Historical discrimination and optimal remediation

Abstract

I consider a society which is jointly committed to ensuring equal opportunity and to increasing aggregate wealth but is faced with the vestiges of past discrimination in the form of a historically skewed distribution of social resources. Focusing on the problem of allocating the existing (fixed) quantity of social inputs, I consider two policy instruments: directly transferring resources from the advantaged to the disadvantaged or affording preferential treatment in employment to the disadvantaged group (affirmative action). After describing the general procedure for determining an optimal policy, I demonstrate by means of an example that either of the instruments might constitute an optimal remediation policy and I identify conditions which favor each.

Appendices

Appendix 1: Proofs

This appendix contains proofs of some of the lemmas in Sect. 4.

Lemma 5

When $\alpha = 0$, $\Gamma$ is convex and exhibits the asymptotic behavior ${lim}_{\varepsilon \rightarrow 0} \frac{dQ/d\varepsilon }{dEV/d\varepsilon }=-\infty$ and ${lim}_{ \varepsilon \rightarrow \widehat{s}}\frac{dQ/d\varepsilon }{ dEV/d\varepsilon }=-\frac{1}{D}$.

Proof

As in the proof of Proposition 4,

$\frac{dQ}{dEV}=\frac{\left( D-\left( \sqrt{\widehat{s}+\varepsilon }- \sqrt{\widehat{s}-\varepsilon }\right) \right) \left( \sqrt{\widehat{s}+\varepsilon }+\sqrt{\widehat{s}-\varepsilon }\right) }{D^{2}\left( \alpha \sqrt{\widehat{ s}+\varepsilon }-\sqrt{\widehat{s}-\varepsilon }\right) +\frac{1}{2} (1-\alpha )\left( D-\left( \sqrt{\widehat{s}+\varepsilon }-\sqrt{\widehat{s} -\varepsilon }\right) \right) ^{2}\left( \sqrt{\widehat{s}+\varepsilon }+\sqrt{ \widehat{s}-\varepsilon }\right) }$.

Hence, at $\alpha =0$, $\frac{dQ}{dEV}=\frac{\left( D-(\sqrt{ \widehat{s}+\varepsilon }-\sqrt{\widehat{s}-\varepsilon })\right) \left( \sqrt{\widehat{s}+\varepsilon }+\sqrt{\widehat{s}-\varepsilon }\right) }{ -D^{2}\sqrt{\widehat{s}-\varepsilon }+\frac{1}{2}\left( D-(\sqrt{\widehat{s} +\varepsilon }-\sqrt{\widehat{s}-\varepsilon })\right) ^{2}\left( \sqrt{ \widehat{s}+\varepsilon }+\sqrt{\widehat{s}-\varepsilon }\right) }$. Upon further simplification, this can be expressed as $\frac{dQ}{dEV}=\frac{ D\left( \sqrt{\widehat{s}+\varepsilon }+\sqrt{\widehat{s}-\varepsilon } \right) -2\varepsilon }{(\frac{1}{2}D^{2}+\varepsilon )(\sqrt{\widehat{s} +\varepsilon }-\sqrt{\widehat{s}-\varepsilon })-2D\varepsilon }$. Using $\widehat{s}=\frac{D^{2}}{2}$, ${lim}_{\varepsilon \rightarrow 0} \frac{dQ/d\varepsilon }{dEV/d\varepsilon }=-\infty$ can be verified directly noting that the denominator is decreasing in $\varepsilon$ at zero. Similarly, ${lim}_{\varepsilon \rightarrow \widehat{s}}\frac{ dQ/d\varepsilon }{dEV/d\varepsilon }=-\frac{1}{D}$ can be shown directly using l’Hôpital’s rule.

