<p>Let <span class="math-container">$(q(\theta), p(\theta))$</span> be the mechanism. Let us show that the optimal mechanism will define a cutoff <span class="math-container">$\theta^\ast$</span> such that <span class="math-container">$q(\theta) = 1$</span> and <span class="math-container">$p(\theta) = \theta^\ast$</span> if and only if <span class="math-container">$\theta \ge \theta^\ast$</span>.</p>
<p>The IC constraint will require that for all <span class="math-container">$\theta$</span> and <span class="math-container">$\theta' \in [0,1]$</span>:</p>
<p><span class="math-container">$$
\theta q(\theta) - p(\theta) \ge \theta q(\theta') - p(\theta'). \tag{IC}
$$</span></p>
<p>The PC (outside option normalized to <span class="math-container">$0$</span>) tells us that:
<span class="math-container">$$
\theta q(\theta) - p(\theta) \ge 0. \tag{PC}
$$</span></p>
<p>Now assume that the mechanism specifies to sell to type <span class="math-container">$\theta$</span>, so <span class="math-container">$q(\theta) = 1$</span></p>
<p>This requires that:
<span class="math-container">$$
\theta - p(\theta) \ge 0 \to p(\theta) \le \theta.
$$</span></p>
<p>Now, take any type <span class="math-container">$\theta' &gt; \theta$</span>. Then the (IC) constraint specifies that:
<span class="math-container">$$
\theta' q(\theta') - p(\theta') \ge \theta'  - p(\theta)\ge \theta' - \theta &gt; 0.
$$</span>
If <span class="math-container">$p(\theta') \ge 0$</span> this requires that <span class="math-container">$q(\theta') = 1$</span>.</p>
<p>Then using the (IC) constraint again, we have:
<span class="math-container">$$
\theta'- p(\theta') \ge \theta' - p(\theta),
$$</span>
which implies that <span class="math-container">$p(\theta') \le p(\theta)$</span>.</p>
<p>If <span class="math-container">$q(\theta) = 0$</span>, then the (PC) constraint tells us that <span class="math-container">$p(\theta) = 0$</span>.</p>
<p>Let <span class="math-container">$\theta^\ast$</span> be the lowest value of <span class="math-container">$\theta$</span> for which <span class="math-container">$q(\theta) = 1$</span>. Summarizing, we have the following:</p>
<ul>
<li>If <span class="math-container">$\theta \ge \theta^\ast$</span>, then <span class="math-container">$q(\theta) = 1$</span> and <span class="math-container">$p(\theta) \le p(\theta^\ast)$</span>.</li>
<li>If <span class="math-container">$\theta &lt; \theta^\ast$</span> then <span class="math-container">$q(\theta) = 0$</span> and <span class="math-container">$p(\theta) = 0$</span>.</li>
<li>For all <span class="math-container">$\theta$</span> with <span class="math-container">$q(\theta) = 1$</span>, <span class="math-container">$p(\theta) \le \theta$</span></li>
</ul>
<p>Note that, as the firm tries to maximize profits, it will decide to put <span class="math-container">$p(\theta) = p(\theta^\ast) = \theta^\ast$</span> for all <span class="math-container">$\theta \ge \theta^\ast$</span>.</p>
<p>This implies the following rule.</p>
<ul>
<li>For all <span class="math-container">$\theta \ge \theta^\ast$</span>, set <span class="math-container">$q(\theta) = 1$</span> and <span class="math-container">$p(\theta) = \theta^\ast$</span>.</li>
<li>For all <span class="math-container">$\theta &lt; \theta^\ast$</span>, set <span class="math-container">$q(\theta) = 0$</span> and <span class="math-container">$p(\theta) = 0$</span>.</li>
</ul>
<p>The profits will then be equal to:
<span class="math-container">$$
\theta^\ast (1 - \theta^\ast) - \frac{1}{2}(1- \theta^\ast)^2.
$$</span>
Taking first order conditions gives:
<span class="math-container">$$
1-2\theta^\ast + (1- \theta^\ast) = 2 - 3\theta^\ast = 0
$$</span>
The second order condition is negative so the objective function is concave.
This gives a value of, <span class="math-container">$\theta^\ast = 2/3$</span>.</p>

