Economics Midterm Answer Key. Q1 èiè In this question we have a Marshallian demand function with arguments Cèp;mè =

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1 Economics Midterm Answer Key PART A Q èiè In this question we have a Marshallian demand function with arguments Cè;mè = Cè; w; w Lè. We can determine this function from the solution to max fc;lg minfc; Lg st C + wl = w L Clearly, in this roblem we have u = C = L. Substituting this into the budget constraint gives, C + wc = w L, and rearranging yields Cè; w; w Lè= w L + w èiiè Substituting the Marshallian demand into the utility function yields indirect utility, è è vè; w; wlè wl = min + w ; w L = w L + w + w èiiiè We know from the ëbasic four" that, in general, vè;mè=vè;eè;uèè = u. So in this case we have vè; w; eè; w; uèè = u, or eè; w; uè + w = u Rearranging, we derive the exenditure function, eè; w; uè =uè + wè èivè The deænition of the money metric indirect utility function is è; q;mè = eè;vèq;mèè. In this case we have èè ;w è; è ;w è;w Lè=eèè ;w è;vè ;w ;w Lèè = vè ;w ;w Lèè +w è= w L + w è +w è Q èiè A cost function, Cèy;wè, derived from some convex monotonic technology has the following four roerties

2 èè hd in w. èè Nondecreasing in w, nondecreasing in y. èè Concave inw. è4è Continuous in w. Now èè says Cèy;twè=tCèy;wè, so we have Cèy;tèw ;w èè = yëatw + btw + cètw è 5=8 ètw 4 è =4 ë=tëyèaw + bw + t,=8 cw 5=8 w =4 4 èë We clearly require c =0. Sowenowhave Cèy;w ;w è=yèaw + bw è. Checking = = yb 0; which imlies a 0, b 0. Now note that we have a function that is linear in w and w,so it is both concave and continuous, satisfying èè and è4è above. èiiè We have Cèy;w ;w è=yèæw + æw è, so from basic duality theory we know that the underlying technology is Leontief, of the form fèx ;x è = min è è x æ ; x æ Arguing backwards, and with an examle, let æ = and æ =. We know that x = x ==y at any cost minimising combination of factors. So to roduce one unit of y we need one unit of x and two units of x. The cost of one unit y is therefore w +w. So we get a linear cost function, Cèy;w ;w è=yèw +w è. It can be seen can see that this argument follows for any æ and æ. Q èiè The consumer will only accet F if her exected utility is higher than if she is not insured. That is, F must satisfy the following inequality, logèw, F è logèw, hè+è, è logèw + hè

3 Since the logarithmic function is monotonically increasingm, the largest such F will be the one satisfying the following equality, logèw, F è= logèw, hè+è, è logèw + hè èiiè From above we know F satisæes logèw, F è = logèw, h 0 W è+è, è logèw + h 0 W è = logèw è, h 0 èè+è, è logèw è + h 0 èè = logèw è+è, è logèw è+ logè, h 0 è+è, è logè + h 0 è = logèw è+ logè, h 0 è+è, è logè + h 0 è Alying the exonential function to both sides and rearranging yields F W =, è, h 0è è + h 0 è è,è Note that F is not exactly ëindeendent" of W. What is true is that F=W is indeendent of W. The second art of the question asks for the eæect of increasing h 0 on F=W. Taking the derivative of the above equation, we have W = è, h 0 è è,è è + h 0 è è,è, è, èè, h 0 è è + h 0 0 This derivative is ositive if è, h 0 è è,è è + h 0 è è,è, è, èè, h 0 è è + h 0 è, 0 Rearranging, we have, è!è! è, h0 è, è + h0 è, è, h 0 è è + h 0 è, 0; or,, h 0 +h 0 So whether F=W varies ositively with h 0 deends on the value of. For examle, if ==, then F=W increases with h 0. But for a small enough robability of loss we have F=W decreasing with h 0.

4 PART B Q4 èaè The consumer's roblem is max q uèw, L, q + qè+è, èuèw, qè The FOC is è, èu 0 èw, L, q + qè, è, èu 0 èw, qè =0 Rearranging, we have è,!, = u0 èw, L, q + qè èè u 0 èw, qè Since =, the LHS of èè is equal to. Therefore u 0 èw, L, q + qè =u 0 èw, qè. We know u 00 èæè é 0, so u 0 èæè is a strictly decreasing function, and therefore one-to-one. So W,L,q+q = W,q, imlying q æ = L. This, of course, is just the Full Insurance Princile. èbè We can rearrange the LHS of èè as,,. Since é,wehave é and, é. This imlies the LHS of èè is greater than, so, u0 èw, L, q + qè éu 0 èw, qè. Using the fact that u 0 èæè is a decreasing function, we have W, L, q + qéw, q; imlying qél So q ææ él= q æ. This makes sense í when insurance remiums are above actuarialy fair rates, risk averse consumers will less than fully insure against otential losses. Q5 èaè ènote that the roof below that aèè must be constant is a little subtle. By assuming indirect utility satisæes hd0 in è;mè, it is shown that indirect utility cannot satisfy nonincreasing in unless aèè is constant. It is not suæcient to simly state that hd0 fails í you must show that hd0 and nonincreasing in cannot simultaneously hold.è 4

5 In order for vè;mè=aèè to satisfy hd0 in è;mè, we must have aèè =aèè for any é0. Diæerentiating both sides of this equation wrt and evaluating at = gives i i = i i = =0 Since all rices are ositive, i é 0 for any good i, to maintain the above equality there must be at least one good j j é 0. But this would violate another of the indirect utility function's roerties nonincreasing in i 0 for all i. So the only ossibility is i = 0 for all i. That is, aèè is a constant function. èbè We have vè ; ;mè= æ æ m. Since vè ; ;mè is nonincreasing in è ; è, we æ æ = æ æ, æ m 0; æ æ = æ æ æ, m 0 This imlies æ 0, æ 0. Now, vè ; ;mè is hd0 in è ; ;mè. So imlying that æ+æ+ =,oræ + æ =,. è è æ è è æ èmè = æ+æ+ æ æ m = æ æ m; ècè We can ænd the direct utility function by solving uèx ;x è = min f ; g vè ; ;mè such that x + x = m. In Lagrangian form this roblem is The FOCs are min L = m,=,= + è x + x, mè f m,4=,= = m,=,5= = x Rearranging gives m,=,= = x ; m,=,= = x Adding together and using the budget constraint yields m,=,= = è x + x è=m; 5

6 so =,=,=. Substituting this back into the FOCs gives m,=,= =,=,= x ; m,=,= =,=,= x ; or Therefore m uèx ;x è=m x,= = m ; = m x x m,= = x,=,= x = x = = x = 4 = x = From duality theory we know that Cobb-Douglas indirect utility, vè ; ;mè=m,æ,è,æè corresonds to Cobb-Douglas direct utility of the form uèx ;x è=kx æ x,æ, so the arametric form of the direct utility function derived above is what we exect. Q6 èaè Since xè; mè =5, 4 is not a function of m, wehave hè; uè =5, 4. Hicksian demand is exactly Marshallian demand. èbè The integrability equations = 5, 4; with boundary condition èq; q;mè = m ècè è; q;mè = Z q è5, 4tèdt + èq; q;mè=5t, t æ ææ t= t=q + m =5, + m, 5q +q èdè Since è; q;mè =eè; vèq;mèè, a sensible guess for vèq;mè is vèq;mè=m, 5q +q We can verify by checking =,,5+4q =5, 4q = xèq;mè 6