Next, computing $\frac{d^{2}Q}{dEV^{2}}$ from the above expression, it can be seen that $\frac{d^{2}Q}{dEV^{2}}>0$ if and only if

$$\begin{aligned}&\begin{array}{c} 2\varepsilon ^{2}\sqrt{\widehat{s}+\varepsilon }-2\widehat{s} D^{3}+2\varepsilon ^{2}\sqrt{\widehat{s}-\varepsilon }-2D\sqrt{\widehat{s} -\varepsilon }\left( \widehat{s}+\varepsilon \right) ^{\frac{3}{2}}+2D\left( \widehat{s}-\varepsilon \right) ^{\frac{3}{2}}\sqrt{\widehat{s}+\varepsilon }\\ +\,6\widehat{s}D^{2}\sqrt{\widehat{s}+\varepsilon }-3D^{2}\varepsilon \sqrt{ \widehat{s}+\varepsilon }+2\widehat{s}D^{2}\sqrt{\widehat{s}-\varepsilon } +D^{2}\varepsilon \sqrt{\widehat{s}-\varepsilon }-4\widehat{s}D\varepsilon >0 \text {.} \end{array} \end{aligned}$$

(31)

Manipulating (31) and using $\widehat{s}=\frac{D^{2}}{2}$, $\frac{d^{2}Q}{dEV^{2}}>0$ if and only if

$$\begin{aligned}&(2\varepsilon ^{2}-3D^{2}\varepsilon +3D^{4})\sqrt{\widehat{s}+\varepsilon } +(2\varepsilon ^{2}+D^{2}\varepsilon +D^{4})\sqrt{\widehat{s}-\varepsilon } -4D\varepsilon \sqrt{\widehat{s}+\varepsilon }\sqrt{\widehat{s}-\varepsilon }\nonumber \\&\quad -D^{5}-2D^{3}\varepsilon >0\text {,} \end{aligned}$$

(32)

or

$$\begin{aligned}&(2\varepsilon ^{2}-3D^{2}\varepsilon +3D^{4})\sqrt{\widehat{s}+\varepsilon } +(2\varepsilon ^{2}+D^{2}\varepsilon +D^{4})\sqrt{\widehat{s}-\varepsilon } -4D\varepsilon \sqrt{\widehat{s}+\varepsilon }\sqrt{\widehat{s}-\varepsilon }\nonumber \\&\quad >D^{5}+2D^{3}\varepsilon \text {.} \end{aligned}$$

(33)

Notice the RHS of (33) is linear in $\varepsilon$. Also, $D^{5}+2D^{3}\varepsilon <2D^{5}$ on the interior of the interval $[0, \widehat{s}]$ and $D^{5}+2D^{3}\varepsilon =2D^{5}$ at $\varepsilon = \widehat{s}$. Therefore, it is sufficient to show the LHS $(2\varepsilon ^{2}-3D^{2}\varepsilon +3D^{4})\sqrt{\widehat{s}+\varepsilon }+(2\varepsilon ^{2}+D^{2}\varepsilon +D^{4})\sqrt{\widehat{s}-\varepsilon }-4D\varepsilon \sqrt{\widehat{s}+\varepsilon }\sqrt{\widehat{s}-\varepsilon }>2D^{5}$ on $(0,\widehat{s})$.

For convenience, let $T_{1}(\varepsilon )=(2\varepsilon ^{2}-3D^{2}\varepsilon +3D^{4})\sqrt{\widehat{s}+\varepsilon }$. It is easily verified that $T_{1}(0)=\frac{3}{\sqrt{2}}D^{5}>2D^{5}$, $T_{1}( \widehat{s})=2D^{5}$. and $\frac{dT_{1}}{d\varepsilon }<0$ on $(0,\widehat{s} )$. Hence, $T_{1}(\varepsilon )>2D^{5}$ uniformly on $(0,\widehat{s})$.

