本来计划做一份所有往年题的答案（已经全部做好了但希望打出一份电子版），但不幸考前🐑了，只完成此三题的誊抄，姑且留个纪念吧。

Problem 1 (stochastic dominance) 

Consider lotteries $L_1\equiv(p_1\circ x_1,p_2\circ x_2)$ and $L_2\equiv(1\circ \bar{x})$, where $\bar{x} > p_1x_1+p_2x_2$, and $\bar{x} < x_2$. Let $x$ denote the outcome of $L_1$ and $\widetilde{x}$ the outcome of $L_2$. ($L_2$ is a trivial lottery. So $\widetilde{x}$ is a random variable that equals $\bar{x}$ with probability one.) Let $z$ denote the outcome of lottery

where $n$ is some finite number greater than 1. Let $F_1$ and $F_2$ denote the cumulative distributions of $x+z$ and $\widetilde{x}+z$, respectively.

1. Suppose $n=2$. Use a diagram to illustrate $F_1$ and $F_2$.
2. Show that $\widetilde{x}+z$ cannot first-order stochastically dominates (f.o.s.d.) $x+z$. [Recall that for any two lotteries $L_1$ and $L_2$ with cdf $F_1$ and $F_2$, respectively, we say that $F_1$ first-order stochastically dominates $L_2$ if for any $x$, $F_1(x)\leq F_2(x)$.]

3. Let $z$ denote the outcome of a lottery

   $$ L_4 \equiv (q_1\circ z_1, \dots, q_i\circ z_i, \dots) $$

   that has a countably infinite number of outcomes. Show that in this case it is possible that $\widetilde{x}+z$ f.o.s.d. $x+z$. That is, come up with an example of $L_4$ such that $\widetilde{x}+z$ f.o.s.d. $x+z$.

Solution 

(a) Omitted.

(b) We need to prove that there is some $x$ such that $F_2(x) > F_1(x)$. WLOG, assume $z_n$ is the largest outcome of $L_3$, i.e., $x_n\geq x_i$ for every $i$. Since $p_1x_1+p_2x_2 < \bar{x} < x_2$, we must have $x_1 < \bar{x} < x_2$. Then $x_2+z_n$ is the rightmost point where $F_1$ puts a positive mass, which means $F_1(x_2+z_n) = 1 > F_1(\bar{x}+z_n)$. It is easy to see $F_2(\bar{x}+z_n) = 1$. So we have shown $F_2(\bar{x}+z_n) = 1 > F_1(\bar{x}+z_n)$.

(c) If there is some $x_m \geq x_i$ for every $i$, then we can use the same argument as in (b) to show that $F_2$ cannot f.o.s.d. $F_1$. Thus we may consider the case in which $\{z_n\}$ is an increasing sequence unbounded from above. We first show the following claim:

Claim. If $\{z_n\}$ is increasing, then $F_2(x)\leq F_1(x)$ if and only if $F_2(\bar{x}+z_n)\leq F_1(\bar{x}+z_n)$ for each $n\in\mathbb{N}$.

Proof. “Only if” is trivial. We show the “if” part. As both $F_1$ and $F_2$ are discrete distributions, it would suffice to show $F_2(x_j+z_n)\leq F(x_j+z_n)$ for $j=1,2$. For convenience, let $z_0=-\infty$. Then $x_1+z_n\in[\bar{x}+z_0,\bar{x}+z_n)$. There must exist some $m\leq n$ such that $x_1+z_n\in[\bar{x}+z_{m-1},\bar{x}+z_m)$. We have

Likewise, there is some $k\geq n$ such that $x_2+z_n\in[\bar{x}+z_k,\bar{x}+z_{k+1})$ and it follows that $F_2(x_2+z_n)$ is no greater than $F_2(x_2+z_n)$. This completes the proof. Q.E.D.

By the claim, we only need to construct $\{z_n\}$ and the corresponding pmf $\{q_n\}$ so that $F_2(\bar{x}+z_n)\leq F_1(\bar{x}+z_n)$ for each $n$.

