Financial aid in college admissions: need-based versus merit-based

Abstract

In college admission, financial aid plays an important role in students’ enrollment decision as well as their preparation for college application. We analyze how different types of financial aid affect these decisions and admission outcomes. We consider two financial aid regimes—need-based and merit-based—in a simple college admission model and characterize respective equilibria. We find that a more competitive college has a higher admission cutoff under a need-based regime than under a merit-based regime. A less competitive college, on the other hand, benefits from a merit-based regime as it admits students with a higher average ability than it does under no aid. We next allow colleges to choose their own financial aid system so as to account for a stylized fact in the US college admissions. We show that when one college is ranked above the other, it is a dominant strategy for the higher-ranked college to offer need-based aid and for the lower-ranked college to offer merit-based aid.

Appendix I: Proofs

We start with two observations that apply to any equilibrium identified in Sect. 3. Let ${\underline{x}}\in {\mathcal {V}}$ be the ability level satisfying $\sum _{k=L,H}[\sigma ( y_i\in {\mathcal {Y}}, v_i<{\underline{x}}-\delta )]=\frac{2}{3}$.

Observation 1

Under any financial aid regime, both selective colleges have their admission cutoffs at least as high as ${\underline{x}}$. Otherwise, the measure of students who enroll in the non-selective college is smaller than $\frac{2}{3}$, violating the capacity constraint of at least one selective college. In addition, at least one selective college should have its admission cutoff exactly at ${\underline{x}}$, again, to satisfy its capacity constraint. If both selective colleges have their admission cutoffs above ${\underline{x}}$, then the measure of students who enroll in the non-selective college is larger than $\frac{2}{3}$, violating the capacity constraint of at least one selective college.

Observation 2

In equilibrium, the measure of L-type students enrolling in any one of the selective colleges is exactly $\frac{2}{3}$; similarly, the measure of H-type students enrolling in any one of the selective colleges is exactly $\frac{2}{3}$. This is because all students find the option of attending the non-selective college least preferable to all other available options of attending a selective college (with/without exerting effort and with/without receiving financial aid).

1.1 Proof of Proposition 1

We show that there exists a unique equilibrium $(e^*, x_A^*,x_B^*)$ satisfying the stated properties. We start with the following claim.

Claim. If an equilibrium $(e^*, x_A^*,x_B^*)$ exists, $x_A^*> x_B^*$.

Proof

Suppose otherwise, namely that $x_A^*\le x_B^*$. We show that the capacity constraint of A is violated. Consider the students who enroll in B. If student i enrolls in B, then $v_i \ge x_B^*-\delta$, so as to receive an admission offer from B.

Suppose first that $v_i\ge x_B^*$. He can get into B even if he exerts no effort. Since $x_A^*\le x_B^*$, he can also get into A without exerting effort. Therefore, he compares the following options:

      exerting no effort and getting into A; this resulting in utility $u(A,y_i)$.

      exerting no effort and getting into B; this resulting in utility $u(B,y_i)$.

Therefore, student i enrolls in A if $y_i\ge 0$ and in B, otherwise. Suppose, on the other hand, that $v_i \in [x_B^*-\delta , x_B^*]$. Student i can get into B only if he exerts effort. As for college A, there are two possibilities depending on where $x_A^*$ is located in relation to $x_A^*-\delta$.

Case 1. $x_A^*\le x_B^*-\delta$. If $v_i\in [x_B^*-\delta , x_B^*]$, he can get into A without exerting effort. Therefore, he compares the following options:

      exerting no effort and getting into A; this resulting in utility $u(A,y_i)$.

      exerting effort and getting into B; this resulting in utility $u(B,y_i)-c_i$.

Therefore, student i enrolls in B if $y_i\le y^{AB}(c_i)$ and in A, otherwise. Altogether, the measure of all students who enroll in B is calculated as

$$\begin{aligned} \sum _{k=L,H}[\sigma (y\le 0, v\ge x_B^*)+ \sigma (y\le y^{AB}(c_k), x_B^*-\delta \le v\le x_B^*)]=\frac{2}{3}. \end{aligned}$$

(1)

Consider next the students whose native abilities are at least as high as $x_B^*-\delta$ but who enroll in A. The measure of these students is calculated as

$$\begin{aligned} \sum _{k=L,H}[\sigma (y>0, v\ge x_B^*)+ \sigma (y>y^{AB}(c_k), x_B^*-\delta \le v_i< x_B^*)] \end{aligned}$$

(2)

Note that $y^{AB}(c_k)<0$ and by our assumption on g, for each $v\in {\mathcal {V}}$, $G(0|v)< \frac{1}{2}$. Therefore, the first term in (1) is smaller than the first term in (2) and the second term in (1) is also smaller than the second term in (2). Altogether, (2) is larger than $\frac{2}{3}$, violating the capacity constraint of A.

Case 2. $x_B^*-\delta < x_A^*$. If $v_i \in [x_A^*, x_B^*]$, student i can get into A without exerting effort. Therefore, he compares the following options:

      exerting no effort and getting into A; this resulting in utility $u(A,y_i)$.

      exerting effort and getting into B; this resulting in utility $u(B,y_i)-c_i$.