Finally, I establish that $(2\varepsilon ^{2}+D^{2}\varepsilon +D^{4})\sqrt{ \widehat{s}-\varepsilon }-4D\varepsilon \sqrt{\widehat{s}+\varepsilon }\sqrt{ \widehat{s}-\varepsilon }>0$, or $(2\varepsilon ^{2}+D^{2}\varepsilon +D^{4})>4D\varepsilon \sqrt{\widehat{s}+\varepsilon }$, on $(0,\widehat{s})$ . Again for convenience, denote $T_{2}(\varepsilon )=(2\varepsilon ^{2}+D^{2}\varepsilon +D^{4})$ and $T_{3}(\varepsilon )=4D\varepsilon \sqrt{ \widehat{s}+\varepsilon }$. It is also easily verified that $T_{2}(0)=D^{4}$ , $T_{3}(0)=0$, $T_{2}(\widehat{s})=T_{3}(\widehat{s})=2D^{4}$, and $\frac{ dT_{3}}{d\varepsilon }>\frac{dT_{2}}{d\varepsilon }>0$ on $(0,\widehat{s})$. The latter implies $T_{2}(\varepsilon )$ and $T_{3}(\varepsilon )$ are monotonic over the interval and uniformly $T_{2}(\varepsilon )>T_{3}(\varepsilon )$. $\square$

Lemma 6

For each $\delta \in (0,D)$, the slope along $\Gamma (\delta )$, $\frac{\partial Q/\partial \varepsilon }{\partial EV/\partial \varepsilon }<0$.

Proof

In this case, from (30) and (28 ) one obtains

$\frac{\partial Q/\partial \varepsilon }{\partial EV/\partial \varepsilon }=\frac{(D+\delta )( \sqrt{\widehat{s}+\varepsilon }+\sqrt{ \widehat{s}-\varepsilon }) -2\varepsilon }{(\widehat{s}+\varepsilon )( \sqrt{\widehat{s}+\varepsilon }-\sqrt{\widehat{s}-\varepsilon })-\frac{1}{2} \delta ^{2}( \sqrt{\widehat{s}+\varepsilon }+\sqrt{\widehat{s} -\varepsilon }) -2D\varepsilon }$, when $\alpha =0$. Let $T(\varepsilon )$ denote the numerator of this expression and $B(\varepsilon )$ denote the denominator. First, $T(0)=(D+\delta )(2\frac{D}{\sqrt{2}})>0$ and $T(\widehat{s})=\delta D>0$. Moreover, $\frac{dT}{d\varepsilon }=$ $-\frac{1}{2\sqrt{\widehat{s}+\varepsilon }\sqrt{\widehat{s} -\varepsilon }}( (D+\delta )(\sqrt{\widehat{s}+\varepsilon }-\sqrt{ \widehat{s}-\varepsilon })+4\sqrt{\widehat{s}+\varepsilon }\sqrt{\widehat{s} -\varepsilon }) <0$, for $\varepsilon \in [0,\widehat{s}]$, or T is monotonic on the interval. Hence, the numerator is uniformly positive. It is sufficient, therefore, to show that $B(\varepsilon )\le 0$. Indeed, since this must be true for all $\delta \in [0,D]$, it is sufficient to show $(\widehat{s}+\varepsilon )(\sqrt{\widehat{s}+\varepsilon }-\sqrt{ \widehat{s}-\varepsilon })-2D\varepsilon \le 0$, or $(\widehat{s} +\varepsilon )(\sqrt{\widehat{s}+\varepsilon }-\sqrt{\widehat{s}-\varepsilon })\le 2D\varepsilon$. Note that $(\widehat{s}+\varepsilon )(\sqrt{\widehat{ s}+\varepsilon }-\sqrt{\widehat{s}-\varepsilon })=2D\varepsilon$ at $\varepsilon =0$ and $\varepsilon =\widehat{s}$. It remains to be shown that $(\widehat{s}+\varepsilon )(\sqrt{\widehat{s}+\varepsilon }-\sqrt{\widehat{s} -\varepsilon })<2D\varepsilon$ for $0<\varepsilon <\widehat{s}$.

