Optimal student allocation with peer effects

Abstract

This paper studies an optimal assignment problem of heterogenous students to schools with a particular kind of preference complementarity: peer effects, defined by the average ability of those in the same school. The tractability of the problem allows us to characterize the optimal assignment mechanism, which has a simple “(stochastic) pass-fail” structure. Its shape is mainly determined by the convexity/concavity of the attainment function, interpreted as the preference for/against having diverse-ability students in different schools. We also provide comparative statics as to when more or less mixture of heterogenous ability types would be desirable.

Proofs

1.1 Proof of Proposition 1

Formally, this is a corollary of Theorem 2, and hence the proof is omitted.

1.2 Proof of Proposition 2

Let $\underline{q},\overline{q}:[0,1]\rightarrow [0,1]$ be such that $\underline{q}(\theta )\equiv \frac{1}{2}$ and $\overline{q}(\theta )=1_{\left\{ \theta >\frac{1}{2}\right\} }$. That is, $\underline{q}(\cdot )$ corresponds to the full pooling mechanism, and $\overline{q}(\cdot )$ corresponds to the maximum segregation mechanism. Note that $x_1=\frac{1}{2}$ given $\underline{q}(\cdot )$, and $x_1=\frac{3}{4}$ given $\overline{q}(\cdot )$.

Given any $x_1^*\in (\frac{1}{2},\frac{3}{4})$ fixed, let $q(\cdot )$ be such that:

$$\begin{aligned} q(\theta )=\omega \underline{q}(\theta )+(1-\omega )\overline{q}(\theta ), \end{aligned}$$

where $\omega$ satisfies $x_1=\frac{1}{2}\omega +\frac{3}{4}(1-\omega )$ (hence $\omega \in (0,1)$). It is readily checked that $q(\cdot )$ is non-decreasing (and hence satisfies the incentive compatibility with an appropriate $p(\cdot )$ function), and $x_1$ given the assignment $q(\cdot )$ is $x_1^*$.

1.3 Proof of Theorem 1

Recall the original problem:

$$\begin{aligned} \max _{q(\cdot )}{} & {} \int _0^1 [q(\theta )u(\theta ,x_1)+(1-q(\theta ))u(\theta ,1-x_1)]d\theta \\ \text {sub. to}{} & {} q(\theta )\le q(\theta '),\ \forall \theta <\theta '.\\{} & {} x_1=2\int _0^1 \theta q(\theta )d\theta ,\quad \frac{1}{2}=\int _0^1 q(\theta )d\theta \end{aligned}$$

Given any $x_1\in [\frac{1}{2},\frac{3}{4}]$ fixed, consider the restricted problem where maximization is only within the class of $q(\cdot )$ that attains $x_1$ as the peer-effect index. Observe that the objective is:

$$\begin{aligned}{} & {} \int _0^1 [q(\theta )u(\theta ,x_1)+(1-q(\theta ))u(\theta ,1-x_1)]d\theta \\{} & {} \quad =(h(x_1)-h(1-x_1))\int _0^1 q(\theta )g(\theta )d\theta + (\text {Const}), \end{aligned}$$

where $(\text {Const})$ is a term which does not depend on $q(\cdot )$ (though it depends on $x_1$; thus, this term can be ignored (only) in the subproblem). Therefore, the subproblem is essentially to maximize $\int _0^1 q(\theta )g(\theta )d\theta$ among all $q(\cdot )$ that attains $x_1$.

Let $q^*(\cdot )$ be given by:

$$\begin{aligned} q^*(\theta )=\left\{ \begin{array}{ccc} \frac{2y-1}{2y} &{}\text {if}&{} \theta <y\\ 1 &{}\text {if}&{} \theta >y, \end{array}\right. \end{aligned}$$

where y is uniquely determined in order to satisfy $x_1=2(\frac{2y-1}{2y}\int _0^y\theta d\theta +\int _y^1\theta d\theta )$.

Take any non-decreasing $q(\cdot )$ which attains $x_1$ as its peer-effect index. It suffices to show that the distribution of the students’ types in School 1 given $q^*(\cdot )$ is a mean-preserving spread of that given $q(\cdot )$, because then, convexity of g implies that $\int _0^1 q^*(\theta )g(\theta )d\theta \ge \int _0^1 q(\theta )g(\theta )d\theta$.

