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Self-Selection and Discrimination in Credit Markets Stanley D. Longhofer, Federal Reserve Bank of Cleveland Stephen R. Peters, University of Illinois Supplemental Proofs LEMMA 3: If θ Am > θ Bm , then q X ( s ) < q A ( s) for all s. Proof of Lemma 3: If no group B applicants attempt to self-select toward bank Y, θ p( s | θ ) 1 q A ( s) − q X ( s) = ∫ θ Am ∝ = ∫θ 1 m A p( s | θ ) dθ 1 1 θ Bm dθ − θ p( s | θ ) 1 ∫ θ Bm ∫θ 1 m B dθ p( s | θ ) dθ 1 1 θ Am θ Am θ Bm θ Am 1 1 θ Am θ Bm θ Am θ Am θ Bm ∫ p(s | θ )dθ ∫θ p( s | θ )dθ − ∫ p( s | θ )dθ ∫ θ p( s | θ )dθ . (1) ∫ p(s | θ )dθ ∫ θ p( s | θ )dθ − ∫ p(s | θ )dθ ∫ θ p( s | θ )dθ . Now, θ Am 1 θ Am 1 1 θ Am θ Bm θ Am θ Am θ Bm ∫ p(s | θ )dθ ∫ θ p(s | θ )dθ > θ ∫ p(s | θ )dθ ∫ p( s | θ )dθ > ∫ p(s | θ )dθ ∫ θ p( s | θ )dθ , m A θ Bm θ Am (2) . implying that q A ( s ) > q X ( s ) for all s. If group low-θ group B applicants do attempt to self-select toward bank X, there are two cases to consider: θ Am < θ c and θ Am > θ c . If θ Am < θ c , θc θ p(s | θ ) 1 q A (s ) − q X (s) = ∫ θ Am ∫θ 1 m A p ( s | θ )dθ dθ − 1 ∫ θ p(s | θ )γ dθ + ∫ θ p(s | θ ) θ Bm θ θc ∫θ m B 1 2 dθ c p ( s | θ )γ dθ + ∫ c p ( s | θ ) 1 2 dθ 1 θ 1 θ ∝ ∫ θ p ( s | θ ) dθ + ∫ θ p ( s | θ ) d θ m θc θA c θc 1 θA × ∫ p ( s | θ )γ dθ + ∫ p ( s | θ )γ dθ + ∫ p ( s | θ ) 1 2 dθ θ m θ Am θc B c 1 θ − ∫ p ( s | θ )dθ + ∫ p ( s | θ )dθ θ m θc A m (3) θc 1 θA × ∫ θ p ( s | θ )γ dθ + ∫ θ p ( s | θ )γ dθ + ∫ θ p ( s | θ ) 1 2 dθ . θ m θ Am θc B m Simplifying this expression yields θc 1 1 θc q A ( s ) − q X ( s ) ∝ (γ − 1 2 ) ∫ p( s | θ )dθ ∫ θ p( s | θ )dθ − ∫ θ p( s | θ )dθ ∫ p( s | θ )dθ θ m θc θ Am θc A m m c c θA θ θ θA + γ ∫ p ( s | θ ) dθ ∫ θ p ( s | θ ) dθ − ∫ θ p ( s | θ ) dθ ∫ p ( s | θ ) dθ θ m θ Am θ Bm θ Am B m m θA 1 1 θA + γ ∫ p( s | θ )dθ ∫ θ p( s | θ )dθ − ∫ θ p( s | θ )dθ ∫ p( s | θ )dθ . θ m θc θ Bm θc B (4) Using the same bounding techniques employed in (2) above, each of the terms in this expression are positive, proving that q A ( s ) > q X ( s ) . Finally, suppose that θ Am > θ c . In this case, θA 1 θc q A ( s ) − q X ( s ) ∝ ∫ θ p ( s | θ )dθ ∫ p( s | θ )γ dθ + ∫ p ( s | θ ) 1 2 dθ + ∫ p ( s | θ ) 1 2 dθ θm θ Am θc θ Am B m c θA 1 1 θ − ∫ p ( s | θ )dθ ∫ θ p ( s | θ )γ dθ + ∫ θ p ( s | θ ) 1 2 dθ + ∫ θ p ( s | θ ) 1 2 dθ θ m θ Am θc θ Am B c c θ θ 1 1 = γ ∫ θ p ( s | θ )dθ ∫ p ( s | θ )dθ − ∫ p ( s | θ )dθ ∫ θ p ( s | θ )dθ θ m θ Bm θ Am θ Bm A m m θA θA 1 1 − 1 2 ∫ θ p ( s | θ )dθ ∫ p ( s | θ )dθ − ∫ p ( s | θ )dθ ∫ θ p ( s | θ )dθ . θ m θc θ Am θc A m 1 (5) Once again, bounding arguments similar to those used in (2) above can be used to verify that this expression must be positive, proving that q A ( s ) > q X ( s ) . ♠ Proof that qY ( s ) > q A ( s) for all s (Proposition 2, part 2): Since θ Bm > θ Am , we can write θc θ p( s | θ ) 1 q A ( s ) − qY ( s ) = ∫ θ Am ∫θ 1 m A p( s | θ )dθ dθ − 1 ∫ θ p( s | θ )(1 − γ )dθ + ∫ θ p(s | θ ) θ Bm θ θc ∫θ m B This can be simplified to 2 dθ p( s | θ )(1 − γ )dθ + ∫ c p ( s | θ ) 1 2 dθ 1 θ θ 1 ∝ ∫ θ p ( s | θ )dθ + ∫ θ p( s | θ )dθ + ∫ θ p ( s | θ )dθ m θ Bm θc θA c 1 θ × ∫ p( s | θ )(1 − γ )dθ + ∫ p ( s | θ ) 1 2 dθ θ m θc B m c θ 1 θB − ∫ p ( s | θ )dθ + ∫ p ( s | θ )dθ + ∫ p( s | θ )dθ θ m θ Bm θc A c 1 θ × ∫ θ p( s | θ )(1 − γ )dθ + ∫ θ p ( s | θ ) 1 2 dθ . θ m θc B θ Bm 1 c c (6) θB θc θc θB q A ( s ) − qY ( s ) ∝ (1 − γ ) ∫ θ p( s | θ )dθ ∫ p( s | θ )dθ − ∫ p( s | θ )dθ ∫ θ p( s | θ )dθ θ m θ Bm θ Am θ Bm A m m θB 1 1 θB + 1 2 ∫ θ p ( s | θ ) dθ ∫ p ( s | θ ) dθ − ∫ p ( s | θ ) dθ ∫ θ p ( s | θ ) dθ θ m θc θ Am θc A c c θ 1 1 θ + (γ − 1 2 ) ∫ θ p( s | θ )dθ ∫ p( s | θ )dθ − ∫ p( s | θ )dθ ∫ θ p( s | θ )dθ . θ m θc θ Bm θc B m m (7) Once again, using the bounding techniques applied in (2) above, we see that this expression must be negative, proving that qY ( s ) > q A ( s) . ♠