Therefore, he enrolls in B if $y_i\le y^{AB}(c_i)$ and in A, otherwise. If $v_i \in [x_B^*-\delta ,x_A^*]$, on the other hand, he can get into A only if he exerts effort. He compares the following options:

      exerting effort and getting into A; this resulting in utility $u(A,y_i)-c_i$

      exerting effort and getting into B; this resulting in utility $u(B,y_i)-c_i$

Therefore, he enrolls in B if $y_i\le 0$ and in A, otherwise. Altogether, the measure of all students who enroll in B is calculated as

$$\begin{aligned}&\sum _{k=L,H}[\sigma ( y\le 0, v\ge x_B^*)+ \sigma (y\le y^{AB}(c_k), x_A^*\le v\le x_B^*) \nonumber \\&\quad +\sigma ( y\le 0, x_B^*-\delta \le v\le x_A^*)]=\frac{2}{3}. \end{aligned}$$

(3)

Consider now the students whose native abilities are at least as high as $x_B^*-\delta$ but enroll in A. The measure of these students is calculated as

$$\begin{aligned}&\sum _{k=L,H}[\sigma ( y>0, v\ge x_B^*)+ \sigma (y> y^{AB}(c_k), x_A^*\le v\le x_A^*)\nonumber \\&\quad +\sigma ( y>0, x_B^*-\delta \le v\le x_A^*)] \end{aligned}$$

(4)

By the same reason as above, the first term in (3) is smaller than the first term in (4); since $y^{AB}(c_k)<0$, the second term in (3) is smaller than the second term in (4); and the last term in (3) is smaller than the last term in (4). Therefore, (4) is larger than $\frac{2}{3}$, violating the capacity constraint of college A. We therefore conclude that $x_A^*>x_B^*$. $\Box$

Since $x_A^*>x_B^*$, for each $k\in \{L,H\}$ and each $i\in N_k$, student i’s effort choice is uniquely determined so as to maximize his utility as follows:

• (1) If $v_i\ge x_A^*$, or $v_i< x_B^*-\delta$, or $x_B^*\le v_i< x_A^*-\delta$, $e^*_k(y_i,v_i)=0$ for all $y_i\in {\mathcal {Y}}$;

• (2) If $x_A^*-\delta \le v_i<x_A^*$ and $v_i\ge x_B^*$, $e^*_k(y_i,v_i)=1$ if and only if $y_i\ge y^{AB}(-c_k)$;

• (3) If $x_A^*-\delta \le v_i<x_A^*$ and $x_B^*-\delta \le v_i< x_B^*$, $e^*_k(y_i,v_i)=1$ for all $y_i\in {\mathcal {Y}}$;

• (4) If $x_B^*-\delta \le v_i<x_B^*$ and $v_i<x_A^*-\delta$, $e^*_k(y_i,v_i)=1$ for all $y_i\in {\mathcal {Y}}$.

On the other hand, $x_B^*<x_A^*$ implies $x_B^*={\underline{x}}$ by Observation 1. Fixing $x_B^*={\underline{x}}$, consider all possible cutoffs of A. Suppose first that A’s cutoff $x_A$ is ${\underline{x}}$. From students’ effort and enrollment choices and by the assumption that for each $v\in {\mathcal {V}}$, $G(0|v)< \frac{1}{2}$, we find that the measure of students enrolling in A is larger than $\frac{2}{3}$. Suppose next that A’s cutoff $x_A$ is ${\overline{v}}$. Trivially, the measure of students enrolling in A is smaller than $\frac{2}{3}$. As the measure of students enrolling in A is continuous and strictly decreasing in $x_A$, by the intermediate value theorem, there exists a unique $x_A^*\in [{\underline{x}}, {\overline{v}}]$ at which the measure of students enrolling in A is exactly $\frac{2}{3}$. $\square$

1.2 Proof of Proposition 2

We show that there exists a unique $(e^*,(x_A^*,F_A^*), (x_B^*,F_B^*))$ satisfying the stated properties. We start with the following clam.

Claim. If an equilibrium $(e^*,(x_A^*,F_A^*), (x_B^*,F_B^*))$ exists, $x_A^*>x_B^*$ and $F_A^*>F_B^*$.

Proof

Suppose otherwise. There are three cases.

Case 1. $x_A^*>x_B^*$ and $F_A^*\le F_B^*$. Consider the set of students enrolling in A. If student i enrolls in A, then $v_i\ge x_A^*-\delta$. To maximize his utility, he makes effort and enrollment choices as follows.

Suppose first that $v_i\ge x_A^*$. If he is an L-type student, he enrolls in A if and only if $y_i\ge y^{AB}(F_A^*-F_B^*)$; if he is an H-type student, he enrolls in A if and only if $y_i\ge 0$. Note that $y^{AB}(F_A^*-F_B^*)>0$, because $F_A^*\le F_B^*$. As shown, at each $v_i\ge x_A^*$, the L-type students’ preference threshold between A and B is larger than that of the H-type students. Therefore, for each $v_i\ge x_A^*$, college A admits more H-type students than L-type students within the set of students with ability $v_i$.

Suppose, on the other hand, that $v_i\in [x_A^*-\delta ,x_A^*]$. We compare the utilities of the available options just as above and find that at each $v_i\in [x_A^*-\delta ,x_A^*]$, the L-type students’ preference threshold between A and B is larger again than that of the H-type students, no matter where $x_B^*$ is located in relation to $x_A^*-\delta$. Therefore, for each $v_i\in [x_A^*-\delta ,x_A^*]$, college A admits more H-type students than L-type students within the set of students with ability $v_i$.

