Creditor Learning and Discrimination in Lending

Abstract

I study the implications of creditor learning for the estimated racial disparity in access to credit. Utilizing a dataset of mortgage lending, I find that the estimated racial disparity in loan approval rates declines with the length of the borrower’s credit history. In addition, minority borrowers improve significantly their chances of obtaining a loan by accumulating longer credit histories, with the improvements being the largest for those with no credit history. Importantly, I find no significant racial disparity among borrowers with long credit histories, suggesting that one cannot reject the null hypothesis of no taste-based discrimination taking place. I also conduct a number of tests to detect statistical discrimination, which yield inconclusive results.

Appendix

Lemma 1

Let (Y, ϵ _t ) be scalar random variables on a probability space $(\Omega, \mathcal{F}, P)$. Assume that ϵ _t is i.i.d., and that both Y and ϵ _t have zero means. Assume also that Y ⊥ ϵ _t for all t. Consider the linear projection $E(Y|H_T)=\sum_{t=1}^{T} b_{Tt} h_t$, where H _T  = {h₁, ⋯ , h _T }, and h _t  = Y + ϵ _t , t = 1, 2,  ⋯ , T. Let $\Delta(T)=\sum_{t=1}^{T} b_{Tt}$. Then (i) b _Tt  = b _T for all t; (ii) $b_T=\frac{b_{T-1}}{1+b_{T-1}}$, and $b_T=\frac{\sigma_e^2}{T \sigma_e^2+\sigma_\epsilon^2}$; (iii) 0 < Δ(T) < 1 and Δ(T) is increasing in T.

Proof

Let B = [b _T1, ⋯ , b _TT ]. Then $B = \Sigma_{yh} \Sigma_{hh}^{-1}$ with $\Sigma_{yh}=\text{COV}(y,H_T)$ and $\Sigma_{hh}=\text{V}(H_T)$. Note that Σ _yh is a 1 × T vector with all elements equal to $\sigma_y^2$ and that the diagonal elements of Σ _hh are all equal to $\sigma_y^2 +\sigma_\epsilon^2$ and the off-diagonal elements are all equal to $\sigma_y^2$. This implies that b _Tt does not depend on t. So $E(Y|H_T)=b_T \sum_{t=1}^{T} h_t$.

By the law of iterated projection,

$$ E(y|H_{T-1})\!=\! E(E(y|H_{T})|H_{T-1}) \!=\! b_T \left(\sum\limits_{t=1}^{T-1} h_t + E(h_T|H_{T-1}) \right) \!=\! b_T(1+b_{T-1}) \sum\limits_{t=1}^{T-1} h_t $$

(10)

But $E(y|H_{T-1})=b_{T-1} \sum_{t=1}^{T-1} h_t$. So (10) implies that b _T − 1 = b _T (1 + b _T − 1), or $b_T=\frac{b_{T-1}}{1+b_{T-1}}$. Because $b_1=\frac{\sigma_y^2}{\sigma_y^2 +\sigma_\epsilon^2}$, compute b _T recursively to obtain $b_T=\frac{\sigma_e^2}{T \sigma_e^2+\sigma_\epsilon^2}$. Therefore, $\Delta(T)=\frac{T\sigma_e^2}{T \sigma_e^2+\sigma_\epsilon^2}$, which is in (0, 1) and increases in T.□

Proof of Proposition 1

Use assumptions on η and Z to rewrite (1) as:

$$ \pi=\alpha_x X+\left(\phi_g+\alpha_z\kappa_g\right)G+\nu_\eta+\alpha_z \nu_z. $$

(11)

By the law of iterated projection, $E(R_T^d|X,G,H_T) = E(E(\pi|X,Z,G,H_T)|X,G, H_T) \!=\! E(\pi|X,G,H_T).$ Therefore, $E(R_T^d|X,G,H_T) \!=\!\alpha_x X\!+\!(\phi_g\!+\!\alpha_z\kappa_g)G\!+\!E(\nu_\eta+ \alpha_z \nu_z|F_T).$ This proves Result 1 because $E(\nu_\eta+\alpha_z \nu_z|F_T)=\sum_{t=0}^{T} \gamma_{Tt} f_t$, and f _t  = h _t  − (α _x X + (φ _g  + α _z κ _g )G). The law of iterated projection also implies $E(R_T^d|X,G)=E(E(R_T^d|X,G,H_T)|X,G)=E(\pi|X,G).$ Project both sides of (11) on (X, G), $E(R_T^d|X,G)=\alpha_x X+(\phi_g+\alpha_z\kappa_g)G.$ This proves Result 2.□