First, it is easy to see $F_2(\bar{x}+z_n) = \sum_{j=1}^nq_j$. In addition, it is for sure that for $j\leq n$, $x_1+z_j<\bar{x}+z_n$ and so $F_1(\bar{x}+z_n)\geq p_1\sum_{j=1}^nq_j$. To make $F_1(\bar{x}+z_n)$ larger, we can let

for each $n$. Note that this amounts to $x_2-\bar{x}\leq z_{n}-z_{n-1}\leq \bar{x}-x_1$, which requires

Now we have

Thus, a sufficient condition for $F_2(\bar{x}+z_n)\leq F_1(\bar{x}+z_n)$ is

This would require $p_1>\frac12$ since otherwise $q_n$ would explode. We can then construct $\{q_n\}$ as a geometric sequence: $q_n = (1-q)q^{n-1}$ where $q=(1-p_1)/p_1$.

Therefore, if $\bar{x}\geq(x_1+x_2)/2$ and $p_1>\frac12$,

would lead $F_2$ to f.o.s.d. $F_1$.

Problem 2 (SPNE) 

Consider the following two-stage game between players 1 and 2. In stage 1, two players decide simultaneously whether to invest on a joint project. The cost of investment is $c>0$ for both players. No cost is incurred if a player chooses not to invest. The project generates a surplus of $(1+R)c$ if one player invests and of $(2+R)c$ if both players invest, where $0 < R < 1$. The game ends after stage 1 if no one invests and each player receives a payoff 0. If at least one player invests, the game enters stage 2, in which the players negotiate how to divide the realized surplus. Note that even a player did not invest, he is still entitled to participate in the surplus negotiation. The negotiation goes as follows: the players simultaneously announce their demands $d_1$ and $d_2$, where $d_i\in[0,1]$ refers to the share of the realized surplus. An agreement is reached if and only if the demands are compatible (i.e., $d_1+d_2\leq1$), in which case each player simply receives his own demand. If no agreement is reached (i.e., $d_1+d_2>1$), a player receives 0 if he did not invest in stage 1 and receives a compensation of $\delta c$ if he invested, where $0 < \delta < 1$. Each player’s payoff, denoted by $U_i$, is equal to the surplus or the compensation that he receives from the negotiation less his investment cost. For example, if player 1 is the only one who invested in stage 1 and an agreement of $(d_1,d_2)$ is reached in stage 2, then $U_1=d_1(1+R)c-c$ and $U_2=d_2(1+R)c$; if player 1 is the only one who invested in stage 1 and no agreement is reached in stage 2, then $U_1=\delta c-c$ and $U_2=0$.

1. Specify a subgame perfect equilibrium in which no one invests in stage 1.
2. Specify a subgame perfect equilibrium in which both players invest in stage 1.

Suppose that in the negotiation in stage 2, one of the two players in randomly selected, with probability $1/2$, to make a take-it-or-leave-it proposal in the form of $(d_1,d_2)$, with $d_i\geq0$ and $d_1+d_2=1$. The other player then decides whether to accept or reject this proposal. An agreement is reached if and only if the proposal accepted. Players’ payoffs are the same as specified in the previous parts. Assume that $R+\delta>1$.

3. Specify a subgame perfect equilibrium. Which player(s) will invest in the equilibrium?
4. Briefly discuss the efficiency of the equilibrium outcomes in the previous parts.

Solution 

(a) There are four (proper) subgames, and we need to find the stage NEs for each of them. Use $Y$ to denote invest and $N$ not invest.

If both players invest in stage 1, denoted $(Y,Y)$, then in stage 2, given player $i$’s demand $d_i$, player $j$’s payoff function:

Then player $j$’s best response:

If $(Y_i,N_j)$, then in stage 2, given player $i$’s demand $d_i$, player $j$’s payoff function:

Then player $j$’s best response:

If $(N_i,Y_j)$, then in stage 2, given player $i$’s demand $d_i$, player $j$’s payoff function:

Then player $j$’s best response:

If $(N,N)$, then in stage 2, given player $i$’s demand $d_i$, player $j$’s payoff function:

Then player $j$’s best response:

In an SPNE where $(N,N)$ is the stage-1 action profile, both players must have zero payoffs. Then, no incentive to deviate means that player $i$ should have non-positive payoff in the stage-2 subgame corresponding to $(Y_i,N_j)$. For the subgame corresponding to $(Y,Y)$, there is actually no such requirement: any subgame NE is fine; this is because a player’s deviation does not lead him to that subgame.

