Economics 2102: Final Solutions

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1 Economics 10: Final Solutions 10 December, Auctions with Correlated Values: Solutions (a) There are initially eight constraints. Clearly, IR hh and IR hl are redundant. Ignoring IC ll and IC lh we obtain, t lh = 0 t ll = v l t hh = v h t hl = v h + v l One can check that IC ll and IC lh are indeed satisfied. Low types make no rents. High types makes rent U H = p hl(v h v l )/ p hl + p hh Observe that this mechanism is different from a second price auction, where we would have t hl = v l. (b) There are now four constraints. Again, IR h is redundant. If we impose t hh = v h / and t hl = v h, then IC l is also redundant. The IR l constraint is The IC h constraint is These constraints are satisfied if p ll ( v l t ll) + p lh ( t lh ) = 0 0 p hl ( v h t ll) + p hh ( t lh ) t ll 1 p ll p hh v l p lh v h p ll p hh p lh Intuitively, t ll is very low, while t lh is very high. Because valuations are positively correlated, this punishes a high type who pretends to be low worse than a low type who is honest. (c) In part (a) the lowest revenue is t ll = v l. In part (b), the lowest revenue is t ll < v l, using 1

2 the above equation. (d) We impose the constraint that t ll v l / From IR l, we have By IC h, the rent of the high type is p ll ( v l t ll) + p lh ( t lh ) 0 (p hh + p hl )U H p hl ( v h t ll) + p hh ( t lh ) = p hl ( v h t ll) p hhp ll ( v l t ll) p hl ( v h v l = 1 p hl(v h v l ) p lh ) p hhp ll p lh ( v l v l ) as required. In this last equation, the second line uses IR l, while the third uses the fact the equation is decreasing in t ll. [Source: Stanford GSB field exam]. Supermodularity: Solution (a) Let v(x 1, x, p 1 ) := u(x 1, x ) p 1 x 1 p x Clearly v(x 1, x, p 1 ) is supermodular in (x 1, x, p 1 ). Hence x (p 1 ) is decreasing, while x 1 (p 1, x ) is increasing in x and decreasing in p 1. The inequality follows. (b) In this case v(x 1, x, p 1 ) is supermodular in (x 1, x, p 1 ). Hence x (p 1 ) is increasing, while x 1 (x, p 1 ) is decreasing in both x and p 1. The inequality follows. (c) Part (a) clearly generalises: monotone comparative statics apply to multidimensional choices. Part (b) only generalises if we divide the goods to x 1 contains the complements and x contains the substitutes. One cannot consider arbitrary divisions. [Source: Milgrom and Roberts (AER, 1996)]

3 3. Theory of A Market Maker: Solution (a) By MLRP, the agent buys if s s H and sells if s s L. These cutoffs are defined by A = E[θ s = s H ] B = E[θ s = s L ] (b) The zero profit conditions imply A = λ(s H )E[θ s s H ] + (1 λ(s H ))E[θ] B = φ(s L )E[θ s s L ] + (1 φ(s L ))E[θ] where λ(s H ) := αe[1 s s H ] (1 α) + αe[1 s s H ] and φ(s L ) := αe[1 s s L] (1 α) + αe[1 s s L]. (c) Since E[θ s s H ] E[θ], so A E[θ]. Similarly, E[θ] B. To see some agent don t trade suppose, by contradiction, that s H = s L. By part (a), we have A = B. Yet is f(s θ) is nondegenerate, this is contradicted by (b). (d) The equilibrium price is determined by the fixed point: λ(s H )E[θ s s H ] + (1 λ(s H ))E[θ] = E[θ s = s H ] We can rewrite this as Y (α, x) = x where x = E[θ s = s H ]. Observe that Y (α, x) is increasing in α. Applying Milgrom and Roberts (AER, 1994, Theorem 1), the smallest fixed point is therefore increasing in α. 1 Hence an increase in α leads to a rise in s H and A. Similarly, an increase in α leads to a fall in s L and B. (e) Suppose α 1. I wish to show that s H s. Suppose, by contradiction, that s H s < s. 1 This can easily be seen by drawing a picture. Note: the smallest fixed point is appealing since is exhibits most trade. Observe the sequence is monotone and thus converges to something. 3

4 Since E[1 s s ] > 0, we have λ(s ) = 1 so that A = E[θ s = s ] and A = E[θ s s ] We must therefore have s = s H, yielding the required contradiction. That is, in the limit, we obtain the no trade theorem. [Source: Glosten and Milgrom (JFE, 1985)] 4. Team problem: Solution (a) The efficient equilibrium is implementable if β i (1 x) c i. Hence a necessary and sufficient condition is (1 x) c 1 + c. (b) Suppose (1 x) c 1 + c. Then the sharing rules β i = c i /(c 1 + c ) can implement the efficient outcome. (c) The efficient outcome can be implemented in dominant strategies if and only if β i (1 x) c i and β i x c i. Summing yields, 1 c 1 c x c 1 + c. (d) Suppose 1 c 1 c x c 1 + c. Then the efficient outcome can be implemented by setting β i = c i /(c 1 + c ). [Source: Inspired by Hosios (1990, ReStud)] 5. Debt Contracts: Solution (a) Suppose E pays D when she succeeds. She chooses p to maximise p[r(p) D]. This yields FOC R(p) D + pr (p) = 0 The investors break even, so Dp = 1 w. Substituting, p R (p) + pr(p) (1 w) = 0 (b) Suppose w = 1. Then p = 5/8. 4

5 Next, suppose w ( 7 3, 1). Then we obtain p = 1 16 [5 ± (5 3(1 w))] Of these roots, the larger maximises E s utility. One can also check that E s utility is maximised by investing all of his wealth. Finally, if w < 7 3, it is impossible to finance the project. (c) E s ex post wealth equals w for w < 7 3. It then jumps up above the 450 line, and is determined by an increasing concave function. The problem is that E does not act in the debt holder s interest and is too risk loving. This agency cost declines as w 1. [Source: Jeff Zwiebel] 5