pt

u θ

1minusαminusθ pg


p2 a

α pa

α θ pt

u θ

1minusαminusθ pg

1minusαminusθ(53)


partha(p u) partpa

= minusα w p2 a

+ α2 w p2 a

(54)


partxa(p w) partpa

= minusα w p2 a

(55)

partxa(p w) partw

= α pa

(56)

Thus

partxa(p w) partw

xa(p w) = α2w p2 a

(57)


7

















8


0 1 1CV (p p w) = w minus e(p u 0) (64)

We compute

0ue(p 1 u 0) = (65)





α1pa (67)= w p0 a

Thus



p0 a


a

9











1 1 3

4 4

BB UB US

2 1 5

6 6

true state


NATURE

15

21 1

21

US

2

2

1 1

BB UB 5

21

1p 1p1p

1p

1p

1p

1p

1p 1p

2p2p 2p

2p 2p2p

2p2p

r s

t u

x y

2p

2

2

(1 p ) (2 p ) (1 p ) (1 p ) (1 p ) (2 p ) (1 p ) (1 p ) (1 p ) (2 p ) (1 p ) (1 p )1 1 1 2 1 1 1 2 1 1 1 2









NATURE

15

21 1

21

US

2

1 1

BB UB 5

21

1p 1p1p

2p2p 2p

r s

1(1 )p 1(1 )p

1(2 )p

(1 p2 )

(2 p1)shy (2 p1)






14

34




singleton nodesshy


4 4 6 6 p 1 p




Jraquo g t ) c t )1 -=3 1-+ 3 2J t ~

J shy


-z ~ s~t g t gt~ ~ - ---= 7-+ -- i ---shyI 3 52 5~_J





2 1 r ~gt~ ~1~0 --egt~~o ~n~ ( r ) 1






2 VL_~ 2- 1 ~ ~-+2 [-c-)2l-

4

L ~ ~~Ji)- 1 ~ --_ 4

W0-vJt lltll



-2 1 bull (c~- I~Y- -I~ rr~1~ ~-- - u~t ~ - t


z tS ~-e_M 0(l

rlt~JJtJ ~~u thv1N1



oJ o _-4 -



0~~ 0~tll - - shy -- - shy-

(11~ o RQ

-1 ~(







+













partλ

1



a






g lt 0 (7)




g


partλ2







which simplify to



Rearranging yields

α paxa = (14)

θ ptxt α paxa



θpaxa xt = (16)


xg = (17)αpg

2



αpt αpg



α α α paxa

= w (20)α


w xa(p w) = α (21)

pa


w xt(p w) = θ (22)

pt w












w = w (27)

3








pa pt pg α α

θ θ


= λ (29) pa pt pg


We have

v(p w) = u(x(p w)) (30)





= w (33) pa pt pg

4


partw



partv(pw)


partw



partpa pa pa pt pg




pa pt pg









5


pa αxt = (42)

pt θxa pa αxg






= (45)α pa pt pg



pa pt pg


θ uht(p u) = (47)






pa pt pg pa pt pg



u= (50)


6








We need to verify


partpa partpa partw

for u = v(p w)


partha(p u) partpa

= minusα p2 a

α pa

α θ pt

u θ

1minusαminusθ pg


p2 a

α pa

α θ pt

u θ

1minusαminusθ pg

1minusαminusθ(53)


partha(p u) partpa

= minusα w p2 a

+ α2 w p2 a

(54)


partxa(p w) partpa

= minusα w p2 a

(55)

partxa(p w) partw

= α pa

(56)

Thus

partxa(p w) partw

xa(p w) = α2w p2 a

(57)


7

















8


0 1 1CV (p p w) = w minus e(p u 0) (64)

We compute

0ue(p 1 u 0) = (65)





α1pa (67)= w p0 a

Thus



p0 a


a

9









NATURE

15

21 1

21

US

2

1 1

BB UB 5

21

1p 1p1p

2p2p 2p

r s

1(1 )p 1(1 )p

1(2 )p

(1 p2 )

(2 p1)shy (2 p1)






