Risk-Robust Mechanism Design for a Prospect-Theoretic Buyer

Abstract

Consider the revenue maximization problem of a risk-neutral seller with m heterogeneous items for sale to a single additive buyer, whose values for the items are drawn from known distributions. If the buyer is also risk-neutral, it is known that a simple and natural mechanism, namely the better of selling separately or pricing only the grand bundle, gives a constant-factor approximation to the optimal revenue. In this paper we study revenue maximization without risk-neutral buyers. Specifically, we adopt cumulative prospect theory, a well established generalization of expected utility theory. Our starting observation is that such preferences give rise to a very rich space of mechanisms, allowing the seller to extract arbitrary revenue. Specifically, a seller can construct extreme lotteries that look attractive to a mildly optimistic buyer, but have arbitrarily negative true expectation. Therefore, giving the seller absolute freedom over the design space results in absurd conclusions; competing with the optimal mechanism is hopeless. Instead, in this paper we study four broad classes of mechanisms, each characterized by a distinct use of randomness. Our goal is twofold: to explore the power of randomness when the buyer is not risk-neutral, and to design simple and attitude-agnostic mechanisms—mechanisms that do not depend on details of the buyer’s risk attitude—which are good approximations of the optimal in-class mechanism, tailored to a specific risk attitude. Our main result is that the same simple and risk-agnostic mechanism (the better of selling separately or pricing only the grand bundle) is a good approximation to the optimal non-agnostic mechanism within three of the mechanism classes we study.

Appendices

Appendix A: Full CPT Example

Example 5

[35] Consider the following game of chance. You roll a die once and observe the result $v = 1, {\dots } , 6$. If v is even, you receive v; if v is odd, you pay v. This defines a random variable X which takes values (− 5,− 3,− 1, 2, 4, 6), each with probability 1/6. Let X⁺ be the random variable which takes value 0 with probability 1/2, and 2, 4, 6, each with probability 1/6. Also, let X⁻ be the random variable which takes the value 0 with probability 1/2, and values − 1,− 3,− 5, each with probability 1/6, correlated such that X⁺ + X⁻ = X. Assuming that r = 0, the weighted expectation of X⁺ is

$$ \mathbb{E}\left[X^{+}\right] = 2 \cdot (w^{+}(1/2) - w^{+}(1/3) ) +4 \cdot (w^{+}(1/3) - w^{+}(1/6) ) +6 \cdot (w^{+} (1/6) - w^{+} (0) ). $$

The intuition here is that the multiplier of value v is equal to the difference between the weighted probabilities of the events “the outcome of the experiment is at least as good as v” and “the outcome is strictly better than v”. Similarly, the weighted expectation of X⁻ is

$$ \mathbb{E}\left[X^{-}\right] = (-1) \cdot (w^{-} (1/2) - w^{-}(1/3) ) +(-3) \cdot (w^{-} (1/3) - w^{-}(1/6) ) +(-5) \cdot (w^{-} (1/6) - w^{-}(0) ), $$

where this time the multiplier of v is equal to the difference between the weighted probabilities of the events “the outcome is at least as bad as v” and “the outcome is strictly worse than v”. Finally, the weighted expectation of X is simply $\mathbb {E}_{w}\left [{X}\right ] = \mathbb {E}_{w^{+}}[{X^{+}}] + \mathbb {E}_{w-}[{X^{-}}]$.

Appendix B: Optimal Max-Min Mechanism

Theorem 6

$$ \underset{\mathcal{M}}{\max} \underset{w}{\min} {\textsc{Rev}}_{\mathcal{M}}(w,\mathcal{D}) = \textsc{DRev}(\mathcal{D}) $$

Proof

Define w_o, the weighting function of a perfectly optimistic buyer, as the following

$$w_{o}(q) = \begin{cases}0 \quad &q =0 \\ 1 \quad& 0< q\leq 1 \end{cases} .$$

We first prove the following lemma.