First, $\frac{\partial }{\partial \varepsilon }(\widehat{s}+\varepsilon )( \sqrt{\widehat{s}+\varepsilon }-\sqrt{\widehat{s}-\varepsilon })=\frac{1}{2 \sqrt{\widehat{s}-\varepsilon }}( 3\varepsilon -\widehat{s}+3\sqrt{ \widehat{s}+\varepsilon }\sqrt{\widehat{s}-\varepsilon })$. At $\varepsilon =0$,³⁴ this equals $\frac{D}{\sqrt{2}}<2D$. Hence, for small $\varepsilon >0$, $(\widehat{s} +\varepsilon )(\sqrt{\widehat{s}+\varepsilon }-\sqrt{\widehat{s}-\varepsilon })<2D\varepsilon$. To show that $(\widehat{s}+\varepsilon )(\sqrt{\widehat{s }+\varepsilon }-\sqrt{\widehat{s}-\varepsilon })<2D\varepsilon$ uniformly for $0<\varepsilon <\widehat{s}$, it is sufficient to show that $\frac{1}{2 \sqrt{\widehat{s}-\varepsilon }}( 3\varepsilon -\widehat{s}+3\sqrt{ \widehat{s}+\varepsilon }\sqrt{\widehat{s}-\varepsilon })$ increases monotonically on $(0,\widehat{s})$. (That is, since $(\widehat{s} +\varepsilon )(\sqrt{\widehat{s}+\varepsilon }-\sqrt{\widehat{s}-\varepsilon })=2D\varepsilon$ at $\varepsilon =0$ and $\varepsilon =\widehat{s}$, in order for there to be an $\varepsilon$ at which $(\widehat{s}+\varepsilon )( \sqrt{\widehat{s}+\varepsilon }-\sqrt{\widehat{s}-\varepsilon } )>2D\varepsilon$, it must be the case that $\frac{1}{2\sqrt{\widehat{s} -\varepsilon }}( 3\varepsilon -\widehat{s}+3\sqrt{\widehat{s} +\varepsilon }\sqrt{\widehat{s}-\varepsilon })$ decreases for some $\varepsilon \in (0,\widehat{s})$.) By direct computation, $\frac{\partial }{ \partial \varepsilon }\frac{1}{2\sqrt{\widehat{s}-\varepsilon }}( 3\varepsilon -\widehat{s}+3\sqrt{\widehat{s}+\varepsilon }\sqrt{\widehat{s} -\varepsilon }) =\frac{1}{4\sqrt{\widehat{s}+\varepsilon }( \widehat{s}-\varepsilon ) ^{\frac{3}{2}}}3( \widehat{s} -\varepsilon ) ^{\frac{3}{2}}+5\widehat{s}\sqrt{\widehat{s} +\varepsilon }-3\varepsilon \sqrt{\widehat{s}+\varepsilon }$. Since $5 \widehat{s}>3\varepsilon$ for all $\varepsilon \in (0,\widehat{s})$, this establishes the result.

$\square$

Lemma 7

$\Theta$ is concave and has extreme values $(EV( \widehat{s},0),Q(\widehat{s},0))=(\widehat{a}^{\circ }+D,0)$ and $(EV( \widehat{s},D),Q(\widehat{s},D))=(\widehat{a}^{\circ }+\frac{2}{3}D,\frac{1}{ 2})$. Also, along $\Theta$, ${lim}_{\delta \rightarrow 0}\frac{ \partial Q/\delta }{\partial EV/\partial \delta }=-\infty$ and ${lim}_{ \delta \rightarrow D}\frac{\partial Q/\delta }{\partial EV/\partial \delta }=-\frac{1}{D}$.