14

34




singleton nodesshy


4 4 6 6 p 1 p




Jraquo g t ) c t )1 -=3 1-+ 3 2J t ~

J shy


-z ~ s~t g t gt~ ~ - ---= 7-+ -- i ---shyI 3 52 5~_J





2 1 r ~gt~ ~1~0 --egt~~o ~n~ ( r ) 1






2 VL_~ 2- 1 ~ ~-+2 [-c-)2l-

4

L ~ ~~Ji)- 1 ~ --_ 4

W0-vJt lltll



-2 1 bull (c~- I~Y- -I~ rr~1~ ~-- - u~t ~ - t


z tS ~-e_M 0(l

rlt~JJtJ ~~u thv1N1



oJ o _-4 -



0~~ 0~tll - - shy -- - shy-

(11~ o RQ

-1 ~(







+













partλ

1



a






g lt 0 (7)




g


partλ2







which simplify to



Rearranging yields

α paxa = (14)

θ ptxt α paxa



θpaxa xt = (16)


xg = (17)αpg

2



αpt αpg



α α α paxa

= w (20)α


w xa(p w) = α (21)

pa


w xt(p w) = θ (22)

pt w












w = w (27)

3








pa pt pg α α

θ θ


= λ (29) pa pt pg


We have

v(p w) = u(x(p w)) (30)





= w (33) pa pt pg

4


partw



partv(pw)


partw



partpa pa pa pt pg




pa pt pg









5


pa αxt = (42)

pt θxa pa αxg






= (45)α pa pt pg



pa pt pg


θ uht(p u) = (47)






pa pt pg pa pt pg



u= (50)


6








We need to verify


partpa partpa partw

for u = v(p w)


partha(p u) partpa

= minusα p2 a

α pa

α θ pt

u θ

1minusαminusθ pg


p2 a

α pa

α θ pt

u θ

1minusαminusθ pg

1minusαminusθ(53)


partha(p u) partpa

= minusα w p2 a

+ α2 w p2 a

(54)


partxa(p w) partpa

= minusα w p2 a

(55)

partxa(p w) partw

= α pa

(56)

Thus

partxa(p w) partw

xa(p w) = α2w p2 a

(57)


7

















8


0 1 1CV (p p w) = w minus e(p u 0) (64)

We compute

0ue(p 1 u 0) = (65)





α1pa (67)= w p0 a

Thus



p0 a


a

9




Jraquo g t ) c t )1 -=3 1-+ 3 2J t ~

J shy


-z ~ s~t g t gt~ ~ - ---= 7-+ -- i ---shyI 3 52 5~_J





2 1 r ~gt~ ~1~0 --egt~~o ~n~ ( r ) 1






2 VL_~ 2- 1 ~ ~-+2 [-c-)2l-

4

L ~ ~~Ji)- 1 ~ --_ 4

W0-vJt lltll



-2 1 bull (c~- I~Y- -I~ rr~1~ ~-- - u~t ~ - t


z tS ~-e_M 0(l

rlt~JJtJ ~~u thv1N1



oJ o _-4 -



0~~ 0~tll - - shy -- - shy-

(11~ o RQ

-1 ~(







+













partλ

1



a






g lt 0 (7)




g


partλ2







which simplify to



Rearranging yields

α paxa = (14)

θ ptxt α paxa



θpaxa xt = (16)


xg = (17)αpg

2



αpt αpg



α α α paxa

= w (20)α


w xa(p w) = α (21)

pa


w xt(p w) = θ (22)

pt w












w = w (27)

3








pa pt pg α α

θ θ


= λ (29) pa pt pg


We have

v(p w) = u(x(p w)) (30)





= w (33) pa pt pg

4


partw



partv(pw)


partw



partpa pa pa pt pg




pa pt pg









5


pa αxt = (42)

pt θxa pa αxg






= (45)α pa pt pg



pa pt pg


θ uht(p u) = (47)






pa pt pg pa pt pg



u= (50)


6








We need to verify


partpa partpa partw

for u = v(p w)


partha(p u) partpa

= minusα p2 a

α pa

α θ pt

u θ

1minusαminusθ pg


p2 a

α pa

α θ pt

u θ

1minusαminusθ pg

1minusαminusθ(53)


partha(p u) partpa

= minusα w p2 a

+ α2 w p2 a

(54)


partxa(p w) partpa

= minusα w p2 a

(55)

partxa(p w) partw

= α pa

(56)

Thus

partxa(p w) partw

xa(p w) = α2w p2 a

(57)