CC BY-SA 4.0
2024-02-01T15:08:10.550Z
Thank you! I really appreciate your help.
Nazarene
CC BY-SA 4.0
2024-01-31T14:53:21.143Z
2024-01-31T14:53:21.143Z
tdm
<hr />
<p>Hi, everyone! I'd appreciate some help with this problem, please.</p>
<p>Suppose that a seller puts a one-unit good on the market, and a buyer comes around with an unknown valuation <span class="math-container">$θ$</span>. This buyer has a quasi-linear utility function <span class="math-container">$θq-p,$</span> and costs are quadratic with <span class="math-container">$c(q) = \dfrac{1}{2}q^2.$</span></p>
<p>Now suppose that <span class="math-container">$θ$</span> is uniformly distributed on <span class="math-container">$[0,1].$</span></p>
<p>Find the optimal direct mechanism <span class="math-container">$(p(·),q(·)).$</span></p>
<p>At the moment, I understand that we would need to maximize the seller's expected revenue, though I am not sure to navigate this problem. All help is appreciated.</p>

CC BY-SA 4.0
2023-12-09T05:16:30.487Z
It seems that no trade may be a feasible outcome.
Nazarene
<game-theory><mechanism-design>
(Mechanism Design | Mailath Exercise 10.5.3) A Question
Nazarene
# (Mechanism Design | Mailath Exercise 10.5.3) A Question

(Asked by: username_0 on 2023-12-08)

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Hi, everyone! I'd appreciate some help with this problem, please.

Suppose that a seller puts a one-unit good on the market, and a buyer comes around with an unknown valuation $θ$. This buyer has a quasi-linear utility function $θq-p,$ and costs are quadratic with $c(q) = \dfrac{1}{2}q^2.$

Now suppose that $θ$ is uniformly distributed on $[0,1].$

Find the optimal direct mechanism $(p(·),q(·)).$

At the moment, I understand that we would need to maximize the seller's expected revenue, though I am not sure to navigate this problem. All help is appreciated.

**Tags:** game-theory, mechanism-design
> **Comments:**
> **username_0**: It seems that no trade may be a feasible outcome.  

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## 1 answers

### Answer by username_1 on 2024-01-31. Score: 2 | Accepted answer

Let $(q(\theta), p(\theta))$ be the mechanism. Let us show that the optimal mechanism will define a cutoff $\theta^\ast$ such that $q(\theta) = 1$ and $p(\theta) = \theta^\ast$ if and only if $\theta \ge \theta^\ast$.

The IC constraint will require that for all $\theta$ and $\theta' \in [0,1]$:

$$
\theta q(\theta) - p(\theta) \ge \theta q(\theta') - p(\theta'). \tag{IC}
$$

The PC (outside option normalized to $0$) tells us that:
$$
\theta q(\theta) - p(\theta) \ge 0. \tag{PC}
$$

Now assume that the mechanism specifies to sell to type $\theta$, so $q(\theta) = 1$

This requires that:
$$
\theta - p(\theta) \ge 0 \to p(\theta) \le \theta.
$$

Now, take any type $\theta' > \theta$. Then the (IC) constraint specifies that:
$$
\theta' q(\theta') - p(\theta') \ge \theta' - p(\theta)\ge \theta' - \theta > 0.
$$
If $p(\theta') \ge 0$ this requires that $q(\theta') = 1$.

Then using the (IC) constraint again, we have:
$$
\theta'- p(\theta') \ge \theta' - p(\theta),
$$
which implies that $p(\theta') \le p(\theta)$.

If $q(\theta) = 0$, then the (PC) constraint tells us that $p(\theta) = 0$.

Let $\theta^\ast$ be the lowest value of $\theta$ for which $q(\theta) = 1$. Summarizing, we have the following:

* If $\theta \ge \theta^\ast$, then $q(\theta) = 1$ and $p(\theta) \le p(\theta^\ast)$.
* If $\theta < \theta^\ast$ then $q(\theta) = 0$ and $p(\theta) = 0$.
* For all $\theta$ with $q(\theta) = 1$, $p(\theta) \le \theta$

Note that, as the firm tries to maximize profits, it will decide to put $p(\theta) = p(\theta^\ast) = \theta^\ast$ for all $\theta \ge \theta^\ast$.

This implies the following rule.

* For all $\theta \ge \theta^\ast$, set $q(\theta) = 1$ and $p(\theta) = \theta^\ast$.
* For all $\theta < \theta^\ast$, set $q(\theta) = 0$ and $p(\theta) = 0$.

The profits will then be equal to:
$$
\theta^\ast (1 - \theta^\ast) - \frac{1}{2}(1- \theta^\ast)^2.
$$
Taking first order conditions gives:
$$
1-2\theta^\ast + (1- \theta^\ast) = 2 - 3\theta^\ast = 0
$$
The second order condition is negative so the objective function is concave.
This gives a value of, $\theta^\ast = 2/3$.

> **Comments:**
> **username_0**: Thank you! I really appreciate your help.  

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