Lemma 10

For every value distribution $\mathcal {D}$, if the buyer’s weighting function is w_o, then there exists a deterministic mechanism that maximizes the seller’s revenue.

Proof

Consider an arbitrary mechanism ${\mathscr{M}}$ and a buyer with type $v = (v_{1},\dots , v_{m})$. Let ${{\mathscr{L}}}$ be a menu item in ${\mathscr{M}}$. ${{\mathscr{L}}}$ defines a distribution over k outcomes: outcome o_i occurs with some probability q_i, where some subset of items S_i is allocated for some payment p_i. Without loss of generality we assume that outcomes are ordered in increasing utility for the buyer. Then the expected utility of the buyer with type v picking ${\mathscr{L}}$ is

$$ \begin{array}{@{}rcl@{}} \mathbb{E}_{w_{o}}[v, {\mathcal{L}}] &=& \sum\limits_{i=1}^{k-1} \left( \underset{j \in S_{i}}{\sum} v_{j} - p_{j}\right) \left( w_{o}\left( \sum\limits_{j=i}^{k} q_{j}\right) - w_{o}\left( \sum\limits_{j=i+1}^{k} q_{j}\right) \right) + \left( \underset{j \in S_{k}}{\sum} v_{j} - p_{k} \right)w_{o}(q_{k}) \\ &=& \underset{j \in S_{k}}{\sum} v_{j} - p_{k}, \end{array} $$

i.e. the expect utility is the same as the highest utility of all the outcomes in ${{\mathscr{L}}}$.

Let $S_{{{\mathscr{L}}},v}$ and $p_{{{\mathscr{L}}},v}$ be the subset of items and price in the favorite outcome of type v in the menu item ${{\mathscr{L}}}$ in ${\mathscr{M}}$. Now we construct a deterministic mechanism ${\mathscr{M}}^{\prime }$: for each type v and menu item ${\mathscr{L}}$ in ${\mathscr{M}}$, add to ${\mathscr{M}}^{\prime }$ the menu item ${\mathscr{L}}^{\prime }_{{\mathscr{L}},v}$ that deterministically sells $S_{{\mathscr{L}},v}$ at a price $p_{{\mathscr{L}},v}$.

It’s not hard to see that a buyer with type v and weighting function w_o will buy menu item ${{\mathscr{L}}}^{\prime }_{{{\mathscr{L}}},v}$ in ${\mathscr{M}}^{\prime }$ if she buys ${{\mathscr{L}}}$ in ${\mathscr{M}}$. Therefore $\textsc {Rev}_{{\mathscr{M}}}(w_{o},\mathcal {D}) = \textsc {Rev}_{{\mathscr{M}}^{\prime }}(w_{o},\mathcal {D})$. □

Let ${\mathscr{M}}_{det}$ be the mechanism of Lemma 10. Subsequently, we get that

$$ \begin{array}{@{}rcl@{}} \underset{\mathcal{M}}{\max} \underset{w}{\min} {\textsc{Rev}}_{\mathcal{M}}(w,\mathcal{D}) &\leq& \underset{\mathcal{M}}{\max} {\textsc{Rev}}_{\mathcal{M}}(w_{o},\mathcal{D}) \\ &=& {\textsc{Rev}}_{\mathcal{M}_{det}}(w_{o},\mathcal{D}) \\ &\leq& \underset{\mathcal{M}\text{ deterministic} }{\max} {\textsc{Rev}}_{\mathcal{M} }(w_{o},\mathcal{D}) \\ &=& \underset{\mathcal{M}\text{ deterministic} }{\max} {\textsc{Rev}}_{\mathcal{M} }({\mathcal{I}},\mathcal{D}) \\ &=& \textsc{DRev}(\mathcal{D}) \end{array} $$