7

















8


0 1 1CV (p p w) = w minus e(p u 0) (64)

We compute

0ue(p 1 u 0) = (65)





α1pa (67)= w p0 a

Thus



p0 a


a

9

2 VL_~ 2- 1 ~ ~-+2 [-c-)2l-

4

L ~ ~~Ji)- 1 ~ --_ 4

W0-vJt lltll



-2 1 bull (c~- I~Y- -I~ rr~1~ ~-- - u~t ~ - t


z tS ~-e_M 0(l

rlt~JJtJ ~~u thv1N1



oJ o _-4 -



0~~ 0~tll - - shy -- - shy-

(11~ o RQ

-1 ~(







+













partλ

1



a






g lt 0 (7)




g


partλ2







which simplify to



Rearranging yields

α paxa = (14)

θ ptxt α paxa



θpaxa xt = (16)


xg = (17)αpg

2



αpt αpg



α α α paxa

= w (20)α


w xa(p w) = α (21)

pa


w xt(p w) = θ (22)

pt w












w = w (27)

3








pa pt pg α α

θ θ


= λ (29) pa pt pg


We have

v(p w) = u(x(p w)) (30)





= w (33) pa pt pg

4


partw



partv(pw)


partw



partpa pa pa pt pg




pa pt pg









5


pa αxt = (42)

pt θxa pa αxg






= (45)α pa pt pg



pa pt pg


θ uht(p u) = (47)






pa pt pg pa pt pg



u= (50)


6








We need to verify


partpa partpa partw

for u = v(p w)


partha(p u) partpa

= minusα p2 a

α pa

α θ pt

u θ

1minusαminusθ pg


p2 a

α pa

α θ pt

u θ

1minusαminusθ pg

1minusαminusθ(53)


partha(p u) partpa

= minusα w p2 a

+ α2 w p2 a

(54)


partxa(p w) partpa

= minusα w p2 a

(55)

partxa(p w) partw

= α pa

(56)

Thus

partxa(p w) partw

xa(p w) = α2w p2 a

(57)


7

















8


0 1 1CV (p p w) = w minus e(p u 0) (64)

We compute

0ue(p 1 u 0) = (65)





α1pa (67)= w p0 a

Thus



p0 a


a

9



oJ o _-4 -



0~~ 0~tll - - shy -- - shy-

(11~ o RQ

-1 ~(







+













partλ

1



a






g lt 0 (7)




g


partλ2







which simplify to



Rearranging yields

α paxa = (14)

θ ptxt α paxa



θpaxa xt = (16)


xg = (17)αpg

2



αpt αpg



α α α paxa

= w (20)α


w xa(p w) = α (21)

pa


w xt(p w) = θ (22)

pt w












w = w (27)

3








pa pt pg α α

θ θ


= λ (29) pa pt pg


We have

v(p w) = u(x(p w)) (30)





= w (33) pa pt pg

4


partw



partv(pw)


partw



partpa pa pa pt pg




pa pt pg









5


pa αxt = (42)

pt θxa pa αxg






= (45)α pa pt pg



pa pt pg


θ uht(p u) = (47)






pa pt pg pa pt pg



u= (50)


6








We need to verify


partpa partpa partw

for u = v(p w)


partha(p u) partpa

= minusα p2 a

α pa

α θ pt

u θ

1minusαminusθ pg


p2 a

α pa

α θ pt

u θ

1minusαminusθ pg

1minusαminusθ(53)


partha(p u) partpa

= minusα w p2 a

+ α2 w p2 a

(54)


partxa(p w) partpa

= minusα w p2 a

(55)

partxa(p w) partw

= α pa

(56)

Thus

partxa(p w) partw

xa(p w) = α2w p2 a

(57)


7

















8


0 1 1CV (p p w) = w minus e(p u 0) (64)

We compute

0ue(p 1 u 0) = (65)





α1pa (67)= w p0 a

Thus



p0 a


a

9







+













partλ

1



a






g lt 0 (7)




g


partλ2







which simplify to



Rearranging yields

α paxa = (14)

θ ptxt α paxa



θpaxa xt = (16)


xg = (17)αpg

2



αpt αpg



α α α paxa

= w (20)α


w xa(p w) = α (21)

pa


w xt(p w) = θ (22)

pt w












w = w (27)

