The heterogeneity of ethnic employment gaps

Abstract

This paper investigates the heterogeneity of ethnic employment gaps using a new single-index based approach. Instead of stratifying our sample by age or education, we study ethnic employment gaps along a continuous measure of employability, the employment probability minority workers would have if their characteristics were priced as in the majority group. We apply this method to French males, comparing those whose parents are North African immigrants and those with native parents. We find that both the raw and the unexplained ethnic employment differentials are larger for low-employability workers than for high-employability ones. We show in a theoretical framework that this heterogeneity can be accounted for by homogeneous underlying mechanisms and is not evidence for, say, heterogeneous discrimination. Finally, we discuss our main empirical findings in the light of simple taste-based vs. statistical discrimination models.

Appendices

Appendix A: Proofs

1.1 A.1 Consequence of the CIA: Y(F)⊥T|p(X)

Rosenbaum and Rubin (1983) prove that:

$$Y_{i}(F) \perp T_{i} |X_{i}, \: \forall i \Rightarrow Y_{i}(F) \perp T_{i} |P(T_{i}=1|X_{i}), \; \forall i. $$

Following exactly their reasoning, it is possible to prove that, for any random variables A _i , B _i taking values in {0,1}:

$$A_{i} \perp B_{i} |X_{i}, \: \forall i \Rightarrow A_{i} \perp B_{i} |P(A_{i}=1|X_{i}), \; \forall i. $$

Rosenbaum and Rubin (1983) consider A _i =T _i and B _i =Y _i (F). The proof finishes by taking A _i =Y _i (F) and B _i =T _i .

1.2 A.2 Employer’s best guess:

$\hat y_{T}(x) \doteq E [y| \tilde y, x, T] = x + \tilde \varepsilon \left (\frac {\omega ^{2}}{{\sigma _{T}^{2}}+\omega ^{2}} \right )$ This point is derived from Aigner and Cain (1977). Employers’ best guess, given x, T and $\tilde {y}$ is:

$$\hat y_{T} (x) = E [y| \tilde y, x, T]= E [x+\varepsilon| \varepsilon+\eta, x, T]=x+ E [\varepsilon|\varepsilon+\eta, T] $$

The last equality holding because x⊥ε.

The result follows then from:

$$E [\varepsilon|\varepsilon+\eta, T]=\frac{\omega^{2}}{\omega^{2}+{\sigma_{T}^{2}}}(\varepsilon+\eta)=\frac{\omega^{2}}{\omega^{2}+{\sigma_{T}^{2}}}(\tilde y-x) $$

This equation implies that $\hat y_{T} (x) \sim N(x, \omega ^{4}/({\sigma _{T}^{2}} + \omega ^{2}))$.

1.3 A.3 Point 1, Section 5, taste-based discrimination

To offset a utility loss δ, employers set up a cutoff $\underline c_{D} > \underline c$ such that:

$$E( \hat y | \hat y > \underline c_{D}, x)- E( \hat y | \hat y > \underline c, x) = \delta $$

Condition $\hat y > \gamma$, ($\gamma =\underline c$, or $\gamma =\underline c_{D}$) is equivalent to:

$$\frac{\sqrt{\sigma^{2} + \omega^{2}}}{\omega^{2}} (\hat y - x ) > \frac{\sqrt{\sigma^{2} + \omega^{2}}}{\omega^{2}} (\gamma - x)\\ $$

We define $c=(\underline c-x)\frac {\sqrt {\omega ^{2}+\sigma ^{2}}}{\omega ^{2}}$, $c_{D}=(\underline c_{D}-x)\frac {\sqrt {\omega ^{2}+\sigma ^{2}}}{\omega ^{2}}$, $u = (\hat y - x)\frac {\sqrt {\omega ^{2}+ \sigma ^{2}}}{\omega ^{2}} \sim \mathcal N (0,1)$ and we denote λ(.)=φ(.)/Φ(.), with φ and Φ corresponding respectively to the probability distribution function and the cumulative distribution function of a $\mathcal {N}(0,1)$. With these notations E(u|u>γ,x)=λ(γ).