Since A has a capacity of $\frac{2}{3}$, the measure of H-type students enrolling in A is larger than $\frac{1}{3}$ and the measure of L-type students enrolling in A is smaller than $\frac{1}{3}$. By Observation 2, in turn, the measure of L-type students enrolling in B is larger than $\frac{1}{3}$. Since $F_A^*\le F_B^*$, the total expenditure of A is smaller than that of B, violating the budget constraint of at least one selective college.

Case 2. $x_A^*\le x_B^*$ and $F_A^*> F_B^*$. Consider the set of students who enroll in B. Their abilities are at least as high as $x_B^*-\delta$. Consider each $v_i\ge x_B^*-\delta$. By a similar argument as in Case 1, we find that for each $v_i\ge x_B^*-\delta$, college B admits more H-type students than L-type students within the set of students with ability $v_i$.

Since B has a capacity of $\frac{2}{3}$, the measure of H-type students enrolling in B is larger than than $\frac{1}{3}$ and the measure of L-type students enrolling in B is smaller than $\frac{1}{3}$. By Observation 2, in turn, the measure of L-type students enrolling in A should be larger than $\frac{1}{3}$. Since $F_A^*> F_B^*$, the total expenditure of A is larger than that of B, violating the budget constraint of at least one selective college.

Case 3. $x_A^*\le x_B^*$ and $F_A^*\le F_B^*$. Consider the equilibrium under no aid, identified in Proposition 1, and denote it by $(x_A^0,x_B^0, e^0)$. By comparing it with the equilibrium under a need-based regime, we will show that the enrollment of the L-type students at A is smaller than the enrollment of the L-type students at B under a need-based regime. Since $F_A^*\le F_B^*$, this violates the budget constraint of at least one selective college, completing the proof of Claim.

Since $x_A^*\le x_B^*$ and $x_A^0>x_B^0$, by Observation 1, $x_A^*=x_B^0={\underline{x}}$ holds. Consider the set of H-type students enrolling in A in each equilibrium. As shown in Proposition 1, under no aid, any H-type student i enrolling in A has $v_i\ge x_A^0-\delta$ and $y_i\ge 0$. Under a need-based regime where $x_A^*\le x_B^*$, on the other hand, any H-type student i such that $v_i\ge x_A^*-\delta$ and $y_i\ge 0$ enrolls in A. Together with $x_A^*={\underline{x}}<x_A^0$, we conclude that the measure of H-type students enrolling in A under a need-based regime is larger than that under no aid (denote this statement by ($\dag$)).

By Proposition 1, the measure of H-type students enrolling in A under no aid is proven to be larger than $\frac{1}{3}$. By $(\dag )$, the measure of H-type student enrolling in A under a need-based regime will be even larger than $\frac{1}{3}$. Since A has a capacity of $\frac{2}{3}$, the measure of L-type students enrolling in A under a need-based regime will then be smaller than $\frac{1}{3}$. By Observation 2, in turn, the measure of L-type students enrolling in B will be larger than $\frac{1}{3}$. This completes the proof of Claim.  $\square$

Existence: Choose any two non-negative numbers $F_A$ and $F_B$ and let them be the respective amounts of need-based aid offered by A and B. Taking $F_A$ and $F_B$ as given, we identify colleges’ cutoffs at which the colleges’ capacity constraints are satisfied, assuming that the students choose their effort and enrollment to maximize their utilities. It is easy to verify that the cutoffs are uniquely determined as in the proof of Proposition 1. Under these cutoffs and students’ effort and enrollment choices, we identify the measures of L-type students enrolling in A and B, respectively, denoting them by $n_A^L(F_A,F_B)$ and $n_B^L(F_A,F_B)$. Note that $n_A^L(\cdot ,\cdot )$ and $n_B^L(\cdot ,\cdot )$ are continuous in $F_A$ and $F_B$ and are bounded in $[0,\frac{2}{3}]$. Now, define the function $\Phi :[0,\frac{2}{3}]\times [0,\frac{2}{3}]$ by setting

$$\begin{aligned} \Phi (l_A,l_B)=\left(n_A^L\left(\frac{M}{l_A},\frac{M}{l_B}\right), n_B^L\left(\frac{M}{l_A},\frac{M}{l_B}\right)\right).\end{aligned}$$

Since $\Phi (\cdot ,\cdot )$ is continuous on $[0,\frac{2}{3}]\times [0,\frac{2}{3}]$ and its domain is compact and convex, there exists a fixed point. At each fixed point, the colleges’ capacity constraints and budget constraints are trivially satisfied by construction. The students’ effort and enrollment choices are uniquely determined as well to maximize their utilities.

Uniqueness: Suppose, by contradiction, that $(e^*,(x_A^*,F_A^*), (x_B^*,F_B^*))$ and $({\bar{e}}^*,({\bar{x}}_A^*,{\bar{F}}_A^*), ({\bar{x}}_B^*,{\bar{F}}_B^*))$ are equilibria of this problem. Denote them by ${\mathcal {E}}$ and $\bar{{\mathcal {E}}}$, respectively. By Claim above, $x_A^*>x_B^*$ and $F_A^*>F_B^*$ and ${\bar{x}}_A^*>{\bar{x}}_B^*$ and ${\bar{F}}_A^*>{\bar{F}}_B^*$. By Observation 1, we have $x_B^*={\bar{x}}_B^*={\underline{x}}$. Without loss of generality, suppose that $x_A^*< {\bar{x}}_A^*$. To satisfy the capacity constraint of A under both equilibria, we should have $F_A^*-F_B^*<{\bar{F}}_A^*-{\bar{F}}_B^*$.

