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Suppose a seller has two distinct indivisible goods, good $A$ and good $B$, for sale. For simplicity, we assume that the seller values the goods at zero and is risk-neutral, so that she seeks to maximize her expected revenue. There is one buyer in this model. The parameters $v_A$ and $v_B$ indicate the buyer's willingness to pay for the two goods. These parameters are known to the buyer but not known to the seller. The seller's belief about these two parameters is given by the uniform distribution $F$ over the unit square $[0,1]^2$. Note that we assume here that $v_A$ and $v_B$ are stochastically independent. We assume that the seller quotes three prices: $p_A, p_B$, and $p_{A B}$. The interpretation is that the buyer can buy good $A$ at price $p_A$, good $B$ at price $p_B$, or goods $A$ and $B$ at price $p_{A B}$. We assume that the seller cannot stop the buyer from buying goods $A$ and $B$ at price $p_A+p_B$, so the price $p_{A B}$, if it is to have any effect, has to satisfy $p_{A B} \leq p_A+p_B$. What is the optimal choice of $p_A, p_B$ and $p_{A B}$?

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  • $\begingroup$ Please pick a more descriptive title. $\endgroup$ Feb 27, 2023 at 7:38

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Perhaps there is an elegant solution, but the problem is simple enough that one can reason graphically. In the case where $$ 0 \leq p_A,p_B \leq 1 $$ and $$ p_{AB} \leq p_A + p_B $$ we have something like this:

enter image description here

In the $(v_A,v_B)$ coordinate system the green area shows where the buyer achieves the highest surplus by buying $A$ only, so where $$ v_A + v_B - p_{AB} \leq v_A - p_A \text{ and } 0 \leq v_A - p_A. $$ Blue area shows where buyer buys $B$ only, and in the red area they buy both products. Since the probability distributions are independent and uniform, the probabilibities are proportional to these areas, thus you can get them as not too complicated formulas of $p_A,p_B,p_{AB}$. Then you can sketch the expected profit $$ P(A \text{ only}) p_A + P(B \text{ only}) p_B + P(A \text{ and } B) p_{AB}, $$ and maximize this w.r.t. the prices. The problem is that these geometric formulas are only valid when the above constraints hold, otherwise you have to rethink them, and there could potentially be many cases; though my guess is that the real solution lies within the above constraints.

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  • $\begingroup$ Your answer is actually as far as I got. I find $P(A$ only $)=1-p_A, P(B$ only $)=1-p_B$, and $p(A$ and $B)=1-\frac{\left(P_{A B}\right)^2}{2}, \mathrm{I}$ believe. However, I am not sure if this expected profit is correct since there is some overlap with these probabilities. For example, you can't buy A and B and then buy A individually. $\endgroup$ Feb 26, 2023 at 19:26
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    $\begingroup$ @AaronZheng $P(A \text{ only}) \neq 1 - p_A$. That is equal to the total area to the right of the green line. The green rectangle - defined by the intersections of the green and red lines - is smaller than this. $\endgroup$
    – Giskard
    Feb 26, 2023 at 19:43
  • $\begingroup$ "there is some overlap with these probabilities" The three events described above are mutually exclusive, notice the words "only" in the descriptions. The inequalities $$ v_A + v_B - p_{AB} \leq v_A - p_A \text{ and } 0 \leq v_A - p_A $$ iny my answer are about the choices of "$A$ only" and "$A$ and $B$" being mutually exclusive. $\endgroup$
    – Giskard
    Feb 26, 2023 at 19:43
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    $\begingroup$ Letting $p(A \text { and } B)=p\left(v_A+v_B-p_{A B} \geq v_A-p_A \wedge v_A+v_B-p_{A B} \geq v_B-p_B \wedge v_A+v_B \geq p_{A B}\right)$, and a few lines of code, we obtain that $p_A,p_B=0.66$ and $p_{AB}=0.862$! This answer is given in Borger's book on page 28. $\endgroup$ Feb 26, 2023 at 21:44
  • $\begingroup$ @AaronZheng If you have a reference for the problem (Book X, problem Y), would you please include that at the top of your question? This will help other users find it in search. $\endgroup$
    – Giskard
    Feb 27, 2023 at 6:49

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