<p>After posting a bad solution yesterday I believe I got a better one:</p>

<p>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
$$
p_1 \cdot \left( 1 - \frac{p_1 - \delta \cdot \frac{p_1}{2 - \delta}}{1 - \delta} \right) + \frac{1}{2} \cdot \frac{p_1}{2 - \delta} \cdot \left( \frac{p_1 - \delta \cdot \frac{p_1}{2 - \delta}}{1 - \delta} - \frac{p_1}{2 - \delta} \right),
$$</p>

<p>You have to maximize this w.r.t. $p_1$. With $\delta = 0.5$ I got
$$
p_1^* = \frac{9}{20}, \hskip 20pt \bar{v} = \frac{3}{5}, \hskip 20pt p_2^* = \frac{3}{10}.
$$</p>

CC BY-SA 4.0
2019-04-14T14:41:56.570Z
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?
Franklin Pezzuti Dyer
CC BY-SA 4.0
2019-04-14T15:26:22.647Z
@Frpzzd Quoting from https://franklin.dyer.me/post/111, "I might write something [...]  for expediency’s sake, even though this is not strictly mathematically tue [sic]." But yes, you are correct, if $\min(p_1;p_2) > v$ then of course both offers are rejected. This cannot happen in equilibrium though.
Giskard
CC BY-SA 4.0
2019-04-14T16:30:07.477Z
@Giskard Forgive me for the questions; this is my first time seeing a model like this. How can we ignore the possibility of $\nu < \min(p_1,p_2)$, when for any nonzero values of $p_1$ and $p_2$ there is a probability that $\nu$ is smaller? Why doesn’t the buyer take this into account when choosing a strategy?
Franklin Pezzuti Dyer
CC BY-SA 4.0
2019-04-14T18:03:40.827Z
@Frpzzd I am not sure what you mean. The buyer does act accordingly: no purchase if $v$ is lower than $\min(p_1;p_2)$. In my earlier comment I misspoke though, the seller does face the possibility that the buyer makes no purchase. This is shown in the calculations though. What I meant to write was that it is impossible that the seller sets a price at which no buyers make a purchase.
Giskard
CC BY-SA 4.0
2019-04-14T18:34:08.820Z
@Giskard Okay, I see what you mean now. Thanks!
Franklin Pezzuti Dyer
CC BY-SA 3.0
2015-05-03T14:48:33.217Z
I feel like this question can also be interpreted as a firm trying to screening  consumers of different valuations represented as the closed unit interval. The optimal pricing scheme is to set two prices so that customers of high valuations will pay at a higher price at first stage, and some of those of low valuations will pay at a lower price at second stage.
Metta World Peace
CC BY-SA 3.0
2015-05-03T15:07:39.430Z
You have to explain why the utilities are different in round 2. For the seller it could be simple discounting, but for the buyer? If the good were durable then types that buy the good would receive some benefits in both rounds.
Giskard
CC BY-SA 3.0
2015-05-03T15:36:00.030Z
I don't quite follow. Why the buyers can't discount the utility derived in second round? This can be interpreted as a two-period price skimming, right?
Metta World Peace
CC BY-SA 3.0
2015-05-03T17:30:53.753Z
Embarrassing but I have never heard of this model until now. You are correct, this describes the above game nicely.
Giskard
CC BY-SA 3.0
2015-05-02T21:50:11.523Z
2015-05-04T10:18:08.847Z
2015-05-04T10:18:08.847Z
Giskard
<p>I've found an interesting question looking at perfect-bayesian-equilibrium. I haven't seen a question where beliefs are not discrete.</p>

<p>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. </p>

<p>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 $p_1$. </p>

<p>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 $\delta = 0.5$. </p>

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

<p><strong>Find a Perfect Bayesian Equilibrium.</strong></p>

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

CC BY-SA 4.0
2021-05-15T00:07:45.223Z
HELP!!! https://economics.stackexchange.com/questions/43951/take-it-or-leave-it-offer
Luis
CC BY-SA 3.0
2015-05-02T21:51:35.410Z
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?
Giskard
CC BY-SA 3.0
2015-05-03T10:52:12.453Z
Of course, feel free!
Brian
<game-theory><academic-graduate><bayesian-game>
Take-it-or-leave-it PBE
Brian
# Take-it-or-leave-it PBE

(Asked by: username_0 on 2015-05-02)

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?

**Tags:** game-theory, academic-graduate, bayesian-game

> **Comments:**
> **username_2**: 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?  
> **username_0**: Of course, feel free!  
> **username_1**: HELP!!! https://economics.stackexchange.com/questions/43951/take-it-or-leave-it-offer  

---

## 1 answers

### Answer by username_2 on 2015-05-02. Score: 6 | Accepted answer

After posting a bad solution yesterday I believe I got a better one:

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
$$
p\_1^\* = \frac{9}{20}, \hskip 20pt \bar{v} = \frac{3}{5}, \hskip 20pt p\_2^\* = \frac{3}{10}.
$$

> **Comments:**
> **username_4**: I feel like this question can also be interpreted as a firm trying to screening  consumers of different valuations represented as the closed unit interval. The optimal pricing scheme is to set two prices so that customers of high valuations will pay at a higher price at first stage, and some of those of low valuations will pay at a lower price at second stage.  
> **username_2**: You have to explain why the utilities are different in round 2. For the seller it could be simple discounting, but for the buyer? If the good were durable then types that buy the good would receive some benefits in both rounds.  
> **username_4**: I don't quite follow. Why the buyers can't discount the utility derived in second round? This can be interpreted as a two-period price skimming, right?  
> **username_2**: Embarrassing but I have never heard of this model until now. You are correct, this describes the above game nicely.  
> **username_3**: 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?  
> **username_2**: @Frpzzd Quoting from https://franklin.dyer.me/post/111, "I might write something [...]  for expediency’s sake, even though this is not strictly mathematically tue [sic]." But yes, you are correct, if $\min(p_1;p_2) > v$ then of course both offers are rejected. This cannot happen in equilibrium though.  
> **username_3**: @username_2 Forgive me for the questions; this is my first time seeing a model like this. How can we ignore the possibility of $\nu < \min(p_1,p_2)$, when for any nonzero values of $p_1$ and $p_2$ there is a probability that $\nu$ is smaller? Why doesn’t the buyer take this into account when choosing a strategy?  
> **username_2**: @Frpzzd I am not sure what you mean. The buyer does act accordingly: no purchase if $v$ is lower than $\min(p_1;p_2)$. In my earlier comment I misspoke though, the seller does face the possibility that the buyer makes no purchase. This is shown in the calculations though. What I meant to write was that it is impossible that the seller sets a price at which no buyers make a purchase.  
> **username_3**: @username_2 Okay, I see what you mean now. Thanks!  

---

