Tailored recommendations

Abstract

Many popular internet platforms use so-called collaborative filtering systems to give personalized recommendations to their users, based on other users who provided similar ratings for some items. We propose a novel approach to such recommendation systems by viewing a recommendation as a way to extend an agent’s expressed preferences, which are typically incomplete, through some aggregate of other agents’ expressed preferences. These extension and aggregation requirements are expressed by an Acceptance and a Pareto principle, respectively. We characterize the recommendation systems satisfying these two principles and contrast them with collaborative filtering systems, which typically violate the Pareto principle.

Appendix: Proofs

We derive Propositions 1 and 2 from slightly more general results. To this end we consider the following weakening of the Acceptance principle.

Axiom

(Weak Acceptance principle) For all $x,y \in X$, if $x \succsim _0 y$ then $x \succsim y$.

Lemma 1

Let U be a representation of $\succsim$ as per Definition 1.

• (a) $\succsim$ satisfies the Weak Acceptance and Pareto principles if and only if, for all $u \in U$, there exist $u_0 \in U_0,u_1 \in U_1, \ldots ,u_N \in U_N,\theta _0,\theta _1,\ldots ,\theta _N \in {\mathbb {R}}_+ , \kappa _0,\kappa \in {\mathbb {R}}$ such that:

  $$\begin{aligned} u = \theta _0 u_0 + \kappa _0 = \sum \nolimits _{n=1}^N \theta _n u_n + \kappa . \end{aligned}$$

  (3)

• (b) $\succsim$ satisfies the Weak Acceptance and Pareto-Indifference principles if and only if, for all $u \in U$, there exist $u_0 \in U_0, u_1,v_1 \in U_1, \ldots ,u_N,v_N \in U_N,\theta _0,\theta _1,\omega _1,\ldots ,\theta _N,\omega _N \in {\mathbb {R}}_+ , \kappa _0,\kappa \in {\mathbb {R}}$ such that:

  $$\begin{aligned} u = \theta _0 u_0 + \kappa _0 = \sum \nolimits _{n=1}^N (\theta _n u_n - \omega _n v_n) + \kappa . \end{aligned}$$

  (4)

Proof

• (a) It is obvious that if (3) holds then $\succsim$ satisfies the Weak Acceptance and Pareto principles. Conversely, assume (3) does not hold. Then at least one $u \in U$ must lie outside at least one of the two sets

  $$\begin{aligned} S&= \{ \theta _0 u_0 + \kappa _0 : u_0 \in U_0, \theta _0 \in {\mathbb {R}}_+ , \kappa _0 \in {\mathbb {R}} \} \ \text {and} \\ T&= \left\{ \sum \nolimits _{n=1}^N \theta _n u_n + \kappa : u_1 \in U_1, \ldots , u_N \in U_N, \theta _1,\ldots ,\theta _N \in {\mathbb {R}}_+, \kappa \in {\mathbb {R}} \right\} . \end{aligned}$$

  Note that both S and T are closed, convex cones in ${\mathbb {R}}^A$.

  Case 1. If $u \notin S$ then, by the separating hyperplane theorem, there exists $h \in {\mathbb {R}}^A$, $h \ne 0$, such that $\sum _{a \in A} h(a)u(a) > 0 \ge \sum _{a \in A} h(a)s(a)$ for all $s \in S$. Since $s+\beta \in S$ for all $s \in S$ and $\beta \in {\mathbb {R}}$, this can only be the case if $\sum _{a \in A} h(a) = 0$, for otherwise we could make $\sum _{a \in A} h(a)(s(a)+\beta ) = \sum _{a \in A} h(a)s(a) + \beta \sum _{a \in A} h(a)$ arbitrarily large by taking $\beta$ sufficiently close to $+\infty$ or $-\infty$. Hence, letting $\xi = \sum _{a \in A}|h(a)|$ and defining $x,y \in {\mathbb {R}}^A$ by $x(a) = \frac{\max \{h(a),0\}}{\xi }$ and $y(a) = \frac{\max \{-h(a),0\}}{\xi }$, we have $\xi >0$, $x,y \in X$, and $h = \xi (x-y)$. Since $\sum _{a \in A} h(a)u(a) > 0$, it follows that ${{\,\mathrm{E}\,}}u(x) > {{\,\mathrm{E}\,}}u(y)$, so it cannot be the case that $y \succsim x$ by Definition 1. But since $U_0 \subset S$, it also follows that $0 \ge \sum _{a \in A} h(a)u_0(a)$ and, hence, ${{\,\mathrm{E}\,}}u_0(y) \ge {{\,\mathrm{E}\,}}u_0(x)$ for all $u_0 \in U_0$, so that $y \succsim _0 x$ by Definition 1. This contradicts the Weak Acceptance principle.