The thresholds c and c _D are such that:

$$\lambda(-c_{D}) - \lambda(-c) = \delta\frac{\sqrt{\sigma^{2}+\omega^{2}}}{\omega^{2}} \doteq \tilde \delta $$

If δ does not depend on x, differentiating this equation with respect to x leads to:

$$ -c_{D}^{\prime}(x)\lambda^{\prime}(-c_{D}(x)) = -c^{\prime}(x)\lambda^{\prime}(-c(x)) $$

(1)

Given that $\lambda ^{\prime }/\lambda$ is decreasing and that c _D >c, we have that:

$$ \frac{\lambda^{\prime}(-c_{D})}{\lambda(-c_{D})} > \frac{\lambda^{\prime}(-c)}{\lambda(-c)} $$

(2)

Combining Eqs. 1 and 2, and given that $\lambda ^{\prime }<0$, we obtain:

$$ \lambda(-c_{D})(-c_{D}^{\prime}(x)) > \lambda(-c)(-c^{\prime}(x)) $$

(3)

The ratio of hiring probabilities is equal to:

$$\frac{h_{D}}{h_{F}} = \frac{ P(u > c_{D})}{ P(u > c)} = \frac{\Phi(-c_{D})}{\Phi(-c)} $$

Differentiating the ratio of employment probabilities by x, we show that the sign of the derivative is the same as the one of:

$$\lambda(-c_{D})(-c_{D}^{\prime}(x)) - \lambda(-c)(-c^{\prime}(x)) $$

From Eq. 3, we find that the ratio h _D /h _F should be increasing.

1.4 A.4 Point 1, Section 5, statistical discrimination in means

In this case, $\hat y_{D} = \hat y_{F} = x + \tilde \varepsilon \frac {\omega ^{2}}{\sigma ^{2}+\omega ^{2}}$. Condition $\hat y > \underline c$ is equivalent to:

$$\frac{\sqrt{\sigma^{2} + \omega^{2}}}{\omega^{2}} (\hat y - x ) > \frac{\sqrt{\sigma^{2} + \omega^{2}}}{\omega^{2}} (\underline c - x)\\ $$

We define $c=(\underline c-x)\frac {\sqrt {\omega ^{2}+\sigma ^{2}}}{\omega ^{2}}$. In this case, transformed unobservables $u = (\hat y - x)\frac {\sqrt {\omega ^{2}+\sigma ^{2}}}{\omega ^{2}}$ are distributed in a $\mathcal N (0,1)$ in group F and $\mathcal N(-\mu ,1)$ in group D. We denote λ(.)=φ(.)/Φ(.), with φ and Φ corresponding to the probability distribution function and the cumulative distribution function of a $\mathcal {N}(0,1)$.

Then:

$$\frac{h_{D}}{h_{F}} = \frac{ P(u_{D} > c(x))}{ P(u_{F} > c(x))} = \frac{\Phi(-c(x)-\mu)}{\Phi(-c(x))} $$

Differentiating the ratio of employment probabilities by x, and using that $c^{\prime }<0$, we find that the sign of the derivative is the same as the one of:

$$\lambda(-c(x)-\mu) - \lambda(-c) $$

Because −c(x)−μ<−c(x), and $\lambda ^{\prime }<0$, we have λ(−c(x)−μ)>λ(−c), so that the ratio h _D /h _F is increasing.

1.5 A.5 Point 2, Section 5

Condition $\hat y_{T} > \underline c$ is equivalent to:

$$\frac{\sqrt{{\sigma^{2}_{T}} + \omega^{2}}}{\omega^{2}} (\hat y_{T} - x ) > \frac{\sqrt{{\sigma^{2}_{T}} + \omega^{2}}}{\omega^{2}} (\underline c - x)\\ $$

or, denoting $u_{T} = \frac {\sqrt {{\sigma ^{2}_{T}} + \omega ^{2}}}{\omega ^{2}} (\hat y_{T} - x)$, with T=D,F, $c(x)=\frac {\sqrt {{\sigma ^{2}_{F}} + \omega ^{2}}}{\omega ^{2}} (\underline c - x)$ and $k = \frac {\sqrt {{\sigma ^{2}_{D}} + \omega ^{2}}}{\sqrt {{\sigma ^{2}_{F}} + \omega ^{2}}}>1$, so that h _F =P(u _F >c) and h _D =P(u _D >k c).

