Appendix A: Proofs
Proof of Proposition 1
To prove the proposition, first note that
$$\begin{aligned} \frac{\partial \Psi (s,a)}{\partial s}=e^{-\frac{a(1-d)}{d}(1-e^{-s d})}\frac{(1-d)(1-e^{-s d})}{d}>0. \end{aligned}$$
If individuals are homogeneous and \(s_i=s\) for all \(i\in N\), the network marriage rate is
$$\begin{aligned} m(s,a)= & {} a(1-d)\Upsilon (s)+(1-a) \Psi (s,a). \end{aligned}$$
Hence,
$$\begin{aligned} \frac{\partial m(s,a)}{\partial s}(s)= a(1-d)d e^{-s d}+(1-a) (1-\Psi (s))\frac{(1-d)(1-e^{-s d})}{d}>0. \end{aligned}$$
It is trivial to see that \(\partial \Psi (s,a)/\partial a >0\), while \(\Upsilon (s)\) is independent of a. Finally,
$$\begin{aligned} \frac{\partial m(s,a)}{\partial a}(s)= & {} (1-d)\Upsilon (s)-\Psi (s,a)+(1-a) \frac{\partial \Psi (s,a)}{\partial a}\\= & {} (1-d)\Upsilon (s)-\Psi (s,a) +(1-a) (1-\Psi (s))\frac{(1-d)(1-e^{-s d})}{d} \end{aligned}$$
which is null when \(a\rightarrow 0\), equal to \((1-s)(1-e^{-sd})\) when \(a\rightarrow 1\). Furthermore, \(\partial ^2 m(s,a)/\partial a\) goes to \((1-d)(1-e^{-sd})(1+1/d)>0\) when \(a\rightarrow 0\), and to \(-d-(1-d)e^{-sd}+(1-\Psi (s,a))<0\) when \(a\rightarrow 1\). So, m(s, a) in first increasing and then decreasing in a. This concludes the proof of Proposition 1. \(\square \)
Proof of Proposition 2
Suppose an interior equilibrium exists. Consider a profile \(\mathbf {s}\) where \(s_{j}=s\), \(\forall j\ne i\). Note that \(\Upsilon (s_i,s)\) does not depend on \(s_i\), while \(\Psi (s_i,s)\) can be written as
$$\begin{aligned} \Psi (s_i,s)= 1-e^{-\frac{a(1-d)h}{d s} s_i \left( 1-e^{-d s}\right) }. \end{aligned}$$
Then, given (1), an interior equilibrium \(s^*\) solves:
$$\begin{aligned} \frac{\partial EU_{i}}{\partial s_{i}}(s^*,s^*)=-d(1-a)\frac{\partial \phi _{i}}{\partial s_{i}}(s^*,s^*)-c=0. \end{aligned}$$
in large marriage markets, \(s^{*}\) must solve (2). The LHS of this expression is decreasing in \(s^{*}\) because both \(\left( 1-e^{-s^{*}d}\right) /s^{*}\) and \(e^{-\frac{a(1-d)}{d}\left( 1-e^{-s^{*}d}\right) }\) are decreasing in \(s^{*}\). Furthermore, when \(s^{*}\) goes to 0, the LHS converges to \(a d(1-a)(1-d)\), while when \(s^{*}\) goes to infinity the LHS converges to 0. Since marginal returns are continuous in \(s_{i}\), it follows that an interior symmetric equilibrium exists if and only if \(c<a d(1-a)(1-d)\), in which case there is only one symmetric interior equilibrium. This concludes the proof of Proposition 2. \(\square \)
Proof of Proposition 3
We first prove part 1. We derive \(\partial s^{*}/\partial a\) by implicit differentiation of (2) and obtain
$$\begin{aligned} \frac{\partial s^{*}}{\partial a}= \frac{(1-e^{-s^{*}d})\left[ 1-2 a-a(1-a)(1-d)\frac{1-e^{-s^{*}d}}{d} \right] }{a(1-a)\left[ (a(1-d)e^{-s^*d}+\frac{1}{s})(1-e^{-s^*d})-de^{-s^*d}\right] }. \end{aligned}$$
(A-1)
First, the denominator is positive. Indeed, its term in the square parenthesis is strictly increasing in a and is equal to \((1-e^{-s^*d})/s-de^{-s^*d}\) when \(a=0\). Taking the derivative of this with respect to d, we get \(sde^{-s^*d}\), which is also strictly positive. Finally, the limit of \((1-e^{-s^*d})/s-de^{-s^*d}\) as \(d\rightarrow 0\) is 0. Hence, the denominator is positive.