  Case 2. If $u \notin T$ then, by the separating hyperplane theorem, there exists $h \in {\mathbb {R}}^A$, $h \ne 0$, such that $\sum _{a \in A} h(a)u(a) > 0 \ge \sum _{a \in A} h(a)t(a)$ for all $t \in T$. By the same argument as in Case 1, we must have $\sum _{a \in A}h(a)=0$ and, hence, we can write $h=\xi (x-y)$ with $\xi >0$ and $x,y \in X$. Since $\sum _{a \in A} h(a)u(a) > 0$, it follows that ${{\,\mathrm{E}\,}}u(x) > {{\,\mathrm{E}\,}}u(y)$, so it cannot be the case that $y \succsim x$ by Definition 1. But for all $n=1,\ldots ,N$, since $U_n \subset T$, it also follows that $0 \ge \sum _{a \in A} h(a)u_n(a)$ and, hence, ${{\,\mathrm{E}\,}}u_n(y) \ge {{\,\mathrm{E}\,}}u_n(x)$ for all $u_n \in U_n$, so that $y \succsim _n x$ by Definition 1. This contradicts the Pareto principle.

• (b) It is obvious that if (4) holds then $\succsim$ satisfies the Weak Acceptance and Pareto-Indifference principles. Conversely, assume (4) does not hold. Then at least one $u \in U$ must lie outside at least one of the two sets S and:

  $$\begin{aligned} T'&= \left\{ \sum \nolimits _{n=1}^N (\theta _n u_n - \omega _n v_n) + \kappa : u_1,v_1 \in U_1, \ldots , u_N,v_N \in U_N, \theta _1,\omega _1,\ldots ,\theta _N,\omega _N \in {\mathbb {R}}_+, \kappa \in {\mathbb {R}} \right\} . \end{aligned}$$

  We then proceed as in the proof of Part (a), except that in Case 2, since $T'$ is a linear subspace of ${\mathbb {R}}^A$, the separating hyperplane theorem now implies that $\sum _{a \in A} h(a)u(a) > 0 = \sum _{a \in A} h(a)t(a)$ for all $t \in T'$. Hence for all $n=1,\ldots ,N$, we now have ${{\,\mathrm{E}\,}}u_n(x) = {{\,\mathrm{E}\,}}u_n(y)$ for all $u_n \in U_n$, so that $x \sim _n y$ by Definition 1, which contradicts the Pareto-Indifference principle.

$\square$

Proof of Proposition 1

It is obvious that if $\succsim$ can be represented as per Definition 1 by some set $U \subseteq V$ of virtual guides, then it satisfies the Acceptance and Pareto principles. Conversely, assume $\succsim$ satisfies these two principles and fix some representation U of $\succsim$ as per Definition 1. Then for all $u \in U$, (3) holds by Lemma 1(a). It is sufficient to show that (3) actually holds with $\theta _0 > 0$ for some $u_0 \in U_0$, since then the closed convex hull of the set $\bigl \{ \frac{u - \kappa _0}{\theta _0} : u \in U \bigr \}$ also represents $\succsim$ as per Definition 1 and is a subset of V. So suppose (3) does not hold with $\theta _0 > 0$ for any $u_0 \in U_0$. Note that this implies that u is a constant function whereas $U_0$ contains no constant function. Hence, by the separating hyperplane theorem, there exists $h \in {\mathbb {R}}^A$, $h \ne 0$, such that $\sum _{a \in A} h(a)u_0(a) > \kappa \sum _{a \in A} h(a)$ for all $u_0 \in U_0$ and $\kappa \in {\mathbb {R}}$. By the same argument as in Case 1 of the proof of Lemma 1(a), we must have $\sum _{a \in A}h(a)=0$ and, hence, we can write $h=\xi (x-y)$ with $\xi >0$ and $x,y \in X$. For all $u_0 \in U_0$, since $\sum _{a \in A} h(a)u_0(a) > 0$, it follows that ${{\,\mathrm{E}\,}}u_0(x) > {{\,\mathrm{E}\,}}u_0(y)$ and, hence, $x \succ \!\!\succ _0 y$. The Acceptance principle then implies that $x \succ \!\!\succ y$, contradicting the fact that U contains a constant function. $\square$