Proof

From (30) and (28), $\frac{ \partial }{\partial \delta }Q(\widehat{s},\delta )=\frac{\delta }{D^{2}}$ while $\frac{\partial }{\partial \delta }EV(\widehat{s},\delta )=-\frac{ \delta ^{2}}{D^{2}}$. Hence, $\frac{\partial Q/\partial \delta }{\partial EV/\partial \delta }=-\frac{1}{\delta }$. The concavity of $\Theta$ follows immediately from the monotonicity of $\frac{\partial Q/\partial \delta }{ \partial EV/\partial \delta }$ in $\delta$. $\square$

Figure 6. depicts the effect of requiring that the distribution of social resources be no more skewed than the historical distribution, i.e., imposing $\varepsilon ^{\circ }$ as an upper bound on the extent of discrimination, as described in footnote 33. In particular, the bold locus indicates the lower terminus of $\Gamma (\varepsilon ,\delta )$ for each value of $\delta$ in the event $\varepsilon =\varepsilon ^{\circ }$ rather than $\varepsilon =\widehat{s}$.

Fig. 6

Optimal remediation with $\varepsilon ^{\circ }$ lower bound

Appendix 2: Investment incentives

This appendix sketches a modification of the model to include incentive effects. First, as explained in footnote 4, if labor supply were endogenous, then since effort is expended after agents have been assigned to jobs the alternative remedial policies could not affect the individual labor supply decisions.³⁵ However, suppose agents invest in human capital or, equivalently, exert effort in school, and this affects the outcome of the education process; that is, suppose the education technology is $\sigma (a_{i}^{\circ },e_{i},s_{i})$, rather than $\sigma (a_{i}^{\circ },s_{i})$, where $e_{i}$ denotes i’s school effort.

Agent i’s preferences are of the form $u(y_{i},e_{i})$, where $y_{i}=w_{i}a_{i}^{\prime }$, and u is increasing in the first argument and decreasing in the second. It is natural to assume that i knows its own innate ability at the time of choosing $e_{i}$ but does not know the ability of agent with whom it will eventually compete for jobs. However, the distribution of $a^{\circ }$, namely f, is assumed to be common knowledge. Hence, i chooses $e_{i}$ to maximize its expected utility which depends on anticipated wage earnings and effort.

Subsequent to the choice of $e_{i}$ by each agent, the matching formula is the same. That is, in the absence of affirmative action, the agent with the higher effective ability is assigned to job h and the agent with the lower ability is assigned to $\ell$. With affirmative action, agent A is assigned to job h if and only if $a_{A}^{\prime }+\delta >a_{B}^{\prime }$ . Since i’s assignment and, hence, payoff is affected by j’s effort choice, this generates a game with incomplete information or a Bayesian game. Here, agent i’s type is given by $a_{i}^{\circ }$, and a strategy is a mapping $t_{i}$ which associates a level of effort with each $a_{i}^{\circ }\in \Lambda ^{\circ }$. (I thus refer to “agent $a_{i}^{\circ }$”.) A Bayesian Nash (BN) equilibrium is a pair of strategies $(t_{A}^{*},t_{B}^{*})$ such that for each $a_{i}^{\circ }$, $t_{i}^{*}(a_{i}^{\circ })$ is a best response to $t_{j}^{*}$ measured with respect to f, for $i,j=A,B,i\ne j$. After associating equilibrium effort levels with each vector of types $(a_{A}^{\circ },a_{B}^{\circ })$, one can then evaluate expected aggregate wealth and inequality as before.³⁶

To demonstrate, consider an extension of the previous example where $a_{i}^{\circ }$ is distributed uniformly on $\Lambda ^{\circ }$, $\sigma (a_{i}^{\circ },e,s_{i})=a_{i}^{\circ }+e_{i}+\sqrt{s_{i}}$ and $u(y_{i},e_{i})=y_{i}-ce_{i}$. Also, to simplify even further, assume $e_{i}\in \{0,1\}$; that is, agents either exert effort or they do not. In this case, a BN equilibrium is described heuristically as follows. First, given agent j’s strategy, $t_{j}$, let $Ew(a_{i}^{\circ },e_{i},t_{j})$ denote the expected wage of agent $a_{i}^{\circ }$ who chooses effort $e_{i}$ . Then the change in this agent’s expected benefit (utility net of costs) from choosing $e_{i}=1$ versus $e_{i}=0$ can be written as

$$\begin{aligned} \Delta EB(a_{i}^{\circ },t_{j})= & {} \sigma (a_{i}^{\circ },1,s_{i})Ew(a_{i}^{\circ },1,t_{j})- \sigma (a_{i}^{\circ },0,s_{i})Ew(a_{i}^{\circ },0,t_{j}) \end{aligned}$$