The fourth equality holds since a prospect theory buyer has the same preferences as a risk-neutral buyer in a deterministic mechanism. On the other hand:

$$ \textsc{DRev}(\mathcal{D}) = \underset{\mathcal{M} \text{ deterministic}}{\max} {\textsc{Rev}}_{\mathcal{M}}({\mathcal{I}},\mathcal{D}) = \underset{\mathcal{M} \text{ deterministic}}{\max} \underset{w}{\min} {\textsc{Rev}}_{\mathcal{M}}(w,\mathcal{D}) \leq \underset{\mathcal{M}}{\max} \underset{w}{\min} {\textsc{Rev}}_{\mathcal{M}}(w,\mathcal{D}), $$

where the second equality holds since a prospect theory buyer has the same preferences as a risk-neutral buyer in a deterministic mechanism. The theorem follows. □

Appendix C: Properties of Weighted Expectations

Proof Proof of Lemma 1

We prove the statements for a discrete random variable Z over k outcomes; the proof for continuous random variables is analogous. The i-th outcome in Z, Z_i, occurs with probability p_i, and without loss of generality Z_i ≤ Z_i+ 1. Notice that for the random variable W = Z + c, the ordering remains the same.

$$ \begin{array}{@{}rcl@{}} \mathbb{E}\left[Z + c\right] &=& \sum\limits_{i=1}^{k-1} (Z_{i} + c) \cdot \left( w\left( \sum\limits_{j=i}^{k} p_{j} \right) - w\left( \sum\limits_{j=i+1}^{k} p_{j} \right) \right) + (Z_{k} + c) w(p_{k}) \\ &=& \sum\limits_{i=1}^{k-1} Z_{i} \cdot \left( w\left( \sum\limits_{j=i}^{k} p_{j} \right) - w\left( \sum\limits_{j=i+1}^{k} p_{j} \right) \right) + Z_{k} w(p_{k}) \\ && +\sum\limits_{i=1}^{k-1} c \cdot \left( w\left( \sum\limits_{j=i}^{k} p_{j} \right) - w\left( \sum\limits_{j=i+1}^{k} p_{j} \right) \right) + c w(p_{k}) \\ &=& \mathbb{E}\left[Z\right] + c \cdot \left( \sum\limits_{i=1}^{k-1} w\left( \sum\limits_{j=i}^{k} p_{j} \right) - w\left( \sum\limits_{j=i+1}^{k} p_{j} \right) + w(p_{k}) \right) \\ &=& \mathbb{E}\left[Z\right] + c \cdot w\left( \sum\limits_{j=1}^{k} p_{j} \right) \\ &=& \mathbb{E}\left[Z\right] + c. \end{array} $$

Similarly,

$$ \begin{array}{@{}rcl@{}} \mathbb{E}\left[cZ\right] &=& \sum\limits_{i=1}^{k-1} (c Z_{i}) \cdot \left( w\left( \sum\limits_{j=i}^{k} p_{j} \right) - w\left( \sum\limits_{j=i+1}^{k} p_{j} \right) \right) + (c Z_{k}) w(p_{k}) \\ &=& c \cdot \left( \sum\limits_{i=1}^{k-1} Z_{i} \cdot \left( w\left( \sum\limits_{j=i}^{k} p_{j} \right) - w\left( \sum\limits_{j=i+1}^{k} p_{j} \right) \right) + Z_{k} w(p_{k}) \right) \\ &=& c \cdot \mathbb{E}\left[Z\right]. \end{array} $$

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Proof Proof of Claim 6

$$ \begin{array}{@{}rcl@{}} \mathbb{E}\left[-X\right] &=& -{\int}_{-\infty}^{0} \left( 1 - w(1-F_{-X}(z))\right)dz + {\int}_{0}^{\infty} w(1-F_{-X}(z)) dz \\ &=& -{\int}_{0}^{\infty} \left( 1-w(F_{X}(z))\right)dz + {\int}_{-\infty}^{0}w(F_{X}(z)) dz \\ &=& -\mathbb{E}_{w^{\dagger}}\!\left[X\right] \end{array} $$

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