Proof of Proposition 2

Similar to the proof of Proposition 1. $\square$

Proof of Proposition 3

Letting $\kappa = \kappa ^+ - \kappa ^-$ with $\kappa ^+,\kappa ^- \in {\mathbb {R}}_+$ in (1), V is non-empty if and only if the following finite linear system has a non-negative solution:

$$\begin{aligned} 0&= \sum \nolimits _{i=1}^{I(0)} \lambda _{0,i}u_{0,i}(a_k) - \sum \nolimits _{n=1}^N \sum \nolimits _{i=1}^{I(n)} \phi _{n,i}u_{n,i}(a_k) - (\kappa ^+ - \kappa ^-),&k = 1,\ldots ,K, \\ 1&= \sum \nolimits _{i=1}^{I(0)} \lambda _{0,i} . \end{aligned}$$

By Farkas’ Lemma, this system has a non-negative solution if and only if there exist no $h \in {\mathbb {R}}^K$ and $\zeta \in {\mathbb {R}}$ such that:

$$\begin{aligned} \begin{array}{ll} 0 \le \sum \nolimits _{k=1}^K h_k u_{0,i}(a_k) + \zeta , &{} i = 1,\ldots ,I(0), \\ 0 \le - \sum \nolimits _{k=1}^K h_k u_{n,i}(a_k), &{} n = 1,\ldots ,N, i = 1,\ldots ,I(n), \\ 0 \le \sum \nolimits _{k=1}^K h_k, &{}\\ 0 \le -\sum \nolimits _{k=1}^K h_k,&{} \\ 0 > \zeta ,&{}\\ \end{array} \end{aligned}$$

or, equivalently, if there exists no $h \in {\mathbb {R}}^K$ such that:

$$\begin{aligned} \begin{array}{ll} 0 < \sum \nolimits _{k=1}^K h_k u_{0,i}(a_k), &{} i = 1,\ldots ,I(0), \\ 0 \ge \sum \nolimits _{k=1}^K h_k u_{0,i}(a_k), &{} n = 1,\ldots ,N, i = 1,\ldots ,I(n), \\ 0 = \sum \nolimits _{k=1}^K h_k.&{}\\ \end{array} \end{aligned}$$

By the same argument as in Case 1 of the proof of Lemma 1(a), this is equivalent to the non-existence of lotteries $x,y \in X$ such that $x \succ \!\!\succ _0 y$ and $y \succsim _n x$ for all $n=1,\ldots ,N$. The fact that V is then the convex hull of finitely many functions simply follows from the finiteness and linearity of the above system as well as the fact that V is a subset of the compact set $U_0$. $\square$

Proof of Proposition 4

Letting $\psi _{n,i}=\psi _{n,i}^+-\psi _{n,i}^-$ and $\kappa = \kappa ^+ - \kappa ^-$ where $\psi _{n,i}^+,\psi _{n,i}^-,\kappa ^+,\kappa ^- \in {\mathbb {R}}_+$ in (2), W is non-empty if and only if the following finite linear system has a non-negative solution:

$$\begin{aligned} 0&= \sum \nolimits _{i=1}^{I(0)} \lambda _{0,i}u_{0,i}(a_k) - \sum \nolimits _{n=1}^N \sum \nolimits _{i=1}^{I(n)} (\psi _{n,i}^+-\psi _{n,i}^-) u_{n,i}(a_k) - (\kappa ^+ - \kappa ^-),&k = 1,\ldots ,K, \\ 1&= \sum \nolimits _{i=1}^{I(0)} \lambda _{0,i} . \end{aligned}$$

By Farkas’ Lemma, this system has a non-negative solution if and only if there exist no $h \in {\mathbb {R}}^K$ and $\zeta \in {\mathbb {R}}$ such that:

$$\begin{aligned} \begin{array}{ll} 0 \le \sum \nolimits _{k=1}^K h_k u_{0,i}(a_k) + \zeta , &{} i = 1,\ldots ,I(0), \\ 0 \le - \sum \nolimits _{k=1}^K h_k u_{n,i}(a_k), &{} n = 1,\ldots ,N, i = 1,\ldots ,I(n), \\ 0 \le \sum \nolimits _{k=1}^K h_k u_{n,i}(a_k), &{} n = 1,\ldots ,N, i = 1,\ldots ,I(n), \\ 0 \le \sum \nolimits _{k=1}^K h_k, &{}\\ 0 \le -\sum \nolimits _{k=1}^K h_k, &{}\\ 0 > \zeta ,&{}\\ \end{array} \end{aligned}$$

or, equivalently, if there exists no $h \in {\mathbb {R}}^K$ such that:

$$\begin{aligned} \begin{array}{ll} 0 < \sum \nolimits _{k=1}^K h_k u_{0,i}(a_k), &{} i = 1,\ldots ,I(0), \\ 0 = \sum \nolimits _{k=1}^K h_k u_{0,i}(a_k), &{} n = 1,\ldots ,N, i = 1,\ldots ,I(n), \\ 0 = \sum \nolimits _{k=1}^K h_k.&{}\\ \end{array} \end{aligned}$$

