Why do young women marry old men?

Abstract

This paper presents an overlapping generations household model with positive assortative matching (richer individuals marry richer partners), incomplete information about partner’s type (it takes time to reveal income-earning capabilities of individuals) and a gender pay gap on the labor market (men are more likely to end up with a high-paying job). In equilibrium, a gender pay gap creates an excess supply of desirable husbands and women marry early to increase their chance of being matched with an ideal partner, which results in a gender age gap on the marriage market. A modified model with asymmetric information yields a similar result. An extended model where individuals have an option to remain single (the marriage market does not necessarily clear in equilibrium) yields a similar result as well.

Appendix

Proof of proposition 1.

Step 1 We prove that there cannot be an equilibrium when $\left( {1 - f} \right)q \ge \left( {1 - m} \right)p$.

In this case, there is an excess supply $\left( {1 - f} \right)q - \left( {1 - m} \right)p \ge 0$ of women who revealed a high type in the second period. Inequality $\left( {1 - f} \right)q \ge \left( {1 - m} \right)p$ implies inequality $m - \left( {1 - f} \right)q + \left( {1 - m} \right)p \ge f$ so that we have an excess supply of unmatched men who did not reveal their type (over women who did not reveal their type). Thus, every woman who revealed a high type is matched either with a man who revealed a high type as well, or—with a man who has not yet revealed his type. Additionally, every woman who has not yet revealed her type is matched with a man who has not yet revealed his type.

In this case, a man who decides to enter the marriage market in the second period is matched with a woman who revealed a high type with an ex ante probability p.⁹ On the other hand, a man who decides to enter the marriage market in the first period is matched with a woman who revealed a high type with probability

$$ \frac{{\left( {1 - f} \right)q - \left( {1 - m} \right)p}}{m}, $$

and with a woman who has not yet revealed her type—with probability f/m.¹⁰ In other words, the strategy of entering the marriage market in the first period is equivalent to a simple lottery that yields a match with a woman who revealed a high type with probability:

$$ r\mathop = \limits^{\text{def}} \frac{{q - \left( {1 - m} \right)p}}{m} < p, $$

and a match with a woman who revealed a low type with probability $1 - r$.

Thus, for men, the strategy of entering the marriage market in the second period (first-order) stochastically dominates the strategy of entering the marriage market in the first period. This implies that m must be zero, in which case inequality $\left( {1 - f} \right)q \ge \left( {1 - m} \right)p$ cannot hold for any f∊[0,1]. Hence, in any equilibrium we must have $\left( {1 - m} \right)p > \left( {1 - f} \right)q$.

Step 2 We prove that there cannot be an equilibrium when $\left( {1 - m} \right)p > \left( {1 - f} \right)q$ and $f - \left( {1 - m} \right)p + \left( {1 - f} \right)q < m$.

In case when $\left( {1 - m} \right)p > \left( {1 - f} \right)q$ we have an excess supply $\left( {1 - m} \right)p - \left( {1 - f} \right)q > 0$ of men who revealed a high type in the second period. Moreover, when $f - \left( {1 - m} \right)p + \left( {1 - f} \right)q < m$ we also have an excess supply of men who did not reveal their type (over unmatched women who did not reveal their type). Thus, a woman who decides to enter the marriage market in the second period is matched with a man who revealed a high type with probability q, with a man who has not yet revealed his type—with probability

$$ s\mathop = \limits^{\text{def}} \frac{{m - f + \left( {1 - m} \right)p - \left( {1 - f} \right)q}}{1 - f}, $$

and with a man who revealed a low type—with the remaining probability $1 - q - s$. In other words, the strategy of entering the marriage market in the second period is equivalent to a simple lottery that yields a match with a man who revealed a high type with probability $q + sp$ and a match with a man who revealed a low type with probability $1 - q - sp$.

On the other hand, a woman who decides to enter the marriage market in the first period is matched with a man who revealed a high type with probability:

$$ t\mathop = \limits^{\text{def}} \frac{{\left( {1 - m} \right)p - \left( {1 - f} \right)q}}{f}, $$

and with a man who has not yet revealed his type—with the remaining probability $1 - t$. In other words, the strategy of entering the marriage market in the first period is equivalent to a simple lottery that yields a match with a man who revealed a high type with probability $t + \left( {1 - t} \right)p$ and a match with a man who revealed a low type with probability $1 - t - \left( {1 - t} \right)p$.

A simple algebra yields $t + \left( {1 - t} \right)p > q + sp$, i.e., for women, the strategy of marrying in the first period (first-order) stochastically dominates the strategy of marrying in the second period. This implies that f must be one, in which case inequality $f - \left( {1 - m} \right)p + \left( {1 - f} \right)q < m$ cannot hold for any m∊[0,1].

Step 3 We prove that f = 1 in equilibrium.

Results from steps 1 and 2 imply that an equilibrium can exist only when $\left( {1 - m} \right)p > \left( {1 - f} \right)q$ and $f - \left( {1 - m} \right)p + \left( {1 - f} \right)q \ge m$. Thus, in equilibrium we must have an excess supply $\left( {1 - m} \right)p - \left( {1 - f} \right)q > 0$ of men who revealed a high type in the second period (over women who revealed a high type) and an excess supply $f - \left( {1 - m} \right)p + \left( {1 - f} \right)q - m \ge 0$ of unmatched women who did not reveal their type (over men who did not reveal their type). In this case, a woman who decides to enter the marriage market in the second period is matched with a man who revealed a high type with probability q and with a man who revealed a low type—with probability $1 - q$.

On the other hand, a woman who decides to enter the marriage market in the first period is matched with a man who revealed a high type with probability t, with a man who has not yet revealed his type—with probability m/f and with a man who revealed a low type—with the remaining probability $1 - t - m/f$. In other words, the strategy of entering the marriage market in the first period is equivalent to a simple lottery that yields a match with a man who revealed a high type with probability:

$$ t + \frac{mp}{f} = \frac{{p - \left( {1 - f} \right)q}}{f} > q, $$

and a match with a man who revealed a low type with probability $1 - t - mp/f$. Therefore, for women, the strategy of marrying in the first period (first-order) stochastically dominates the strategy of marrying in the second period. This implies that f must be one in equilibrium.

When all women enter the marriage market already in the first period, men are exactly indifferent between entering the marriage market in the first or the second period (in both cases they are matched for sure with a partner who has not yet revealed her type). Thus, the only pure strategy equilibrium on the marriage market is given by f = 1 and m∊[0,1]. □