Because u _D and $u_{F} \sim \mathcal N (0,1)$,

$$\frac{h_{D}}{h_{F}} = \frac{ P(u_{D} > kc(x))}{ P(u_{F} > c(x))} = \frac{\Phi(-kc(x))}{\Phi(-c(x))} $$

First consider the situation when σ _D and σ _F do not vary with x. The derivative of $\frac {h_{D}}{h_{F}}$ with respect to x is positive if:

$$\Phi(-kc(x)) \varphi(-c(x)) c^{\prime}(x)> k \Phi(-c(x)) \varphi(-kc(x)) c^{\prime}(x) $$

and as $c^{\prime }(x)=-\frac {\sqrt {{\sigma _{F}^{2}}+\omega ^{2}}}{\omega ^{2}}<0$, this is equivalent to:

$$\Phi(-kc(x)) \varphi(-c(x)) < k \Phi(-c(x)) \varphi(-kc(x)) $$

Noting λ(.)=φ(.)/Φ(.), this is itself equivalent to:

$$\lambda(-c(x)) < k \lambda(-kc(x)) $$

If c>0, that is $x< \underline c$, we have as k>1, −k c(x)<−c(x), and as λ(.) is positive and decreasing, λ(−c(x))<k λ(−k c(x)). Therefore, $\frac {h_{D}}{h_{F}}$ is increasing in x.

If c<0, that is if $x> \underline c$, conclusion depends on the value of k: $\frac {h_{D}}{h_{F}}$ increases in x iff λ(−c(x))<k λ(−k c(x)) and $\frac {h_{D}}{h_{F}}$ decreases in x if λ(−c(x))>k λ(−k c(x)). Simulations show that ∀k>1 there exists a (unique) −c ₀ such that ∀−c<−c ₀,λ(−c(x))<k λ(−k c(x)) and ∀−c>−c ₀, λ(−c(x))>k λ(−k c(x)) (more details available upon request). So, $\frac {h_{D}}{h_{F}}$ increases with x up to a certain threshold and then decreases. The threshold depends on the employer cut-off $\underline c$ and on the screening error variance ratio k.

1.6 A.6 Point 3, Section 5

Consider the inflow-outflow equation with e, h, and q being functions of x:

$$e(x) = \frac{h(x)}{h(x)+q(x)}. $$

Taking the derivative with respect to x leads to:

$$e^{\prime}(x) = \frac{h^{\prime}(x)q(x) - q^{\prime}(x)h(x)}{(h(x)+q(x))^{2}} $$

with the previous notations, h(x)=P(u>c(x))=Φ(−c(x)) which is increasing in x. Therefore, it suffices for e to be increasing in x, that q be non increasing in x.

1.7 A.7 Two-stage screening model

The two-stage screening model corresponds to drawing u ₁, u ₂ in a bivariate normal distribution such that $u_{1}, u_{2} \sim \mathcal {N}(0,1)$ and $cov(u_{1}, u_{2}) = \rho _{T} = \frac {\omega ^{2}}{\omega ^{2} + {\sigma ^{2}_{T}}}$.