By the same argument as in Case 1 of the proof of Lemma 1(a), this is equivalent to the non-existence of lotteries $x,y \in X$ such that $x \succ \!\!\succ _0 y$ and $x \sim _n y$ for all $n=1,\ldots ,N$. The fact that W is then the convex hull of finitely many functions simply follows from the finiteness and linearity of the above system as well as the fact that W is a subset of the compact set $U_0$. $\square$

Proof of Proposition 5

It is obvious that the recommendation rule on $D_c$ consisting in selecting all virtual guides for all profiles satisfies the Acceptance, Pareto, Internal-Robustness, Anonymity, and External-Robustness principles. Conversely, let f be a recommendation rule on $D_c$ satisfying the Acceptance, Pareto, and Internal-Robustness principles. Consider a profile $\bigl (\succsim _0,\succsim _1,\ldots ,\succsim _N\bigr ) \in D_c$ and let $\succsim \ = f \bigl (\succsim _0,\succsim _1,\ldots ,\succsim _N\bigr )$. Fixing arbitrary representations $U_0,U_1,\ldots ,U_N$ of $\succsim _0,\succsim _1,\ldots ,\succsim _N$, respectively, let V denote the corresponding set of virtual guides. Then by Proposition 1, $\succsim$ can be represented as per Definition 1 by some set $U \subseteq V$.

Suppose $\succsim$ cannot be represented as per Definition 1 by V. By the uniqueness of expected multi-utility representation and our finiteness assumption, we must then have $\{ \theta u + \kappa : u \in U, \theta \in {\mathbb {R}}_+, \kappa \in {\mathbb {R}} \} \ne \{ \theta v + \kappa : v \in V, \theta \in {\mathbb {R}}_+, \kappa \in {\mathbb {R}} \}$ and, hence, there must exist some $u \in U$ that does not belong to $\{ \theta v + \kappa : v \in V, \theta \in {\mathbb {R}}_+, \kappa \in {\mathbb {R}} \}$. By the same argument as in Case 1 of the proof of Lemma 1(a), it follows that there exist $x,y \in X$ such that ${{\,\mathrm{E}\,}}u(x) > E u(y)$ whereas ${{\,\mathrm{E}\,}}v(y) \ge {{\,\mathrm{E}\,}}v(x)$ for all $v \in V$. Hence, letting $\succsim '_0$ be represented as per Definition 1 by $\bigl \{u_0\bigr \}$, we have $x \succ \!\!\succ '_0 y$. On the other hand, letting $\succsim '\ = f(\bigl (\succsim '_0,\succsim _1,\ldots ,\succsim _N\bigr )$, we have $y \succsim ' x$ by the Internal-Robustness principle. This contradicts the Acceptance principle. $\square$

Proof of Proposition 6

Similar to the proof of Proposition 5. $\square$

Proof of Proposition 7

Assume A contains at least three alternatives a, b, c, let $w,w' : A \rightarrow {\mathbb {R}}$ be such that $w(a)=w'(b)=1$ and $w(b)=w'(a)=w(c)=w'(c)=0$, and let U denote the convex hull of $\{w,w'\}$. Note that for all $u,v \in U$, if $u \ne v$ then there exist $x,y \in X$ such that ${{\,\mathrm{E}\,}}u(x) > {{\,\mathrm{E}\,}}u(y)$ whereas ${{\,\mathrm{E}\,}}v(y) > {{\,\mathrm{E}\,}}v(x)$. Now let f be a recommendation rule on $D_c$ (resp. $D_{wc}$) satisfying the Acceptance and Pareto (resp. Pareto-Indifference) principles that yields a complete recommendation for all profiles. Let $\succsim$ be represented as per Definition 1 by U and $\succsim '\ = f\bigl (\succsim ,\succsim ,\ldots ,\succsim \bigr )$. Then by Proposition 1 (resp. 2), $\succsim '$ can be represented as per Definition 1 by $\{u\}$ for some $u \in U$. Let $\succsim ''$ be represented by some $v \in U \setminus \{u\}$ and $\succsim '''\ = f\bigl (\succsim ,\succsim '',\ldots ,\succsim ''\bigr )$. Again by Proposition 1 (resp. 2), $\succsim '''$ can be represented as per Definition 1 by $\{v\}$. Hence $\succsim '''$ does not refine $\succsim '$, so f does not satisfy the External-Robustness principle. $\square$