Consider the H-type students who enroll in A under the two equilibria. Since they are not eligible for aid and $x_A^*< {\bar{x}}_A^*$, the measure of H-type students enrolling in A under ${\mathcal {E}}$ is larger than the measure of H-type students enrolling in A under $\bar{{\mathcal {E}}}$. Since A has a capacity of $\frac{2}{3}$, in turn, the enrollment of the L-type students at A under $\bar{{\mathcal {E}}}$ is larger than the enrollment of the L-type students at A under ${\mathcal {E}}$. This implies that (1) $F_A^*>{\bar{F}}_A^*$ to satisfy college A’s budget constraint and (2) together with Observation 2, the enrollment of the L-type students at B under $\bar{{\mathcal {E}}}$ is smaller than the enrollment of the L-type students at B under ${\mathcal {E}}$. Note that $F_A^*-F_B^*<{\bar{F}}_A^*-{\bar{F}}_B^*$ and (1) jointly imply that ${\bar{F}}_B^*<F_B^*$. Together with (2), college B’s budget constraint in at least one of the equilibria is violated, a contradiction.

Next, let $(e^*, x_A^*,x_B^*)$ be the equilibrium under no aid identified in Proposition 1 and let $(e^*, (x_A^*,F^*_A),$ $(x_B^*,F^*_B))$ be the equilibrium under a need-based regime identified in Proposition 2. Note that ${\bar{x}}_B^*<{\bar{x}}_A^*$ and $x_B^*<x_A^*$. By Observation 1, $x_B^*={\bar{x}}_B^*={\underline{x}}$. We now prove that $x_A^*< {\bar{x}}_A^*$. Suppose otherwise, namely that $x_A^*\ge {\bar{x}}_A^*$. We will prove that the measure of students enrolling in A under a need-based regime is larger than the measure of students enrolling in A under no aid, violating its capacity constraint of $\frac{2}{3}$.

Consider any L-type student i. (1) Suppose that $v_i\ge x_A^*$. He enrolls in A under no aid if $y_i\ge 0$; enrolls in A under a need-based regime if $y_i\ge y^{AB}(F_A^*-F_B^*)$. (2) Suppose next that $v_i\in ([x_A^*-\delta ,x_A^*]\setminus [{\bar{x}}_A^*-\delta , {\bar{x}}_A^*])\setminus [{\underline{x}}-\delta , {\underline{x}}]$. He enrolls in A under no aid if $y_i\ge y^{AB}(-c_L)$; enrolls in A under a need-based regime if $y_i\ge y^{AB}(F_A^*-F_B^*)$. (3) Suppose that $v_i\in ([x_A^*-\delta ,x_A^*]\cap [{\bar{x}}_A^*-\delta , {\bar{x}}_A^*]){\setminus } [{\underline{x}}-\delta , {\underline{x}}]$. He enrolls in A under no aid if $y_i\ge y^{AB}(-c_L)$; enrolls in A under a need-based regime if $y_i\ge y^{AB}(F_A^*-F_B^*-c_L)$. (4) Suppose that $v_i\in ([x_A^*-\delta ,x_A^*]\cap [{\bar{x}}_A^*-\delta , {\bar{x}}_A^*])\cap [{\underline{x}}-\delta , {\underline{x}}]$. He enrolls in A under no aid if $y_i\ge 0$; enrolls in A under a need-based regime if $y_i\ge y^{AB}(F_A^*-F_B^*)$.¹⁵ In (1) to (4), note that $y^{AB}(F_A^*-F_B^*)< 0<y^{AB}(-c_L)$ and $y^{AB}(F_A^*-F_B^*-c_L)<y^{AB}(-c_L)$. This implies that for L-type students, the preference threshold between A and B under no aid is larger than that under a need-based regime at each $v_i\ge x_A^*-\delta$. Therefore, for each $v_i\ge x_A^*-\delta$, college A admits more students under a need-based regime than under no aid within the set of L-type students with ability $v_i$. Next, consider any H-type student i. By the same argument applied to the L-type students, we again find that for each $v_i\ge x_A^*-\delta$, college A admits more students under a need-based regime than under no aid within the set of H-type students with ability $v_i$.

Note that all students enrolling in A should have abilities at least as high as $x_A^*-\delta$ and $x_A^*$ is chosen so that the measure of students enrolling in A under no aid is exactly $\frac{2}{3}$. Since ${\bar{x}}_A^*\le x_A^*$ and college A admits more students under a need-based regime than under no aid over the ability range of $[x_A^*-\delta ,{\overline{v}}]$, the measure of students enrolling in A under a need-based regime exceeds $\frac{2}{3}$, violating the capacity constraint of A. Therefore, $x_A^*< {\bar{x}}_A^*$ should hold. The other statements of the proposition immediately follow from this cutoff change.

1.3 Proof of Proposition 3

We show that there exists a unique equilibrium $(e^*, (x_A^*, m_A^*),(x_B^*, m_B^*))$ satisfying the stated properties. We start with the following claim.