This implies that the sign of this derivative depends on the sign of the numerator, we see that the term in square parenthesis, which we denote by
$$\begin{aligned} \left( a(1-d)e^{-s^*d}+\frac{1}{s} \right) (1-e^{-s^*d})-de^{-s^*d} \end{aligned}$$
(A-2)
is 1 when \(a=0\), while it is equal to \(-1\) when \(a=1\). Defining \(x=(1-d)\left( 1-e^{-s^{*}d}\right) /d\), we rewrite the expression (A-2) as \(1-2a-a(1-a) x\). While \(1-2a-a(1-a) x=0\) admits two solutions, only \((2 + x - \sqrt{4 + x^2})/(2 x)\in [0,1]\). Hence, there is a unique \(\bar{a}\) such that if \(a\le \bar{a}\), (A-1) is positive, while if \(a>\bar{a}\), (A-1) is negative. This concludes the first part of the proof of Proposition 3.
As for the second part of Proposition 3, the change in the matching rate \(\Psi \) when a changes is described by:
$$\begin{aligned} \frac{\partial \Psi }{\partial a}= (1-\Psi )(1-d)\left[ \frac{1-e^{-s^{*}d}}{d}+ \frac{\partial s^{*}}{\partial a} a e^{-s^{*}d} \right] . \end{aligned}$$
(A-3)
While the first term in the square parenthesis is always positive, the sign of the second one depends on the sign of \(\partial s^{*}/\partial a\). Using part 1, \(\partial s^{*}/\partial a\) is positive when a is sufficiently low, and hence \(\partial \Psi /\partial a\) would also be positive. When a tends to 1, inspecting (A-1), we see that the numerator goes to \(-(1-e^{-s^{*}d})<0\) when \(a\rightarrow 1\), while the denominator goes to 0 from above (as we have just shown that it is always positive). Hence, if a is sufficiently large, s goes to zero, so that the first term in the square brackets of (A-3) goes to zero, while the second part is negative; hence (A-3) is negative.
The change in the matching rate \(\Upsilon \) when a changes is described by:
$$\begin{aligned} \frac{\partial \Upsilon }{\partial a}= d(1-d) \left[ (1-e^{-s^{*}d})+ad e^{-s^{*}d} \frac{\partial s^{*}}{\partial a} \right] . \end{aligned}$$
(A-4)
Following a similar argument as for (A-3), we can see that (A-4) is positive for a low enough a and negative for a is large enough.
Given that \(m(s,a)=\Upsilon (s,a)+\Psi (s,a)\), also m(s, a) is increasing in a for a low enough a and decreasing for a is large enough. This concludes the proof of Proposition 3. \(\square \)
Proof of Proposition 5
To prove existence in the socialization effort game in a large society with heterogeneous individuals, we first analyze separately on low types and high types, respectively. We derive conditions for an interior equilibrium to exists in each scenario, and then we combine these results in order to prove existence in the market with two types. We focus on symmetric equilibria in which \(s_i=s_h\) for all \(i\in M_H\), and \(s_j=s_l\) for all \(j\in M_L\).
Part 1: Low types Taking the first order conditions of the utility function with respect to \(s_i\) for \(i\in M_L\), and focusing on the symmetric equilibrium by setting \(s_i=s_l\), we obtain the following:
$$\begin{aligned}&\frac{\partial MU_{i,l}}{\partial s_{i,l}}:=-\frac{(a-1) a (d-1)}{(s_l (h (s_h-s_l)+s_l))} e^{-\frac{\frac{a (d-1) s_l \left( e^{\frac{d (h-1) s_l^2}{h s_h-h s_l+s_l}}-1\right) }{s_l}+\frac{a (1-d) h s_i \left( 1-e^{-d s_h}\right) }{h (s_h-s_l)+s_l}+d^2 s_h}{d}} \nonumber \\&\left[ -h s_l (Y-1) e^{\frac{a (d-1) s_l \left( e^{\frac{d (h-1) s_l^2}{h s_h-h s_l+s_l}}-1\right) }{d s_l}+d s_h}\right. +h s_l (Y-1) e^{\frac{a (d-1) s_l \left( e^{\frac{d (h-1) s_l^2}{h s_h-h s_l+s_l}}-1\right) }{d s_l}} \nonumber \\&\left. +(h s_h-h s_l+s_l) e^{d \left( \frac{(h-1) s_l^2}{h s_h-h s_l+s_l}+s_h\right) }-e^{d s_h} (h s_h+s_l)+h s_l\right] =c. \end{aligned}$$
(A-5)
Taking the limits of \(\partial MU_{i,l}/\partial s_{i,l}\) for \(s_l\rightarrow 0\) and \(s_l\rightarrow \infty \) we obtain the following:
$$\begin{aligned} \lim \limits _{s_l\rightarrow 0} \frac{\partial MU_{i,l}}{\partial s_{i,l}}= & {} \frac{(a-1) a (d-1) Y e^{-d s_h} \left( e^{d s_h}-1\right) }{s_h},\\ \lim \limits _{s_l\rightarrow \infty }\frac{\partial MU_{i,l}}{\partial s_{i,l}}= & {} 0. \end{aligned}$$
The LHS of equation (A-5), \(\partial MU_{i,l}/\partial s_{i,l}\), converges to \((a-1) a (d-1) Y e^{-d s_h} \left( e^{d s_h}-1\right) /s_h\) when taking the limit for \(s_l\rightarrow 0\). Note that, if \(h\rightarrow 0\), the problem for the low types becomes the same as with one type (see proof of Proposition 2). Since the marginal returns are always positive and continuous in \(s_l\), a symmetric interior equilibrium for the low types exists if and only if \(c<\bar{c}_l=(a-1) a (d-1) Y e^{-d s_h} \left( e^{d s_h}-1\right) /s_h\), which is decreasing in \(s_h\).