According to the above best response functions, we can easily find the stage-2 NEs. For example, $(d_1=d_2=1)$ in the subgame corresponding to $(Y,Y)$; $(d_i=\frac{1}{1+R},d_j=\frac{R}{1+R})$ corresponding to $(Y_i,N_j)$; $(d_1=d_2=1)$ corresponding to $(N,N)$. If player $i$ deviates from his current actions to $Y_i$ and other $d_i$’s, he would not be strictly better off. Thus, we indeed find an NE, and it is, as all stage-2 action profiles form NEs, an SPNE.

(b) First, let us find an NE in the subgame corresponding to $(Y,Y)$. A straightforward one is $(d_1=d_2=\frac12)$ in which both players have payoff $\frac12Rc$. Now, no incentive to deviate requires that player $i$ should have payoff no greater than $\frac12Rc$ in the stage-2 subgame corresponding to $(N_i,Y_j)$; easy to see $(d_i=0,d_j=1)$ is an NE that gives player $i$ zero payoff in this subgame. For the subgame corresponding to $(N,N)$, we can still use $(d_1=d_2=1)$.

(c) Analyze the four stage-2 subgames one by one. In the subgame following $(Y,Y)$: If player $i$ is selected as the proposer, then player $j$ would accept if $d_j(2+R)\geq\delta$, and so player $i$ would propose $d_i=1-\frac{\delta}{2+R}$. Player $i$ and $j$’s expected payoff are both

In the subgame following $(Y_i,N_j)$: If player $i$ is selected as the proposer, then player $j$ would always accept, so player $i$ proposes $d_i=1$; if player $j$ is selected as the proposer, then player $i$ would accept if $d_i(1+R)\geq\delta$, so player $j$ would propose $d_i = \frac{\delta}{1+R}$. Players $i$ and $j$’s expected payoffs are thus

In the subgame following $(N,N)$: Any proposal would be accepted and $U_i=U_j=0$.

The above analysis indicates an SPNE where one player invests while the other does not. Suppose $(Y_i,N_j)$. Clearly, player $i$ has no incentive to deviate to $N_i$ and player $j$ has no incentive to deviate to $Y_j$.

(d) The SPNE in which neither invests is inefficient. The other SPNEs in which at least one invests are efficient since the total (expected) payoff is $Rc$ and there is no room for any further improvement.

Problem 3 (mechanism design) 

Consider a bilateral trade setting. A seller owns an object, which a buyer wants to buy. The object quality $v$ is drawn from $[0,1]$ according to the distribution $F(v) = v^a$, where $a > 0$ is a parameter. The seller knows $v$, while the buyer does not. When the object quality is $v$, its value is $v$ to the seller and $bv$ to the buyer with $b > 1$. If a trade happens at the price $p$, then the seller gets $p−v$ and the buyer gets $bv − p$. When there is no trade, both players get 0.

For each $(a,b)$, present an incentive–compatible individually–rational mechanism that maximizes the (ex-ante) probability of trade. Denote the maximal probability by $H(a,b)$. Derive the function $H$.

Solution 

A mechanism consists of $q\colon [0,1]\to[0,1]$ and $p\colon [0,1]\to\mathbb{R}$. More specifically, $q(v)$ is the probability of the seller accepting the trade (allocation rule) and $p(v)$ stipulates the payment (transfer rule).

The seller of type $v$, if he tells the truth (reporting his type as is), has payoff

where we let $t(v):=q(v)p(v)$; it is the expected transfer. Now, instead of directly finding the optimal $q$ and $p$, we would determine $q$ and $t$.

Incentive compatibility implies

Thus, we have

This means $q(v)\geq q(v')$ whenever $v < v'$. In other words, $q(\cdot)$ is decreasing.

Further, IC must imply

By the envelope theorem (assuming $u_s$ is differentiable), $u_s'(v) = -q(v)\leq0$. Since $q(v)$ is decreasing, $u_s(v)$ is a convex, decreasing function. (In fact, the convexity can be proved directly from the formula above without imposing differentiability assumption.) Then, we have

Now, given IC, IR would imply

If an IC and IR mechanism maximizes the probability of trade, then it maximizes

subject to

Doing some algebra, the constraint reduces to

If $a(b-1)<1$, then any $q(v)\geq 0$ would not make trade happen with a positive probability. If $a(b-1)\geq 1$, then $q(v) = 1$ with $u_s(1) = 0$ would make trade happen for sure and so is the optimal allocation rule. Hence, we derive $H(a,b) = \mathbb{1}\{a(b-1)\geq1\}$.

最后修改于 2024-09-05