Claim. If an equilibrium $(e^*, (x_A^*, m_A^*),(x_B^*, m_B^*))$ exists, for each $k\in \{A,B\}$, $m_k^*-\delta >x_k^*$, $m_A^*> m_B^*$, and $x_A^*>x_B^*$.

Proof

We first show that $m_A^*-\delta >x_A^*$. Suppose otherwise, that $x_A^*\ge m_A^*-\delta$. Since ${K\le \frac{1}{2}}$, the measure of students who enroll in A receiving merit-based aid is $\frac{2}{3}K\le \frac{1}{3}$. Therefore, the measure of students enrolling in A without receiving merit-based aid should be at least $\frac{1}{3}$ (denote this statement by $(*)$). Note that the abilities of these students belong to $[x_A^*-\delta , m_A^*-\delta ]$, a subset of $[x_A^*-\delta ,x_A^*]$, since $x_A^*\ge m_A^*-\delta$. By our assumption that for each $v_i\in {\mathcal {V}}$, $\sigma (y\in {\mathcal {Y}},v_i-\delta<v<v_i)<\frac{1}{6}$, we have $\sum _{k\in \{L,H\}}[\sigma (y\in {\mathcal {Y}}, v\in [x_A^*-\delta ,x_A^*])]<\frac{1}{3}$. Therefore, the measure of students enrolling in A without receiving aid will also be smaller than $\frac{1}{3}$, a contradiction to $(*)$. Therefore, $m_A^*-\delta >x_A^*$ should hold. By the same argument, $x_B^*\le m_B^*-\delta$.

Next, we show that $m_A^*>m_B^*$. Suppose otherwise, namely, that $m_A^*\le m_B^*$. Denote the measure of students who enroll in A (or in B) receiving merit-based aid in equilibrium by $e_A^m$ (or $e_B^m$). All students who enroll in B receiving merit-based aid have their ability levels at least as high as $m_B^*-\delta$. Since $m_A^*\le m_B^*$, all students enrolling in B with merit-based aid should have non-positive preference parameters, irrespective of whether they are L-type or H-type. In addition, any student i of each type such that $v_i\ge m_B^*-\delta$ and $y_i>0$ enrolls in A, receiving merit-based aid (denote this statement by $(\S )$). Altogether,

$$\begin{aligned} \begin{array}{lll}e_B^m\le \sum _{k\in \{L,H\}}[\sigma (y\le 0, v\ge m_B^*-\delta )]< & {} \sum _{k\in \{L,H\}}[\sigma (y>0, v\ge m_B^*-\delta )]\le e_A^m,\end{array} \end{aligned}$$

where the second inequality comes from the assumption that for each $v\in {\mathcal {V}}$, $G(0|v)< \frac{1}{2}$ and the last inequality comes from $(\S )$. Therefore, we obtain a contradiction to $e_A^m=e_B^m=\frac{2}{3}K$.

Lastly, we prove $x_A^*>x_B^*$. Suppose otherwise, namely, that $x_A^*\le x_B^*$. Together with the inequalities shown above, we have $m_A^*>m_B^*>m_B^*-\delta \ge x_B^*\ge x_A^*$. We claim that the measure of students who enroll in A without receiving merit-based aid is larger than $\frac{2}{3}-\frac{2}{3}K$, violating the capacity constraint of A. The abilities of these student belong to $[x_A^*-\delta ,m_A^*-\delta ]$. Consider, on the other hand, the students who enroll in B without receiving merit-based aid. The measure of these students is also $\frac{2}{3}-\frac{2}{3}K$ and their abilities belong to $[x_B^*-\delta , m_B^*-\delta ]$, which is a subset of $[x_A^*-\delta ,m_A^*-\delta ]$. In addition, the preference parameters of these students are non-positive: consider any student i among them. If $v_i\in [x_B^*,m_B^*-\delta ]$, then he can get into both selective colleges without exerting effort, but he cannot receive aid from both. Therefore, he enrolls in B if $y_i\le 0$. If $v_i\in [x_B^*-\delta ,x_B^*]$, then he can get into B if he exerts effort, but he cannot receive aid from both selective colleges. He can get into A, on the other hand, with or without exerting effort, depending on where $x_A^*$ is located in relation to $x_B^*-\delta$: his preference threshold between A and B is either 0 or $y^{AB}(c_i)(\le 0)$.

Since we assume that for each $v\in {\mathcal {V}}$, $G(0|v)< \frac{1}{2}$, altogether, the measure of students enrolling in A over the ability range $[x_A^*-\delta ,m_A^*-\delta ]$ is larger than that enrolling in B over the same range, where the latter is exactly  $\frac{2}{3}-\frac{2}{3}K$. Altogether, the measure of students enrolling in A without receiving aid exceeds $\frac{2}{3}-\frac{2}{3}K$, a contradiction.  $\square$

Existence: Let ${\overline{m}}, {\underline{m}}\in {\mathcal {V}}$ be the numbers such that

$$\begin{aligned}&\sum _{k\in \{L,H\}}\sigma ({y\ge 0}, {v\ge {\overline{m}}-\delta })=\frac{2}{3}K;\\&\sum _{k\in \{L,H\}}\sigma ({y\in {\mathcal {Y}}}, {v\ge {\underline{m}}-\delta })=\frac{4}{3}K. \end{aligned}$$