3








pa pt pg α α

θ θ


= λ (29) pa pt pg


We have

v(p w) = u(x(p w)) (30)





= w (33) pa pt pg

4


partw



partv(pw)


partw



partpa pa pa pt pg




pa pt pg









5


pa αxt = (42)

pt θxa pa αxg






= (45)α pa pt pg



pa pt pg


θ uht(p u) = (47)






pa pt pg pa pt pg



u= (50)


6








We need to verify


partpa partpa partw

for u = v(p w)


partha(p u) partpa

= minusα p2 a

α pa

α θ pt

u θ

1minusαminusθ pg


p2 a

α pa

α θ pt

u θ

1minusαminusθ pg

1minusαminusθ(53)


partha(p u) partpa

= minusα w p2 a

+ α2 w p2 a

(54)


partxa(p w) partpa

= minusα w p2 a

(55)

partxa(p w) partw

= α pa

(56)

Thus

partxa(p w) partw

xa(p w) = α2w p2 a

(57)


7

















8


0 1 1CV (p p w) = w minus e(p u 0) (64)

We compute

0ue(p 1 u 0) = (65)





α1pa (67)= w p0 a

Thus



p0 a


a

9



a






g lt 0 (7)




g


partλ2







which simplify to



Rearranging yields

α paxa = (14)

θ ptxt α paxa



θpaxa xt = (16)


xg = (17)αpg

2



αpt αpg



α α α paxa

= w (20)α


w xa(p w) = α (21)

pa


w xt(p w) = θ (22)

pt w












w = w (27)

3








pa pt pg α α

θ θ


= λ (29) pa pt pg


We have

v(p w) = u(x(p w)) (30)





= w (33) pa pt pg

4


partw



partv(pw)


partw



partpa pa pa pt pg




pa pt pg









5


pa αxt = (42)

pt θxa pa αxg






= (45)α pa pt pg



pa pt pg


θ uht(p u) = (47)






pa pt pg pa pt pg



u= (50)


6








We need to verify


partpa partpa partw

for u = v(p w)


partha(p u) partpa

= minusα p2 a

α pa

α θ pt

u θ

1minusαminusθ pg


p2 a

α pa

α θ pt

u θ

1minusαminusθ pg

1minusαminusθ(53)


partha(p u) partpa

= minusα w p2 a

+ α2 w p2 a

(54)


partxa(p w) partpa

= minusα w p2 a

(55)

partxa(p w) partw

= α pa

(56)

Thus

partxa(p w) partw

xa(p w) = α2w p2 a

(57)


7

















8


0 1 1CV (p p w) = w minus e(p u 0) (64)

We compute

0ue(p 1 u 0) = (65)





α1pa (67)= w p0 a

Thus



p0 a


a

9



αpt αpg



α α α paxa

= w (20)α


w xa(p w) = α (21)

pa


w xt(p w) = θ (22)

pt w












w = w (27)

3








pa pt pg α α

θ θ


= λ (29) pa pt pg


We have

v(p w) = u(x(p w)) (30)





= w (33) pa pt pg

4


partw



partv(pw)


partw



partpa pa pa pt pg




pa pt pg









5


pa αxt = (42)

pt θxa pa αxg






= (45)α pa pt pg



pa pt pg


θ uht(p u) = (47)






pa pt pg pa pt pg



u= (50)


6








We need to verify


partpa partpa partw

for u = v(p w)


partha(p u) partpa

= minusα p2 a

α pa

α θ pt

u θ

1minusαminusθ pg


p2 a

α pa

α θ pt

u θ

1minusαminusθ pg

1minusαminusθ(53)


partha(p u) partpa

= minusα w p2 a

+ α2 w p2 a

(54)


partxa(p w) partpa

= minusα w p2 a

(55)

partxa(p w) partw

= α pa

(56)

Thus

partxa(p w) partw

xa(p w) = α2w p2 a

(57)


7

















8


0 1 1CV (p p w) = w minus e(p u 0) (64)

We compute

0ue(p 1 u 0) = (65)





α1pa (67)= w p0 a

Thus



p0 a


a

9








pa pt pg α α

θ θ


= λ (29) pa pt pg


We have

v(p w) = u(x(p w)) (30)