Let $\Pi ^*(\theta )$ be the cdf of the students’ types in School 1 given $q^*(\cdot )$:

$$\begin{aligned} \Pi ^*(\theta )=\frac{\int _0^\theta q^*(\theta )d\theta }{\int _0^1 q^*(\theta )d\theta }=2\int _0^\theta q^*(\theta )d\theta , \end{aligned}$$

and similarly, let $\Pi (\theta )$ be the one given $q(\cdot )$:

$$\begin{aligned} \Pi (\theta )=2\int _0^\theta q(\theta )d\theta . \end{aligned}$$

Applying the definition of mean-preserving spread (Rothschild and Stiglitz, Machina and Pratt), it suffices to show:

$$\begin{aligned} \int _0^y (\Pi ^*(\theta )-\Pi (\theta ))d\theta \ge 0, \end{aligned}$$

for all $y\in [0,1]$, where the equality hods with $y=1$. First, if $y=1$, then:

$$\begin{aligned} \int _0^1 (\Pi ^*(\theta )-\Pi (\theta ))d\theta =0\ \Leftrightarrow \ \int _0^1 (\theta q^*(\theta )-\theta q(\theta ))d\theta =0, \end{aligned}$$

which is indeed the case, because both $q^*(\cdot )$ and $q(\cdot )$ attain the same peer-effect index.

It suffices to show that there exists $z\in [0,1]$ such that (i) $\Pi ^*(\theta )-\Pi (\theta )\ge 0$ for all $\theta <z$, which implies:

$$\begin{aligned} \int _0^y (\Pi ^*(\theta )-\Pi (\theta ))d\theta \ge 0,\ \forall y\le z, \end{aligned}$$

and (ii) $\Pi ^*(\theta )-\Pi (\theta )\le 0$ for all $\theta >z$, which implies:

$$\begin{aligned} \int _0^y (\Pi ^*(\theta )-\Pi (\theta ))d\theta =-\int _y^1 (\Pi ^*(\theta )-\Pi (\theta ))d\theta \ge 0,\ \forall y>z. \end{aligned}$$

For the sake of contradiction, suppose that there exist $z'<z''$ such that $\Pi ^*(z')< \Pi (z')$ and $\Pi ^*(z'')> \Pi (z'')$. First, that $\Pi ^*(z'')> \Pi (z'')$ implies $z''\le y$, because for any $\theta >y$, we have:

$$\begin{aligned} \Pi ^*(\theta )=2\left( \frac{1}{2}-\int _{\theta }^1 1d\theta \right) \le 2\left( \frac{1}{2}-\int _{\theta }^1 q(\theta )d\theta \right) = \Pi (\theta ). \end{aligned}$$

Because $\Pi ^*(\theta )$ is linear for $\theta <y$, we obtain:

$$\begin{aligned} \frac{\Pi (z'')-\Pi (z')}{z''-z'}<\frac{\Pi ^*(z'')-\Pi ^*(z')}{z''-z'}=\frac{\Pi ^*(z')-\Pi ^*(0)}{z'-0}<\frac{\Pi (z')-\Pi (0)}{z'-0}, \end{aligned}$$

where $\Pi ^*(0)=\Pi (0)=0$. This means that $\Pi$ is not globally convex, which is a contradiction, because $\Pi$ is an integral of a non-decreasing function (which must be globally convex).

Now, given that $q^*(\cdot )$ is optimal in the subproblem for each given level of $x_1$, the optimal choice of y must clearly satisfy the condition in the statement, and hence we omit this part of the proof.

1.4 Proof of Theorem 2

The case where $y^*=1$ (i.e., $q^*(\cdot )$ exhibits full pooling) is trivial, and thus, we focus on the case where $y^*<1$.

In what follows, first, consider the case where $y^*$ is interior (i.e., $y^*\in (\frac{1}{2},1)$). The first-order condition with respect to y implies:

$$\begin{aligned} \frac{\partial }{\partial y}f(x_1(y),y)+x_1'(y^*)\frac{\partial }{\partial x_1}f(x_1(y),y)=0 \end{aligned}$$

at $y=y^*$, where:

$$\begin{aligned} f(x_1,y)=\int _0^{y} g(\theta )(\frac{2y-1}{2y}h(x_1)+\frac{1}{2y}h(1-x_1))d\theta +\int _{y}^1g(\theta )h(x_1)d\theta , \end{aligned}$$

and $x_1(y)=2(\int _0^y\frac{2y-1}{2y}\theta d\theta +\int _y^1\theta d\theta )$. Note that $x_1'(y)\le 0$.