Note that if there exists an equilibrium $(e^*, (x_A^*, m_A^*),(x_B^*, m_B^*))$, then ${\underline{m}}\le m_A^*\le {\overline{m}}$ should hold. Suppose otherwise. If $m_A^*>{\overline{m}}$, then the measure of students who enroll in college A receiving merit-based aid is smaller than $\frac{2}{3}K$, a violation of the budget constraint of A. If $m_A^*<{\underline{m}}$, on the other hand, then $m_B^*<{\underline{m}}$, since $m_B^*< m_A^*$ in equilibrium. Altogether, the measure of students who enroll in A or B receiving merit-based aid is larger than $\frac{4}{3}K$. This is a violation of the budget constraint of at least one selective college. Therefore, $m_A^*\in [{\underline{m}},{\overline{m}}]$ if the equilibrium exists.

Now, consider any pair $(m_A,x_A)\in {\mathcal {V}}\times {\mathcal {V}}$ such that ${\underline{m}}\le m_A\le {\overline{m}}$ and ${\underline{x}}\le x_A\le m_A$ and assume that $m_A$ and $x_A$ are cutoffs for merit-based aid and admission at A. Taking $(m_A,x_A$) as given, we show that there exists a unique cutoff for merit-based aid of B, $m_B\in {\mathcal {V}}$, at which the measure of students who enroll in B receiving merit-based aid is exactly $\frac{2}{3}K$. First, if $m_B={\overline{v}}$, the measure of students who enroll in B receiving merit-based aid is trivially smaller than $\frac{2}{3}K$. Next, if $m_B={\underline{v}}$, the measure of students who enroll in B receiving aid is larger than $\frac{2}{3}K$. Lastly, the measure of students who enroll in B receiving aid is continuous and decreasing in $m_B$. Therefore, by the intermediate value theorem, there exists a unique cutoff $m_B$. We write it as $m_B(m_A,x_A)$.

Next, taking $(m_A,x_A)$ and $m_B(m_A,x_A)$ as given, we show that there exists a unique cutoff for admission of B, $x_B\in {\mathcal {V}}$, at which the measure of students enrolling in B is exactly $\frac{2}{3}$. First, if $x_B=m_B(m_A,x_A)$, the measure of students enrolling in B is trivially smaller than $\frac{2}{3}$. Next, if $x_B={\underline{v}}$, then the measure of students enrolling in B is larger than $\frac{2}{3}$. Lastly, the measure of students enrolling in B is decreasing and continuous in $x_B$. Therefore, by the intermediate value theorem, there exists a unique cutoff $x_B$. We write it as $x_B(m_A,x_A)$. Note that $m_B(\cdot )$ and $x_B(\cdot )$ are continuous in $(m_A,x_A)$.

We now return to $(m_A, x_A)$ to consider college A’s capacity and budget constraints. First, choose any $m_A\in [{\underline{m}}, {\overline{m}}]$ and assume that $m_A$ is a cutoff for merit-based aid at A. Taking $m_A$ as given, we show that there exists $x_A\in {\mathcal {V}}$ at which the measure of students who enroll in A is exactly $\frac{2}{3}$. First, if $x_A=m_A$, the measure of students enrolling in A is smaller than $\frac{2}{3}$ trivially. Next, if $x_A={\underline{x}}$, the measure of these students is at least $\frac{2}{3}$. Lastly, $m_B(m_A,\cdot )$, $x_B(m_A,\cdot )$, and the measure of students enrolling in A is continuous in $x_A$. Therefore, by the intermediate value theorem, there exists such a cutoff $x_A$. We write it as $x_A(m_A)$ and we write $m_B(\cdot )$ and $x_B(\cdot )$ as functions of $m_A$ as well. Note that $x_A(\cdot )$, $m_B(\cdot )$, $x_B(\cdot )$ are continuous in $m_A$.

Lastly, we identify $m_A$ at which the budget constraint of A is satisfied. By a similar argument as above, we show that there exists such $m_A$: If $m_A={\overline{m}}$, the measure of students who enroll in A receiving merit-based aid is smaller than $\frac{2}{3}K$; if $m_A={\underline{m}}$, the measure of these students is larger than $\frac{2}{3}K$; lastly, $x_A(\cdot )$, $m_B(\cdot )$, and $x_B(\cdot )$ are continuous in $m_A$. Therefore, by the intermediate value theorem, there exists such a cutoff $m_A$. We write it as $m_A^*$. We also write the values of $x_A(\cdot )$, $m_B(\cdot )$, and $x_B(\cdot )$ at $m_A^*$ as $x_A^*$, $m_B^*$, and $x_B^*$, respectively. Under these cutoffs, students make effort choices, $e^*$, just as described in Proposition 3; the two colleges’ capacity and budget constraints are also satisfied.