(34a)

$$\begin{aligned}= & {} [\sigma (a_{i}^{\circ },1,s_{i})-\sigma (a_{i}^{\circ },0,s_{i})]Ew(a_{i}^{\circ },1,t_{j}) \nonumber \\&\qquad +\,\sigma (a_{i}^{\circ },0,s_{i})[Ew(a_{i}^{\circ },1,t_{j})-Ew(a_{i}^{\circ },0,t_{j})] \end{aligned}$$

(34b)

$$\begin{aligned}= & {} Ew(a_{i}^{\circ },1,t_{j}) \nonumber \\&\qquad +\,(a_{i}^{\circ }+\sqrt{s_{i}})[Ew(a_{i}^{\circ },1,t_{j})-Ew(a_{i}^{\circ },0,t_{j})]\text {.} \end{aligned}$$

(34c)

One can distinguish two effects of effort on expected benefits corresponding to the two components in (34b). The first is the productivity effect: since effort increases productivity and, hence, earnings in either job, it may be worth choosing $e=1$ on this basis alone, even if $[Ew(a_{i}^{\circ },1,t_{j})-Ew(a_{i}^{\circ },0,t_{j})]=0$. The second is the competitive effect, where exerting greater effort increases the likelihood of obtaining job h.

The best response of agent $a_{i}^{\circ }$ to $t_{j}$ is determined by comparing $\Delta EB(a_{i}^{\circ },t_{j})$ to c: if $\Delta EB(a_{i}^{\circ },t_{j})>c$, the agent should choose $e_{i}=1$; and if $\Delta EB(a_{i}^{\circ },t_{j})<c$, then it should choose $e_{i}=0$. Indeed, if c is very high, namely, if $c>\Delta EB(a_{i}^{\circ },t_{j})$ for all $a_{i}^{\circ }\in \Lambda ^{\circ }$, then i will always choose $e_{i}=0$; and if $c<Ew(a_{i}^{\circ },1,t_{j})$ for all $a_{i}^{\circ }$, then it would choose $e_{i}=1$ on the basis of the productivity effect alone. Therefore, the interesting case is that in which $Ew(a_{i}^{\circ },1,t_{j})<c<\Delta EB(a_{i}^{\circ },t_{j})$ for some $a_{i}^{\circ }$. For such values, one can see from (34c) that if the competitive effect is sufficiently small, then $\Delta EB(a_{i}^{\circ },t_{j})<c$ and the optimal choice is $e_{i}=0$. For A, this would occur when $a_{A}^{\circ }$ is so low that it is unlikely to be assigned job h even if it were to choose $e_{A}=1$, and for B this would occur when $a_{B}^{\circ }$ is so high that it is likely to be assigned job h even if it were to choose $e_{B}=0$.³⁷ These are most likely to occur when the overlap between $\Lambda ^{\prime }(s_{A})$ and $\Lambda ^{\prime }(s_{B})$ is small,³⁸ that is, when $\sqrt{s_{B}}-\sqrt{ s_{A}}$ is large.

