(i) Let X, X and Y , Y ’ be independent samples of size 2 from continuous distributions F and G, respectively. Then p = P{max(X, X

) < min(Y, Y

)} + P{max(Y, Y

) < min(X, X

)}

= 1 3 + 2, where  = 

(F − G)2 d[(F + G)/2].

(ii)  = 0 if and only if F = G.

[(i): p = 

(1 − F)2 dG2 + 

(1 − G)2 d F2 which after some computation reduces to the stated form.

(ii):  = 0 implies F(x) = G(x) except on a set N which has measure zero both under F and G. Suppose that G(x1) − F(x1) = η > 0. Then there exists x0 such that G(x0) = F(x0) + 1 2 η and F(x) < G(x)for x0 ≤ x ≤ x1. Since G(x1) − G(x0) > 0, it follows that  > 0.]