Take-it-or-leave-it PBE

I've found an interesting question looking at perfect-bayesian-equilibrium. I haven't seen a question where beliefs are not discrete.

There is a single potential buyer of an object which has zero value to the seller. This buyer’s valuation v is uniformly distributed on [0, 1] and is private information. The seller names a price $p_1$ which the buyer accepts or rejects.

If he accepts, the object is traded at the agreed price and the buyer’s payoff is $v − p_1$ and the seller’s is $p_1$.

If he rejects then the seller makes another price offer, p2. If the buyer accepts this, his payoff is $\delta_(v − p_2)$ and the seller’s is $\delta p_2$, where $\delta = 0.5$.

If he rejects, both players get zero (there are no further ￼￼￼offers).

Find a Perfect Bayesian Equilibrium.

My usual approach is to fix beliefs, but I don't quite know how to do this with continuous beliefs. Any advice?

• Sorry, I could not think of an easy way to give partial advice. This is a nice exercise. Would you (or the creator) mind if I used it in class? May 2 '15 at 21:51
• Of course, feel free! May 3 '15 at 10:52
• – Luis
May 15 '21 at 0:07

The strategy of the buyer consists of two functions, $(f_1(v,p_1),f_2(v,p_1,p_2))$ where both functions map to $\left\{A,R\right\}$ (where $A$ stands for Accept, $R$ for Reject). The strategy of the seller is $(p_1,p_2(f_1(v,p_1)))$. You get the solution via backward induction. In PBE $f_2(v,p_1,p_2)$ maps to $A$ if and only if $v \geq p_2$. (There is inconsequential leeway at equality.) In PBE the seller believes that there is a set $H$ of types for which the buyer refused her offer $p_1$. Then $$p_2^* = \arg\max_{p_2} p_2 \cdot Prob(f_2(v,p_1,p_2) = A | f_1(v,p_1) = R).$$ The buyer will accept offer $p_1$ if and only if $$v - p_1 \geq \delta \cdot (v - p_2).$$ From this you get $$v \cdot (1 - \delta) \geq p_1 - \delta \cdot p_2.$$ The left hand side of this equation is increasing in $v$, so types with high valuation will Accept. This means that in PBE the set $H$ is such that $$H = [0, \bar{v}).$$ From this we get the optimal $p_2$ given $\bar{v}$: $$p_2^* = \arg\max_{p_2} p_2 \cdot Prob(v \geq p_2 | v \in [0, \bar{v})) = \frac{\bar{v}}{2}.$$ In PBE $\bar{v}$ is a function of $p_1$: $$\bar{v} \cdot (1 - \delta) = p_1 - \delta \cdot \frac{\bar{v}}{2},$$ so $$\bar{v} = \frac{p_1}{1 - \frac{\delta}{2}}.$$ We have determined all the PBE strategies but $p_1$. The expected payoff of the seller is $$p_1 \cdot \left( 1 - \frac{p_1 - \delta \cdot p_2(\bar{v}(p_1))}{1 - \delta} \right) + \frac{1}{2} \cdot p_2(\bar{v}(p_1)) \cdot \left( \frac{p_1 - \delta \cdot p_2(\bar{v}(p_1))}{1 - \delta} - p_2(\bar{v}(p_1)) \right),$$ where $$p_2(\bar{v}(p_1)) = \frac{\bar{v}(p_1)}{2} = \frac{\frac{p_1}{1 - \frac{\delta}{2}}}{2} = \frac{p_1}{2 - \delta}.$$ Substituting this we get $$p_1 \cdot \left( 1 - \frac{p_1 - \delta \cdot \frac{p_1}{2 - \delta}}{1 - \delta} \right) + \frac{1}{2} \cdot \frac{p_1}{2 - \delta} \cdot \left( \frac{p_1 - \delta \cdot \frac{p_1}{2 - \delta}}{1 - \delta} - \frac{p_1}{2 - \delta} \right),$$
You have to maximize this w.r.t. $p_1$. With $\delta = 0.5$ I got $$p_1^* = \frac{9}{20}, \hskip 20pt \bar{v} = \frac{3}{5}, \hskip 20pt p_2^* = \frac{3}{10}.$$
• You said that the buyer will accept $p_1$ if and only if $$v-p_1\ge \delta(v-p_2)$$ but won’t the buyer reject if both $p_1$ and $p_2$ are greater than $v$, regardless of whether the above inequality is satisfied? Apr 14 '19 at 14:41