Writing u ₂=ρ _T u ₁+ν, with $V(\nu ) = \sqrt {1-{\rho _{T}^{2}}}$ leads to

$$P(u_{2}>c | u_{1}>c) = \frac{P\left( \frac{\nu}{\sqrt{1-{\rho_{T}^{2}}}} > \frac{c-\rho_{T} u_{1}}{\sqrt{1-{\rho_{T}^{2}}}}\text{~~\&~~} u_{1}>c\right) }{P(u_{1}>c)}. $$

With the previous notations, it follows that:

$$P(u_{2}>c | u_{1}>c) = \frac{\int_{c}^{\infty} \Phi\left( \frac{\rho_{T} u - c}{\sqrt{1-{\rho_{T}^{2}}}}\right)\varphi(u)}{\Phi(-c)}. $$

The denominator does not depend on ρ _T , and $\frac {\rho _{T} u - c}{\sqrt {1-{\rho _{T}^{2}}}}$ is increasing in ρ _T as long as u>ρ _T c (which is the case here). P(u ₂>c|u ₁>c) is thus increasing in ρ _T , and therefore decreasing in σ _T . Minority workers are more likely to be dismissed than majority ones.

Appendix B: Similarities and differences between the propensity and employability scores

The employability score shares similarities with the propensity score but it differs from it. Note first that the employability score is not a balancing score in the sense defined by Rosenbaum and Rubin (1983). In general, we do not have X⊥T|e(X). To see this, just consider two populations T=0 and T=1, and a unique explanatory variable X with values 0 and 1, and taking value 1 with probability q if T=0 and probability 1−q if T=1 ( q≠1−q). Assume also that employment Y is such that P(Y|X,T)=1/2 independent of T and X. It follows that T X|P(Y|X)=1/2.

Even if the employability score is not a balancing score, Y(F)⊥T|e(X) entails that conditional treatment effects are identified at any value of e(X). So the employability score provides a different dimension of analysis that is not redundant with nor cannot be summarized in general by the propensity score.

Furthermore, applying the same reasoning as Rosenbaum and Rubin (1983) on e(X) instead of on the propensity score, we can define balancing scores relative to Y, instead of balancing scores relative to T. Let b _Y be a balancing score relative to Y, b _Y is such that X⊥Y|b _Y (X). Theorem 2 of Rosenbaum and Rubin (1983) says that the propensity score p(X)=P(T=1|X) is the coarsest balancing score in the sense that if b _T is a balancing score (relative to T), then p = f(b _T ) for some function f. Considering now Y instead of T, it follows that e(X)=P(Y=1|X) is the coarsest balancing score relative to Y.

Theorem 3 of Rosenbaum and Rubin (1983) says that if treatment assignment is strongly ignorable given X, then it is strongly ignorable given any balancing score b _T (X), which holds in particular for the propensity score p(X). Considering again Y instead of T, treatment assignment is also ignorable given any balancing score relative to Y, b _Y (X), in particular given the employability e(X).

To justify even more the use of the employability, we show next that it is, with the propensity score, the only other unidimensional score that could lead to the previous results in a general way. It may happen, that in specific situations, other unidimensional scores could summarize the CIA and be good candidates for a conditional analysis, but the only ones that can work on a general basis are the propensity score and the employability. To see that, it is sufficient to find an example in which they are the only valid scores (in the above sense).

Assume that there is one single covariate X, and that Y is such that P(Y=1|X)=Λ(X), with $\Lambda (x)=\exp (x)/(1+ \exp (x))$, and T is such that P(T=1|X)=1−Λ(X).

Imagine that there is some function g such that Y⊥T|g(X) but g is neither a balancing score relative to Y nor to T: X T|g(X) and X Y|g(X). This means that there exist x ₁<x ₂, such that g(x ₁)=g(x ₂)=γ but P(Y=1|x ₁)≠P(Y=1|x ₂). Given the specific form of Y and T, this also means that P(T=1|x ₁)≠P(T=1|x ₂).

Assume without loss of generality that $g=\gamma \Rightarrow x \in (x_{1}, x_{2})$ and that X follows a non informative distribution. It follows that P(Y=1|T=0,g=γ)<P(Y=1|T=1,g=γ). Indeed, with T=0, it is more likely that x=x ₁ than x=x ₂. This contradicts the fact that Y⊥T|g(X). Therefore, in general, the only scores b that are such that Y⊥T|b(X) are balancing scores relative to Y or T.