Proposition 2

Assume that statistical discrimination does not exist. Assume also that G = E(G|X, Z) + ν _g  = β _x X + β _z Z + ν _g . Let $E(\nu_z|F_T)=\sum_{t=0}^{T} \theta_{Tt} f_t$ with f _t and F _T defined in Proposition 1 and $\Theta(T)=\sum_{t=0}^{T} \theta_{Tt}$. Let $E(\nu_\eta+\phi_g \nu_g |K_T)=\sum_{t=0}^{T} \lambda_{Tt} k_t$ and $\Lambda (T)=\sum_{t=0}^{T}\lambda_{Tt}$, where K _T  = { k₀, k₁, ..., k _T } with k _t  = h _t  − E(π|X, Z). Denote s _i  = α _i  + ϕ _g β _i , i = x,z. Then,

• 1. $E(R_T^{nd}|X,Z,G,H_T)\!=\! R_T^{nd}\!=\!(\alpha_x\!+\!\phi_g\beta_x)(1\!-\!\Lambda(T))X\!+\!(\alpha_z\!+\!\phi_g \beta_z) (1\!-\!\Lambda(T)) Z+ \sum_{t=0}^{T} \lambda_{t} h_t.$

• 2. $E(R_T^{nd}|X,Z,G)=b_{xT}X+b_{zT}Z+b_{gT}G$ with b _iT  = s _i  − ϕ _g β _i Λ(T), i = x, z, and b _gT  = ϕ _g Λ(T); also $\frac{b_{iT}-b_{i0}}{b_{gT}-b_{g0}}=-\beta_i$, i = x, z.

• 3. $E(R_T^{nd}|X,G,H_T)\!=\!b_{xT} X+b_{gT} G+\sum_{t=0}^{T} b_{hT}(t) h_t$, with b _xT  = (s _x  − s _z α _x Θ(T)) (1 − Λ(T)), b _gT  = s _z (κ _g  − (ϕ _g  + α _z κ _g )Θ(T))(1 − Λ(T)), and b _hT (t) = λ _Tt  + s _z (1 − Λ(T))θ _Tt .

• 4. $E(R_T^{nd}|X,G) = b_{xT} X + b_{gT} G,$ with b _xT  = s _x  − ϕ _g β _x Λ(T) and b _gT  = s _z κ _g  + ϕ _g (1 − β _z κ _g )Λ(T); also $\frac{b_{xT}-b_{x0}}{b_{gT}-b_{g0}}=-b_{gx}$, where E(G|X) = b _gx X.

Proof

Substituting G in (2) and projecting both sides on (X, Z, H _T ), I obtain a non-discriminating lender’s expected profits:

$$ R_T^{nd} = \left(\alpha_x X+\alpha_z Z\right)+\phi_g \left(\beta_x X + \beta_z Z\right) + E\left(\nu_\eta + \phi_g \nu_g |K_T\right). $$

(12)

Comparing to (3), the above equation shows that while a discriminating lender is able to use G to identify directly the entire race-correlated private information ϕ _g G, the non-discriminating lender can predict it using (X, Z) up to ϕ _g (β _x X + β _z Z) and pretend the residual ϕ _g ν _g were unpredictable and try to “learn” it from the borrower’s credit history.□

Solve (12) in terms of the underwriting data (X, Z, H _T ):

$$ R_T^{nd}=\left(\alpha_x+\phi_g\beta_x\right)\left(1-\Lambda(T)\right)X+(\alpha_z+\phi_g \beta_z) (1-\Lambda(T)) Z+\sum\limits_{t=0}^{T} \lambda_{t} h_t, $$

(13)

For a researcher with the data on (X, Z, G, H _T ), a logit or probit regression of approval on (X, Z, G, H _T ) will provide a consistent estimate of $E(R_T^{nd}|X,Z,G,H_T)$. By the law of iterated projection,

$$ E\left(R_T^{nd}|X,Z,G,H_T\right)\!=E\left(E(\pi|X,Z,H_T)|X,Z,G,H_T\right)\!=E\left(\pi|X,Z,H_T\right) \!= R_T^{nd}. $$

(14)

Thus, the coefficient of G in the approval regression on (X, Z, G, H _T ) should be zero. This proves Result 1.

