Why can high variance help you in Heckman (1998) but not Aigner and Cain (1977)?

Question

Heckman (1998) argues that, if the bar for acceptance is high (so most applicants are rejected), then belonging to a group who quality has high variance tends to help you. The argument is something like this: if employers only hire people whose quality exceeds a certain cutoff, but most people fail the cutoff, then if your group's distribution of quality has higher variance it will have more mass above the cutoff.

Aigner and Cain (1977) also discuss variance based discrimination. Assuming that you send a signal equal to your group's mean, the only effect of higher variance is to dissuade risk averse employers to hire you. There is no Heckman style mechanism where high variance can help you if most people get rejected.

I would be very grateful if someone could explain what leads to this contrast.

My (possibly confused) thoughts on this: the Heckman model makes sense if employers decide whether to grant an interview based on whether the probability that someone's quality is above a certain level exceeds a certain cutoff. But that is a crazy hiring rule! Instead, they should just look at whether expected quality exceeds a certain cutoff (to be fair to Heckman, this is supposed to be the hiring rule.)

OK suppose that this is the hiring rule: can high variance still help you (in the context of an audit/correspondence study)? Well, I can see how this works if the researcher perfectly controls on one dimension (e.g. what the auditors say) but doesn't control on another (e.g. attractiveness), and one group has higher variance on that second dimension, which means more of them get over the bar. But that logic totally breaks down in correspondence studies (supposedly also a target). Here the employer does not get to see some realisation from a distribution of quality that is unobserved by the researcher. At best, the employer can say: that group's high variance, so more likely to exceed the cutoff. But that brings us back to the crazy hiring strategy already discussed.

Answer 1

Say you have two distributions, $G$ and $H$ with equal means supported on (say) $[0,1]$ and $H$ is second-order stochastically dominated by $G$. Then, for any concave increasing $u$, we'll have $$\int udG \geq \int udH$$ This would correspond to Aigner and Cain, where the risk averse firm would prefer to hire an agent from distribution G.

However, now suppose that $G$ and $H$ are distributions of talent (with the same stochastic ordering i.e. $H \succ_{C} G$). One equivalent notion to this idea is that distribution $G$ can be obtained by ``fusing" together part of the mass of $H$, i.e. by collapsing part of the mass to its respective barycenter. Given this, it is easy to see that there is always a cutoff $\eta \in [0,1]$ such that $H(\eta) \geq G(\eta)$.

This corresponds to Heckman's result--if the job is sufficiently demanding, then it will be (at least weakly better) to be part of the high variance population. Moreover, note that this by no means depends on the expected quality of the agent. There are simply more of the highest quality people in the group with distribution $H$ than in the group with distribution $G$. They are not just more likely to exceed the cutoff, there are more of them that exceed the cutoff.

Another way to think about the hiring problem is as follows: say a firm wants to hire one person but receives applications from $10$ applicants, $5$ from group $A$ and $5$ from group $B$. A candidate's quality is unknown and it it ranges from $[0,1]$ continuously. Suppose also for the sake of simplicity that the firm may observe the candidate's group (and so knows what distribution he/she is drawn from): the distribution of quality for group $A$ is given by $G$ and the distribution of quality for group $B$ is given by $H$. Interviewing a candidate fully reveals a candidate's type, but the firm has the budget to interview only $4$ candidates. Whom should they interview?

The answer is simple: all four candidates interviewed should be from group $B$ (which has distribution $H$). Why? Because we can think about a candidate's quality as a random variable $X_{i}$ and thus the firm wants to maximize $\mathbb{E}[Z]$ where $$Z:=\max\big\{X_{1},X_{2},X_{3},X_{4}\big\}$$ Since $H \succ_{C} G$, $$\mathbb{E}_{H^{4}}[Z] \geq \mathbb{E}_{G^{4}}[Z]$$

Note also that $H \succ_{C} G$ $\Rightarrow$ $Var(H) \geq Var(G)$ but not $\Leftarrow$. In this sort of thing, I think it is stochastic dominance and not variance that is the apposite metric.