# (Mechanism Design | Mailath Exercise 10.5.3) A Question

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.

• It seems that no trade may be a feasible outcome. Dec 9, 2023 at 5:16

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, $$\theta^\ast = 2/3$$.

• Thank you! I really appreciate your help. Feb 1 at 15:08