Uniqueness: Suppose, by contradiction, that $(e^*, (x_A^*, m_A^*),(x_B^*, m_B^*))$ and $({\overline{e}}^*, ({\overline{x}}_A^*, {\overline{m}}_A^*),({\overline{x}}_B^*, {\overline{m}}_B^*))$ are equilibria of this problem. By Claim above, $m_B^*<m_A^*$ and ${\overline{m}}_B^*<{\overline{m}}_A^*$. In addition, since $x_A^*>x_B^*$ and ${\overline{x}}_A^*>{\overline{x}}_B^*$, by Observation 1, $x_B^*={\overline{x}}_B^*={\underline{x}}$. We first show that $m_A^*={\overline{m}}_A^*$. Suppose otherwise, without loss of generality, that $m_A^*>{\overline{m}}_A^*$. Then, $m_B^*>m_A^*-\delta$ and $m_B^*>{\overline{m}}_B^*$ should hold. Otherwise, we identify the students’ choices to maximize their utilities and find that the measure of students who enroll in A receiving merit-based aid at $(e^*, (x_A^*, m_A^*),(x_B^*, m_B^*))$ is smaller than that at $({\overline{e}}^*, ({\overline{x}}_A^*, {\overline{m}}_A^*),({\overline{x}}_B^*, {\overline{m}}_B^*))$, a contradiction. Therefore, $m_A^*>{\overline{m}}_A^*$ and $m_B^*>{\overline{m}}_B^*$.

Next, the measure of students who enroll in either one of A and B, receiving merit-based aid is $\frac{4}{3}K$ in each equilibrium. Given $m_A^*>{\overline{m}}_A^*$ and $m_B^*>{\overline{m}}_B^*$, we should have $x_A^*>m_B^*-\delta$ and $x_A^*>{\overline{x}}_A^*$. Otherwise, again, it is easy to check that at least one selective college necessarily violates the joint capacity/budget constraint of $\frac{4}{3}K$. Therefore, $m_A^*>{\overline{m}}_A^*$, $m_B^*>{\overline{m}}_B^*$, and $x_A^*>{\overline{x}}_A^*$.

Lastly, consider the students who enroll in A (with or without merit-based aid) and the students who enroll in B receiving merit-based aid. The measure of all these students should be $\frac{2}{3}+\frac{2}{3}K$ in each equilibrium. Given $x_A^*>{\overline{x}}_A^*$, we claim that $x_B^*>x_A^*-\delta$ and $x_B^*>{\overline{x}}_B^*$ should hold. Otherwise, again, at least one selective college necessarily violates the aforementioned joint capacity/budget constraint of $\frac{2}{3}+\frac{2}{3}K$. Therefore, $x_B^*>x_A^*-\delta$ and $x_B^*>{\overline{x}}_B^*$. However, this contradicts $x_B^*={\overline{x}}_B^*={\underline{x}}$. Therefore, $m_A^*={\overline{m}}_A^*$.

We next prove $m_B^*={\overline{m}}_B^*$. Suppose otherwise, without loss of generality, that $m_B^*>{\overline{m}}_B^*$. By a similar argument as above, we should have $x_A^*>{\overline{x}}_A^*$ and $x_B^*>{\overline{x}}_B^*$, contradicting $x_B^*={\overline{x}}_B^*={\underline{x}}$. The same argument applies to show that $x_A^*={\overline{x}}_A^*$, so we omit it.

Next, let $(e^*, x_A^*,x_B^*)$ be the equilibrium with no financial aid and let $({\bar{e}}^*, ({\bar{m}}_A^*,{\bar{x}}_A^*),({\bar{x}}_B^*, {\bar{m}}_B^*))$ be the equilibrium under a merit-based regime. Note that ${\bar{m}}_A^*>{\bar{m}}_B^*$. Therefore, as can be shown in Proposition 3, there are some students who enroll in B, even though their ability levels are above $x_A^*-\delta$ and their preference parameters are non-negative.

Since ${\bar{x}}_B^*<{\bar{x}}_A^*$ and $x_B^*<x_A^*$, by Observation 1, $x_B^*={\bar{x}}_B^*={\underline{x}}$ should hold. We prove that ${\bar{x}}_A^*<x_A^*$. Suppose otherwise, that ${\bar{x}}_A^*\ge x_A^*$. By a similar argument as in the proof of Proposition 2, we find that at each $v_i\ge {\bar{x}}_A^*-\delta$, each student’s preference threshold between A and B under a merit-based regime is at least as large as that under no aid. Therefore, college A admits more students students under no aid than under a merit-based regime over the ability range of $[{\bar{x}}_A^*-\delta ,{\bar{v}}]$. Note that ${\bar{x}}_A^*\ge x_A^*$ and ${\bar{x}}_A^*$ is chosen so that the measure of students enrolling in A under a merit-based regime is exactly $\frac{2}{3}$. Therefore, the measure of students enrolling in A under no aid would exceed $\frac{2}{3}$, violating the capacity constraint of A. Therefore, ${\bar{x}}_A^*<x_A^*$. The other statements of the proposition immediately follow from this cutoff change.

1.4 Proof of Proposition 4

For each $(w_A,w_B)$ and each $s\in \{A,B\}$, let $x_s^*(w_A,w_B)$ and $m_s^*(w_A,w_B)$ be the equilibrium cutoffs for admission and merit-based aid of college s in equilibrium, respectively (whichever applicable); let $F_s^*(w_A,w_B)$ be the amount of need-based aid per student at college s and let $e^L_s(w_A,w_B)$ be the total enrollment of L-type students at college s in equilibrium. Recall that for each $v_i\in {\mathcal {V}}$, $\sigma (y\in {\mathcal {Y}},v_i-\delta<v<v_i)\le \frac{1}{12}$. Then, $\Delta \equiv \max _{v_i\in {\mathcal {V}}}\sigma (y\in {\mathcal {Y}},v_i-\delta<v<v_i)\le \frac{1}{12}$. Also, recall that we assume that $F\ge 3M$ and $c_L,c_H\le \frac{3M}{2}$. There are four possibilities.