Let $\tilde{f}$ be the same as f except that h is replaced by $\tilde{h}$:

$$\begin{aligned} \int _0^{y} g(\theta )(\frac{2y-1}{2y}\tilde{h}(x_1)+\frac{1}{2y}\tilde{h}(1-x_1))d\theta +\int _{y}^1g(\theta )\tilde{h}(x_1)d\theta . \end{aligned}$$

It suffices to show:

$$\begin{aligned} \frac{\partial }{\partial y}\tilde{f}(x_1(y),y)+x_1'(y)\frac{\partial }{\partial x_1}\tilde{f}(x_1(y),y)\le 0 \end{aligned}$$

at $y=y^*$, because this implies that there is some $\hat{y}\le y^*$ that solves the maximization problem with $\tilde{h}$, as desired.

We have:

$$\begin{aligned}{} & {} \frac{\partial }{\partial y}\tilde{f}(x_1(y),y)+x_1'(y)\frac{\partial }{\partial x_1}\tilde{f}(x_1(y),y)\\{} & {} \quad = \frac{\partial }{\partial y}[\tilde{f}(x_1(y^*),y^*)-f(x_1(y^*),y^*)\psi '(h(1-x_1(y^*)))]\\{} & {} \qquad +x_1'(y^*)\frac{\partial }{\partial x_1}[\tilde{f}(x_1(y^*),y^*)-f(x_1(y^*),y^*)\psi '(h(1-x_1(y^*)))]. \end{aligned}$$

For the first term:

$$\begin{aligned}{} & {} \frac{\partial }{\partial y}[\tilde{f}(x_1(y^*),y^*)-f(x_1(y^*),y^*)\psi '(h(1-x_1(y^*)))]\\{} & {} \quad =\frac{\partial }{\partial y}\left[ \frac{2y^*-1}{2y^*}\int _0^{y^*} g(\theta )(\tilde{h}(x_1^*)-\tilde{h}(1-x_1^*))d\theta +\int _{y^*}^1g(\theta )(\tilde{h}(x_1^*)-\tilde{h}(1-x_1^*))d\theta \right] \\{} & {} \qquad -\psi '(h(1-x_1))\frac{\partial }{\partial y} \Bigg [ \frac{2y^*-1}{2y^*}\int _0^{y^*} g(\theta )(h(x_1^*)-h(1-x_1^*))d\theta \\{} & {} \qquad +\int _{y^*}^1g(\theta )(h(x_1^*)-h(1-x_1^*))d\theta \Bigg ]\\{} & {} \quad =\frac{\partial }{\partial y} \left[ \frac{2y^*-1}{2y^*}\int _0^{y^*} g(\theta )d\theta +\int _{y^*}^1g(\theta )d\theta \right] \Delta _1, \end{aligned}$$

where

$$\begin{aligned} \Delta _1= & {} \tilde{h}(x_1^*)-\tilde{h}(1-x_1^*)-\psi '(h(1-x_1^*))(h(x_1^*)-h(1-x_1^*))\\\ge & {} \psi '(h(1-x_1^*))(h(x_1^*)-h(1-x_1^*))-\psi '(h(1-x_1^*))(h(x_1^*)-h(1-x_1^*))=0, \end{aligned}$$

by the mean-value theorem and convexity of $\psi$. Noticing that $\frac{\partial }{\partial y} \Big [ \frac{2y^*-1}{2y^*}\int _0^{y^*} g(\theta )d\theta +\int _{y^*}^1\,g(\theta )d\theta \Big ]\le 0$, we conclude the first term is non-positive.

For the second term:

$$\begin{aligned}{} & {} x_1'(y^*)\frac{\partial }{\partial x_1}[\tilde{f}(x_1(y^*),y^*)-f(x_1(y^*),y^*)\psi '(h(1-x_1(y^*)))]\\{} & {} \quad =x_1'(y^*)\left[ \frac{2y^*-1}{2y^*}\int _0^{y^*} g(\theta )d\theta +\int _{y^*}^1g(\theta )d\theta \right] \Delta _2, \end{aligned}$$

where

$$\begin{aligned} \Delta _2= & {} \psi '(h(x_1^*))h'(x_1^*)-\psi '(h(1-x_1^*))h'(x_1^*)\ge 0 \end{aligned}$$

because of convexity of $\psi$. Because $x_1'(y^*)\le 0$ and $\frac{2y^*-1}{2y^*}\int _0^{y^*} g(\theta )d\theta +\int _{y^*}^1\,g(\theta )d\theta \ge 0$, we conclude that the second term is non-positive.

