Consider an example of a risk-neutral seller that has two distinct indivisible goods for sale. The seller wants to maximize the expected revenue. The buyer's utility is $$I_av_a+I_bv_b-t,$$ where $I_a,I_b$ are indicators of whether the buyer gets good $a$ or $b$ and $t$ is the monetary transfer. Both $v_a,v_b$ are the types of the buyer (willingness to buy the good). Let $v_a$ and $v_b$ be i.i.d. draws from a uniform distribution on $[0,1]$.

Assume a seller quotes three prices: $p_a,p_b,p_{ab}$, where $p_{ab}\leq p_a+p_b$. What is the optimal choice of $p_a,p_b,p_{ab}$?

The textbook (T. Borgers, An Intro to the theory of Mechanism Design) I am studying simply says that this is a simple calculus problem and gives the solution as $$p_a=p_b=2/3\\ p_{ab}=\frac{4-\sqrt{2}}{3}.$$

I am not sure how to do that. I'd appreciate any help or hint. Thank you.

  • 2
    $\begingroup$ I suggest not to close this question, because it is rather a conceptual one than a simple homework question. Moreover, it is more advanced than the usual undergrad textbook question. We should encourage everyone who is interested in economics on that level. Let me add a missing assumption in your question (in italics). $\endgroup$
    – Bayesian
    Commented Mar 20, 2019 at 13:31

1 Answer 1


Consider the incentive and participation constraints. That is, look at which buyer types would choose which option.

A buyer does not buy anything if $$v_a < p_a \quad v_b < p_b \quad v_a +v_b < p_{ab}.$$ A buyer chooses only good $i \in \{a,b\}$ with $j\neq i$ if $$v_i - p_i \geq 0 \quad v_i - p_i \geq v_i + v_j - p_{ab}.$$ Otherwise she chooses the bundle. With the assumption of independent uniformly distributed values, it then requires some caution to derive the probabilities of each event.

Finally, you want to maximize your profit given by \begin{align} (1-p_a)(p_{ab}-p_a) p_a + (1-p_b)(p_{ab}-p_b) p_b + \\ \left( \frac{(p_a + p_b - p_{ab})^2}{2} + (1 - p_a) (1 + p_a - p_{ab}) + (1 - p_b) (1 + p_b - p_{ab}) \\ - (1 - p_a) (1 - p_b) \right) p_{ab}, \end{align} which is super annoying, but Mathematica suggests that your solution is correct.


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