Case 1: $(w_A,w_B)=(nb,nb)$. As analyzed in Proposition 2, there exists a unique equilibrium: $e^L_A(nb,nb)=e^L_B(nb,nb)=\frac{1}{3}$ and $F_A^*(nb,nb)=F_B^*(nb,nb)=3M$ by the budget and capacity constraints.

Case 2: $(w_A,w_B)=(nb,mb)$. Consider a student who receives an admission offer from A.

• (1) Suppose that student i has $v_i\ge x_A^*(nb,mb)$ and $v_i\ge m^*_B(nb,mb)$. Then, A offers $F_A^*(nb,mb)$ and B offers F. Since $y^{AB}(-F)={\underline{y}}$, he chooses to enroll in A, no matter what his type is.

• (2) Suppose that student i has $v_i\in [x_A^*(nb,mb)-\delta , x_A^*(nb,mb)]$ and $v_i\in [m^*_B(nb,mb)-\delta , m^*_B(nb,mb)]$. Then, A offers $F_A^*(nb,mb)$ and he has to exert an effort to get into A; and B offers F and he has to exert an effort to receive the merit-based aid. By the same argument as (1), he chooses to enroll in A, no matter what his type is.

• (3) Suppose that student i has $v_i\in [x_A^*(nb,mb)-\delta , x_A^*(nb,mb)]$ and $v_i\ge m^*_B(nb,mb)$. Then, A offers $F_A^*(nb,mb)$ and he has to exert an effort to get into A, while he receives F even without exerting an effort. If he is L-type, he chooses to enroll in A for all $y_i$, because $F_A^*(nb,mb)-c_L>0$ and $y^{AB}(-F)={\underline{y}}$; if he is H-type, he chooses to enroll in A if and only if $y_i\ge y^{AB}(-F-c_H)$.

By (1) to (3), the minimal enrollment of L-type students at A is $\frac{1}{3}$, in which case, the budget constraint implies that $F^*_A(nb,mb)\ge 3M$ and therefore, we conclude that $c_L,c_H< F^*_A(nb,mb)$. Going back to (2), in turn, $F_A^*(nb,mb)-c_L\ge 0$ and thus, $y^{AB}(F_A^*(nb,mb)-c_L-F)\le y^{AB}(-F)={\underline{y}}$, implying that all L-type students who have to exert an effort to get into A eventually enroll in A, receiving need-based aid, even if college B offers F. Note that, in contrast, $y^{AB}(-F-c_H)\ge {\underline{y}}$, implying that some H-type students could enroll in B after receiving F as he has to exert an effort to get into A.

Case 3: $(w_A,w_B)=(mb,nb)$. By a similar argument as above, if student i is H-type and $v_i\ge x_A^*(mb,nb)-\delta$, then he always enrolls in A, because he is not eligible for need-based aid that B offers. Given this observation and the fact that $x^*_B(mb,nb)$ is fixed constant in all equilibria, $e_B^L(mb,nb)$ is minimal if and only if $e_A^L(mb,nb)$ is maximal. This is when all L-type students with $v_i\ge x_A^*(mb,nb)-\delta$ enroll in A, namely, $e_A^L(mb,nb)\le \frac{1}{3}\le e_B^L(mb,nb)$. By the budget constraint, in turn, $F^*_B(mb,nb)\cdot e_B^L(mb,nb)=M$ and therefore, $F^*_B(mb,nb)\le 3M<F$. Since $F_B^*(mb,nb)\le F$, any L-type student i with $v_i\ge x_A^*(mb,nb)$ enroll in college A, even if college B offers need-based aid; and an L-type student i with $v_i\in [x_A^*(mb,nb)-\delta , x_A^*(mb,nb)]$ enrolls in A if and only if $y_i\ge y^{AB}(-F_B^*(mb,nb)-c_L)$.

Case 4: $(w_A,w_B)=(mb,mb)$. Suppose that student i has $v_i\ge x_A^*(mb,mb)$. Then, he enrolls in A, no matter what his type is. Suppose that student i has $v_i\in [x_A^*(mb,mb)-\delta ,x_A^*(mb,mb)]$. When he is L-type, he enrolls in B if and only if he receives merit-based aid from B without exerting any effort and $y_i\le y^{AB}(-F-c_L)$; when he is H-type, he enrolls in B if and only if he receives merit-based aid from B without exerting any effort and $y_i\le y^{AB}(-F-c_H)$.

We first observe that (nb, nb) and (mb, mb) do not constitute an equilibrium, because at (nb, nb), college B has an incentive to deviate to mb, switching to Case 2, as this deviation increases the average ability of incoming students. Similarly, at (mb, mb), college A has an incentive to deviate to nb, switching to Case 2, as this deviation increases the average ability of incoming students. A similar argument applies to prove that (mb, nb) is not an equilibrium: at (mb, nb), college A has an incentive to deviate to nb, as this deviation increases the average ability of incoming students. From these arguments, we conclude that (nb, mb) is the unique dominant equilibrium, regardless of the distribution g.  $\square$