Part 2: High types As for individuals of education level h, we consider the maximization problem of an individual \(i\in M_H\), keeping \(s_h\) and \(s_l\) fixed. Taking the first order conditions of the utility function with respect to \(s_i\), and setting \(s_i=s_h\) we obtain the following:
$$\begin{aligned} \frac{\partial MU_{i,h}}{\partial s_{i,h}}:=\frac{(1-a) a (1-d) h Y \left( 1-e^{-d s_h}\right) e^{-\frac{a (1-d) h s_h \left( 1-e^{-d s_h}\right) }{d (h (s_h-s_l)+s_l)}}}{h (s_h-s_l)+s_l}=c. \end{aligned}$$
(A-6)
Note that \(s_h>0\) and \(s_l>0\), LHS of equation A-6, \(\partial MU_{i,h}/\partial s_{i,h}\), is always positive and continuous in both \(s_h\) and \(s_l\). Moreover:
$$\begin{aligned} \lim \limits _{s_h\rightarrow 0}\frac{\partial MU_{i,h}}{\partial s_{h}}= & {} 0 \\ \lim \limits _{s_h\rightarrow \infty }\frac{\partial MU_{i,h}}{\partial s_{h}}= & {} 0 \end{aligned}$$
Hence, we can conclude that there exists a threshold \(\bar{c}_h\) such that for any \(c<\bar{c}_h\) a solution \(s_h\) of \(\partial MU_{i,h}/\partial s_{i,h}=c\) in a symmetric equilibrium exists.
Part 3: From part 1 of the proof, we have obtained that a symmetric interior equilibrium for low types exists if and only if \(c<\bar{c}_l\), and that \(\bar{c}_l\) is decreasing in \(s_h\). Additionally, from part 2 of the proof, we have obtained that a symmetric interior equilibrium for high types exists if and only if \(c<\bar{c}_h\). Since \(\partial MU_{i,l}/\partial s_{i,l}\) and \(\partial MU_{i,h}/\partial s_{i,h}\) in a symmetric equilibrium are both continuous in \(s_h\) and \(s_l\), we conclude that there exists a threshold \(\bar{c}\) such that for all \(c<\bar{c}\) both (A-5) and (A-5) are simultaneously satisfied, i.e., at least one interior equilibrium exists. This concludes the proof of Proposition 5. \(\square \)
Proof of Proposition 4
We report for convenience the equations for the network matching rates.
$$\begin{aligned} \Psi _{h,h}(s)= & {} 1-\lim _{n \rightarrow \infty } \phi _{h,h}(s) = 1-e^{-\frac{a(1-d)h s_h}{d (h s_h+(1-h)s_l)}\left( 1-e^{-d s_h}\right) }, \\ \Psi _{l,h}(s)= & {} 1-e^{-\frac{a(1-d) h s_l}{d (h s_h+(1-h)s_l)}\left( 1-e^{-d s_h}\right) }, \\ \Psi _{l,l}(s)= & {} 1-e^{-\frac{a(1-d)(1-h)s_l}{d (h s_h+(1-h)s_l)}\left( 1-e^{-d (1-h) s_l\frac{sl}{h s_h+(1-h) s_l} }\right) } . \end{aligned}$$
Inspecting these expressions, it is immediate to see that, given any positive value of \(s_l\) and \(s_h\), the network matching rates are increasing in the arrival rate a. This concludes the proof of Proposition 4. \(\square \)