1.5 Proof of Theorem 3

Let $\hat{q}(\cdot )$ be as follows:

$$\begin{aligned} \hat{q}(\theta )=\left\{ \begin{array}{ccc} 1 &{}\text {if}&{} \theta <y_1\\ 0 &{}\text {if}&{} \theta \in (y_1,y_2)\\ 1 &{}\text {if}&{} \theta >y_2, \end{array}\right. \end{aligned}$$

where $y_1,y_2$ are chosen so that $y_1< y_2$ and:

$$\begin{aligned}{} & {} \int _0^{y_1}\hat{q}(\theta )d\theta +\int _{y_2}^1\hat{q}(\theta )d\theta =y_1+1-y_2=\frac{1}{2}\\{} & {} \int _0^{y_1}\theta \hat{q}(\theta )d\theta +\int _{y_2}^1\theta \hat{q}(\theta )d\theta =\frac{y_1^2}{2}+\frac{1-y_2^2}{2}=2x_1^*. \end{aligned}$$

Note that $y_1,y_2\in (0,1)$ exist as long as $x_1^*<\frac{3}{4}$ (i.e., $q^*(\cdot )$ is not maximally segregating).

The planner’s payoff difference between under $\hat{q}(\cdot )$ and under $q^*(\cdot )$ is:

$$\begin{aligned} \int _0^1 (\hat{q}(\theta )-q^*(\theta ))g(\theta )(h(x_1)-h(1-x_1))d\theta . \end{aligned}$$

Because $g''>0$ and $h(x_1)\ge h(1-x_1)$, it suffices to show that $\hat{q}(\cdot )$ is a mean-preserving spread of $q^*(\cdot )$.

Let $\Pi ^*(\theta )$ be the cdf of the students’ types in School 1 given $q^*(\cdot )$:

$$\begin{aligned} \Pi ^*(\theta )=2\int _0^\theta q^*(\theta )d\theta , \end{aligned}$$

and similarly, let $\hat{\Pi }(\theta )$ be the one given $\hat{q}(\cdot )$.

Applying the definition of a mean-preserving spread, it suffices to show:

$$\begin{aligned} \int _0^y (\hat{\Pi }(\theta )-\Pi ^*(\theta ))d\theta \ge 0, \end{aligned}$$

for all $y\in [0,1]$, where the equality hods with $y=1$. First, if $y=1$, then:

$$\begin{aligned} \int _0^1 (\hat{\Pi }(\theta )-\Pi ^*(\theta ))d\theta =0 \ \Leftrightarrow \ \int _0^1 (\theta \hat{q}(\theta )-\theta q^*(\theta ))d\theta =0, \end{aligned}$$

which is indeed the case, because both $\hat{q}(\cdot )$ and $q^*(\cdot )$ attain the same peer-effect index.

For any $y\le y_1$, we have:

$$\begin{aligned} \int _0^y (\hat{\Pi }(\theta )-\Pi ^*(\theta ))d\theta >0, \end{aligned}$$

because for $\theta \le y_1$, we have $\hat{\Pi }(\theta )-\Pi ^*(\theta )=\int _0^\theta (1-q^*(z))dz>0$.

For any $y\ge y_2$, we have:

$$\begin{aligned} \int _0^y (\hat{\Pi }(\theta )-\Pi ^*(\theta ))d\theta =-\int _y^1 (\hat{\Pi }(\theta )-\Pi ^*(\theta ))d\theta \ge 0, \end{aligned}$$

because for $\theta \ge y_2$, we have $\hat{\Pi }(\theta )=1-2\int _\theta ^1 1dz$ and $\Pi ^*(\theta )=1-2\int _\theta ^1 q^*(z)dz\ge 1-2\int _\theta ^1 1dz$.

For any $y\in (y_1,y_2)$, in order to show that

$$\begin{aligned} \int _0^y (\hat{\Pi }(\theta )-\Pi ^*(\theta ))d\theta \ge 0, \end{aligned}$$

it suffices to show that there exists $z\in [y_1,y_2]$ such that $\int _0^y (\hat{\Pi }(\theta )-\Pi ^*(\theta ))d\theta$ is non-decreasing in y for $y<z$ and non-increasing in y for $y>z$ (recall that the expression is non-negative for $y=y_1$ and $y=y_2$).¹⁵

Indeed:

$$\begin{aligned} \frac{\partial }{\partial y}\left[ \int _0^y (\hat{\Pi }(\theta )-\Pi ^*(\theta ))d\theta \right] =\hat{\Pi }(y)-\Pi ^*(y), \end{aligned}$$

where, for $y\in (y_1,y_2)$, $\hat{\Pi }(y)$ is constant and $\Pi ^*(y)$ is non-decreasing. Therefore, there exists $z\in [y_1,y_2]$ such that $\int _0^y (\hat{\Pi }(\theta )-\Pi ^*(\theta ))d\theta$ is non-decreasing in y for $y<z$ and non-increasing in y for $y>z$.