Answers to Problem Set #4

Transcription

1 Economics 201b 1. As a preliminary, note b(x, θ) = x 0 p(t, θ)dt = θ( kx 1 2 x2). Of course, b/ x = p(x, θ). (a) Expression (12) can be written 0=p ( x(θ),θ ) ( c + 1 F (θ) p ( x(θ),θ ) ) θ = θ ( k x(θ) ) ( c + 1 F (θ) ( ) ) k x(θ) Solving for x( ) yields: c ˆx(θ) =k θ 1 F (θ). (b) Observe, from the last expression, that ˆx( ) is monotonic only if θ 1 F (θ) is non-decreasing in θ (at least for θ such that ˆx(θ) > 0). This would be not be feasible if the derivative of the Mills ratio was greater than one. (c) Observe this means F (θ) = θ θ 0 (so 1 F (θ) = θ 1 θ 1 )and =. θ 1 θ 0 θ 1 θ 0 θ 1 θ 0 Hence, c ˆx(θ) =k 2θ θ 1 for θ ( θ 1 + c/k ) /2 θ; and T (θ) =b (ˆx(θ),θ ) θ ) b (ˆx(t),t dt (from (9)) θ 0 θ = θ ( kˆx(θ) 1 θ 2 ˆx(θ)2) ( 1 kˆx(t) θ 2 ˆx(t)2) dt = k2 (2 θ θ 1 ) θ(2θ θ 1 ) 2 + c 2( ) θ 1 θ 2θ(2 θ θ), 2(θ 1 2 θ)(2θ θ 1 ) 2 where θ = max{θ 0, θ}. Copyright c 2005 Benjamin E. Hermalin. All rights reserved.

2 2. There are a number of ways to answer this question. This is just one way. Let the seller s profit be t cx (i.e., s = t and w(x, θ) = cx). The assumption of a constant marginal cost, c, because it makes it feasible to work with a single customer without worrying about the overall cost of total output. Let the customer s utility be u(x, θ) t, where θ [θ L,θ H ] Θis type. Further assume that u(,θ) is strictly concave for all θ; that u(0,θ)/ x > c for all θ (θ L,θ H ]; and there exists an x F (θ) for all θ Θ such that u ( x F (θ),θ ) / x = c. Let θ F, where F is differentiable with density f and > 0 θ Θ. Assume u(0,θ) = 0 for all θ Θ. Note the reservation utility is, thus, 0. Assume the Spence-Mirrlees condition: 2 u θ x > 0. The above assumptions are sufficient for the standard framework. If you further wished to guarantee that the optimal allocation, x ( ), could be found by point-wise maximization of Σ(x, θ) u(x, θ) cx 1 F (θ) u(x, θ), θ then you need to further assume: For all x, u/ θ 0; The function Σ(,θ) is strictly quasi-concave for all θ Θ; and The cross partial 2 Σ θ x is non-negative for all x. 3. (a) This is without loss of generality because if f n = 0, then it is as if the nth type doesn t exist (i.e., could be ignored). (b) This represents the information rent that a type n + 1 agent would get from mimicking a type n agent. (c) The Spence-Mirrlees (sm) condition in this context is that φ/ x be monotonic in n. The screening condition certainly implies that: R n ( ) strictly increasing means that 0 <R n(x) = φ(x, n) x φ(x, n +1) x hence, φ/ x is decreasing in n. So sc sm. Clearly, sm implies that R n ( ) is increasing. Given the assumption that φ(0,n)=0for all n, it follows that sm also implies that R n (x) > 0 for all x>0. Hence, the only property of sc not implied by sm is that R n ( ) be convex. ; 2

3 (d) Following the hints: i. As usual, it is easier to work with utility than payments. Set u n = s n φ(x n,n). Observe, then, s n = u n + φ(x n,n). Using this substitution, revealed preference tells us for any m and n, { s m }} { u n u m + φ(x m,m) φ(x m,n) (1) u m u n + φ(x n,n) }{{} s n φ(x n,m) (2) Assume n>m. Rearrange (1) and (2) as φ(x n,m) φ(x n,n) u n u m φ(x m,m) φ(x m,n). (3) Now, we can proceed in one of two ways. First, using part (c) specifically, that sc sm we could use the fact that x ( φ(z,m) φ(x, m) φ(x, n) = φ(z,n) ) dz 0 z z and the fact that φ(z,m)/ z > φ(z,n) n (the Spence-Mirrlees condition) to conclude that the left-most term of (3) can exceed the right-most term of (3) only if x n x m. Alternatively, observe that φ(x, m) φ(x, n) = hence (3) can be rewritten as R j (x n ) u n u m R j (x); R j (x m ). (4) The screening-condition requires that R j ( ) be strictly increasing for all j; therefore, (4) can hold only if x n x m. Note, as a bonus, we ve also established, given that R j (x) > 0forx>0, that u n u m. In other words, equilibrium utilities must be non-deceasing in type. ii. We need to establish that the adjacent ic constraints imply (1) and (2) for all n and m given that x j is non-decreasing in j. Fix an n and consider any m<n 1 (obviously (1) and (2) hold if m = n 1 because those are just the adjacent ic constraints). Observe u n u m = (u j+1 u j ) R j (x j ) R j (x m ) = φ(x m,m) φ(x m,n), (5) 3

4 where the first inequality follows from the adjacent ic constraints and the second inequality follows because x is non-decreasing in type, so x m x j if j m. Observe that (5) can be rearranged to yield (1). Similarly, we have u n u m = (u j+1 u j ) R j (x j+1 ) R j (x n ) = φ(x n,m) φ(x n,n), where the first inequality follows from the adjacent ic constraints and the second follows because x is non-deceasing in type. Expression (6) can be rearranged to yield (2). If m>n+ 1, then just reverse the definitions of m and n in the above. Hence, we ve established that a non-decreasing allocation profile and the adjacent ic constraints are sufficient to have all the ic constraints hold. iii. Fix an n. By assumption x n+1 x n. By construction, u n = j=1 R j(x j ) (where we utilize the convention that 0 j=1 is zero). Consider one adjacent ic constraint: which is equivalent to u n u n+1 R n (x n+1 ) 0 R n (x n ) R n (x n+1 ), which holds because x is non-decreasing in type and R n ( ) isan increasing function. Now consider the other: which is equivalent to u n+1 u n + R n (x n ) R n (x n ) R n (x n ), which holds trivially. iv. The program (P) given the constraints (IR) and (IC) is obviously what the principal wishes to maximize. We ve established that the allocation profile must be non-decreasing, so we re done if we can show that N N f n u n = (1 F n )R n (x n ). (7) n=1 n=1 Simple algebra reveals that (7) follows if u is such that u 1 =0andu n = R j (x j )ifn>1. (8) j=1 (6) 4

5 But clearly, the principal wants to make the u s as small as possible. Clearly, u 1 = 0 is the smallest possible value of u 1. And, as just shown, the smallest ic values of the u n, n>1, are those defined by (8). This completes the proof. 5