This suggests that a BN equilibrium would have the form

$$\begin{aligned} t_{A}^{*}(a_{A}^{\circ })= & {} \left\{ \begin{array}{cc} 0 &{}\quad \text {if }\,\,a_{A}^{\circ }<\widetilde{a}_{A}^{\circ } \\ 1 &{}\quad \text {if }\,\,a_{A}^{\circ }\ge \widetilde{a}_{A}^{\circ } \end{array} \right. , \end{aligned}$$

(35)

$$\begin{aligned} t_{B}^{*}(a_{B}^{\circ })= & {} \left\{ \begin{array}{cc} 1 &{} \quad \text {if }\,a_{B}^{\circ }<\widetilde{a}_{B}^{\circ } \\ 0 &{} \quad \text {if }\,a_{B}^{\circ }\ge \widetilde{a}_{B}^{\circ } \end{array} \right. , \end{aligned}$$

(36)

where $\widetilde{a}_{i}^{\circ }\in \Lambda ^{\circ }$ is determined by the equation

$$\begin{aligned} c=\Delta EB(\widetilde{a}_{i}^{\circ },t_{j}^{*}) \end{aligned}$$

(37)

if such a solution exists. Otherwise, if $c<\Delta EB(a_{i}^{\circ },t_{j}^{*})$ for all $a_{i}^{\circ }\in \Lambda ^{\circ }$, then $\widetilde{a}_{A}^{\circ }=\underline{a}^{\circ }$ and $\widetilde{a} _{B}^{\circ }=\overline{a}^{\circ }$; and if $c>\Delta EB(a_{i}^{\circ },t_{j}^{*})$ for all $a_{i}^{\circ }\in \Lambda ^{\circ }$, then $\widetilde{a}_{A}^{\circ }=\overline{a}^{\circ }$ and $\widetilde{a} _{B}^{\circ }=\underline{a}^{\circ }$.

Turning to the evaluation of welfare, a BN equilibrium of the above form (endogenously) partitions the type space $\Lambda ^{\circ }\times \Lambda ^{\circ }$ into four quadrants: $[\underline{a}^{\circ },\widetilde{a} _{A}^{\circ })\times [\underline{a}^{\circ },\widetilde{a}_{B}^{\circ })$, $[\widetilde{a}_{A}^{\circ },\overline{a}^{\circ }]\times [\underline{a}^{\circ },\widetilde{a}_{B}^{\circ })$, $[\underline{a}^{\circ },\widetilde{a}_{A}^{\circ })\times [\widetilde{a}_{B}^{\circ }, \overline{a}^{\circ }]$, and $=[\widetilde{a}_{A}^{\circ },\overline{a} ^{\circ }]\times [\widetilde{a}_{B}^{\circ },\overline{a}^{\circ }]$. EV can be determined by computing the expected value of $V(\mathbf {a} ^{\circ },\mathbf {s})=a_{H}^{\prime }w^{h}+a_{L}^{\prime }w^{\ell }$ over the partition, where $a_{H}^{\prime }=\max \{a_{A}^{\prime },a_{B}^{\prime }\}$, $a_{L}^{\prime }=\min \{a_{A}^{\prime },a_{B}^{\prime }\}$, and $a_{i}^{\prime }=\sigma (a_{i}^{\circ },t_{i}^{*}(a_{i}^{\circ }),s_{i})$ for $i=A,B$.

Regarding the likelihood of inversions, if the agents choose the same level of e, then the evaluation of I is straightforward; that is, if they both choose $e=0$ or both choose $e=1$, then the analysis would be the same as in the main text, although the values of $\sigma (a_{A}^{\circ },e,s_{A})$ and $\sigma (a_{B}^{\circ },e,s_{B})$ would differ in the two cases. Also, in the event $a_{A}^{\circ }>a_{B}^{\circ }$ and $\sigma (a_{A}^{\circ },1,s_{A})<\sigma (a_{B}^{\circ },0,s_{B})$, then it is clear that the resulting job assignment is the result of discrimination, although the criterion for inversion is significantly weaker.³⁹ The difficulty lies in the case in which $a_{A}^{\circ }>a_{B}^{\circ }$ and $\sigma (a_{A}^{\circ },0,s_{A})<\sigma (a_{B}^{\circ },1,s_{B})$. Then it is not clear whether B would obtain job h because of the asymmetry in s or because B exerted greater effort. However, it is easy to accommodate this as well as the previous case in which $a_{A}^{\circ }>a_{B}^{\circ }$ and $\sigma (a_{A}^{\circ },1,s_{A})<\sigma (a_{B}^{\circ },0,s_{B})$.