= w (33) pa pt pg

4


partw



partv(pw)


partw



partpa pa pa pt pg




pa pt pg









5


pa αxt = (42)

pt θxa pa αxg






= (45)α pa pt pg



pa pt pg


θ uht(p u) = (47)






pa pt pg pa pt pg



u= (50)


6








We need to verify


partpa partpa partw

for u = v(p w)


partha(p u) partpa

= minusα p2 a

α pa

α θ pt

u θ

1minusαminusθ pg


p2 a

α pa

α θ pt

u θ

1minusαminusθ pg

1minusαminusθ(53)


partha(p u) partpa

= minusα w p2 a

+ α2 w p2 a

(54)


partxa(p w) partpa

= minusα w p2 a

(55)

partxa(p w) partw

= α pa

(56)

Thus

partxa(p w) partw

xa(p w) = α2w p2 a

(57)


7

















8


0 1 1CV (p p w) = w minus e(p u 0) (64)

We compute

0ue(p 1 u 0) = (65)





α1pa (67)= w p0 a

Thus



p0 a


a

9


partw



partv(pw)


partw



partpa pa pa pt pg




pa pt pg









5


pa αxt = (42)

pt θxa pa αxg






= (45)α pa pt pg



pa pt pg


θ uht(p u) = (47)






pa pt pg pa pt pg



u= (50)


6








We need to verify


partpa partpa partw

for u = v(p w)


partha(p u) partpa

= minusα p2 a

α pa

α θ pt

u θ

1minusαminusθ pg


p2 a

α pa

α θ pt

u θ

1minusαminusθ pg

1minusαminusθ(53)


partha(p u) partpa

= minusα w p2 a

+ α2 w p2 a

(54)


partxa(p w) partpa

= minusα w p2 a

(55)

partxa(p w) partw

= α pa

(56)

Thus

partxa(p w) partw

xa(p w) = α2w p2 a

(57)


7

















8


0 1 1CV (p p w) = w minus e(p u 0) (64)

We compute

0ue(p 1 u 0) = (65)





α1pa (67)= w p0 a

Thus



p0 a


a

9


pa αxt = (42)

pt θxa pa αxg






= (45)α pa pt pg



pa pt pg


θ uht(p u) = (47)






pa pt pg pa pt pg



u= (50)


6








We need to verify


partpa partpa partw

for u = v(p w)


partha(p u) partpa

= minusα p2 a

α pa

α θ pt

u θ

1minusαminusθ pg


p2 a

α pa

α θ pt

u θ

1minusαminusθ pg

1minusαminusθ(53)


partha(p u) partpa

= minusα w p2 a

+ α2 w p2 a

(54)


partxa(p w) partpa

= minusα w p2 a

(55)

partxa(p w) partw

= α pa

(56)

Thus

partxa(p w) partw

xa(p w) = α2w p2 a

(57)


7

















8


0 1 1CV (p p w) = w minus e(p u 0) (64)

We compute

0ue(p 1 u 0) = (65)





α1pa (67)= w p0 a

Thus



p0 a


a

9








We need to verify


partpa partpa partw

for u = v(p w)


partha(p u) partpa

= minusα p2 a

α pa

α θ pt

u θ

1minusαminusθ pg


p2 a

α pa

α θ pt

u θ

1minusαminusθ pg

1minusαminusθ(53)


partha(p u) partpa

= minusα w p2 a

+ α2 w p2 a

(54)


partxa(p w) partpa

= minusα w p2 a

(55)

partxa(p w) partw

= α pa

(56)

Thus

partxa(p w) partw

xa(p w) = α2w p2 a

(57)


7

















8


0 1 1CV (p p w) = w minus e(p u 0) (64)

We compute

0ue(p 1 u 0) = (65)





α1pa (67)= w p0 a

Thus



p0 a


a

9

















8


0 1 1CV (p p w) = w minus e(p u 0) (64)

We compute

0ue(p 1 u 0) = (65)





α1pa (67)= w p0 a

Thus



p0 a


a

9


0 1 1CV (p p w) = w minus e(p u 0) (64)

We compute

0ue(p 1 u 0) = (65)





α1pa (67)= w p0 a

Thus



p0 a


a

9

Micro Prelim June 2016 – ANSWER KEYS QUESTION 1 If€¦ · Micro Prelim June 2016 – ANSWER KEYS...

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