By the law of iterated projection, (2) implies that E(h _t |X, Z, G) = E(π|X, Z, G) = α _x X + α _z Z + ϕ _g G. Use this and (13) to solve for $E(R_T^{nd}|X,Z,G)$;

$$\begin{array}{rll} E\left(R_T^{nd}|X,Z,G\right) &=&s_x (1-\Lambda(T))X+s_z (1-\Lambda(T))Z+\sum\limits_{t=0}^{T} \lambda_{Tt} E\left(h_t|X,Z,G\right) \\ &=&\left(s_x-\phi_g\beta_x\Lambda(T)\right)X+\left(s_z-\phi_g\beta_z\Lambda(T)\right)Z+\phi_g\Lambda(T)G \end{array} $$

(15)

That is, in the probit regression of approval on (X, Z, G), the coefficient of G is ϕ _g Λ(T), which increases in T in absolute value (by Lemma 1). Moreover, because Λ(0) = 0, $\frac{b_{iT}-b_{i0}}{b_{gT}-b_{g0}} = -\beta_i, i=x,z$. This proves Result 2.

To solve for $E(R_t^{nd}|X,G,H_T)$, project both sides of (13) on (X, G, H _T ):

$$ E\left(R_T^{nd}|X,G,H_T\right)=s_x(1-\Lambda(T))X+s_z(1-\Lambda(T))E\left(Z|X,G,H_T\right) +\sum\limits_{t=0}^{T} \lambda_{Tt} h_t. $$

(16)

Because f _t  = h _t  − E(π|X, G), f _t is orthogonal to (X, G). Also, (11) implies E(π|X, G) = α _x X + (ϕ _g  + α _z κ _g )G. Thus, assumption of Z implies:

$$\begin{array}{rll} E(Z|X,G,H_T) & =&E(Z|X,G,F_T)=\kappa_g G+E\left(\nu_z|F_T\right) = \kappa_g G + \sum\limits_{t=0}^{T} \theta_{Tt} f_t \\ & =&-\alpha_x \Theta(T) X +\left(\kappa_g - \left(\phi_g+\alpha_z\kappa_g\right) \Theta(T)\right)G+ \sum\limits_{t=0}^{T} \theta_{Tt} h_t. \end{array} $$

(17)

Substitute (17) for E(Z|X, G, H _T ) in (16) and rearrange to obtain Result 3.¹⁰

Finally, because $E(R_T^{nd}|X,G) = E(E(R_T^{nd}|X,Z,G)|X,G)$, project both sides of (15) on (X, G) to solve for $E(R_T^{nd}|X,G)$:

$$\begin{array}{rll} E\left(R_T^{nd}|X,G\right) &=&\left[s_x-\phi_g\beta_x\Lambda(T)\right]X+\left[s_z-\phi_g\beta_z\Lambda(T)\right]E(Z|X,G)+\phi_g\Lambda(T)G \\ &=&\left[s_x-\phi_g\beta_x\Lambda(T)\right]X+\left[s_z\kappa_g+\phi_g(1-\beta_z\kappa_g) \Lambda(T)\right]G. \end{array} $$

(18)

Thus, $\frac{b_{xT}-b_{x0}}{b_{gT}-b_{g0}}=-\frac{\beta_x}{1-\beta_z\kappa_g}$. But by assumptions on Z and G, $G = \frac{\beta_x}{1-\beta_z \kappa_g} X + \frac{\beta_z \nu_z+\nu_g}{1-\beta_z \kappa_g}.$ Because X is orthogonal to the residue, the coefficient of the linear projection of G on X is $\frac{\beta_x}{1-\beta_z\kappa_g}$. This proves Result 4.