Since the education technology $\sigma (a_{i}^{\circ },e_{i},s_{i})$ is assumed to be known, one can accurately isolate and measure the effect of s by evaluating the effective productivities if the agents had chosen the same effort level. That is, even if it were the case that $e_{A}\ne e_{B}$, it is possible to consider what $\sigma$ would have been had they chosen the same level, i.e., to compare $\sigma (a_{A}^{\circ },0,s_{A})$ to $\sigma (a_{B}^{\circ },0,s_{B})$ or $\sigma (a_{A}^{\circ },1,s_{A})$ to $\sigma (a_{B}^{\circ },1,s_{B})$ in order to render the comparison meaningful.⁴⁰ In this case, the measure of inversions is the same as in the main section, namely,

$$\begin{aligned} I(\mathbf {s})=\int _{\underline{a}^{\circ }}^{\overline{a}^{\circ }}\int _{a_{A}^{\prime }}^{\sigma (a_{A}^{\circ },e,s_{B})}f_{B}(a_{B}^{\prime })f(a_{A}^{\circ })da_{B}^{\prime }da_{A}^{\circ }\text {,} \end{aligned}$$

where e is the common effort level.

The next question is how including school effort is likely to affect the policy instruments and the determination of an optimal policy. First, as mentioned above, when the agents choose the same level of effort, either 0 or 1, then the policy analysis will be the same. However, aggregate wealth would be greater if the agents were to choose $e=1$ than if they were to choose $e=0$. Generally, as discussed above, the agents are more likely to choose $e=0$ (in equilibrium) when the overlap $\Lambda ^{\prime }(s_{A})\cap \Lambda ^{\prime }(s_{B})$ is small or $\sqrt{s_{B}}-\sqrt{s_{A} }$ is large. Since resource transfers directly increase the overlap, they will increase the degree of competition between A and B and thus induce greater effort. Affirmative action will also induce greater effort, but via a different mechanism. Rather than changing the support of the distributions of ex post ability, the effect of AA would be to lower $\widetilde{a} _{A}^{\circ }$ and to raise $\widetilde{a}_{B}^{\circ }$, thus increasing the proportion of both types who choose $e=1$. In effect, AA will render the outcome more uncertain for the marginal A and B agents. For the marginal A agent, AA increases the probability that it will be assigned job h, thereby placing h “within reach” and thus increasing the competitive effect in (34c). For the marginal B agent, AA decreases the probability of h, thus putting the assignment “at risk”, and requiring greater effort to maintain.

In summary, both policies will lead to increased effort. When both agents do choose $e=1$, the analysis of inversions will be similar to that in the main text. It will also increase expected aggregate wealth relative to the case in which both choose $e=0$. However, the overall effect on expected aggregate wealth is again likely to be highly dependent on the parameters of the problem and worthy of systematic study.

A full analysis of investment incentive effects and the determination of an optimal policy is beyond the scope of the present paper. Before undertaking such an investigation, however, there are several fundamental issues to consider. First, as mentioned earlier in footnote 36, the approach is non-welfarist in that the social goals are unrelated to the preferences of the agents. Next, when effort is endogenous, it is implicitly assumed that greater opportunities resulting from higher levels of effort are well-deserved and need not be subject to remediation. However, if one interprets the model as describing the effect of primary and secondary education on job opportunities, then one might ask how young is too young to be fully responsible for the consequences of one’s decisions? And finally, both here and in the general discussion of responsibility, there is the presumption that effort is independent of circumstances, but that may not be the case; it may be that effort is influenced by external resources and education culture.⁴¹ Exploring these issues as well as developing the model more thoroughly and verifying the previous conjectures is left for future research.