14 CHAPTER 7.STATISTICAL FUNCTIONALS AND THE DELTA METHOD Example 4.6 For distribution functions F,G on R,define T by T(F,G)= FdG=P(X≤Y) where X~F,Y~G are independent.Let F-Flloo sup F(z)-F(x)=F-FllF where F={1(-oe,:t∈R} Proposition 4.1 A.T(F,G)is Hadamard differentiable with respect to at every pair of df's (F,G)with derivative T given by ③)t(E,G:a,)=adG-8r B.T(F,G)is not Frechet differentiable with respect to lllloo Proof.The following proof of A is basically Gill's(1989),lemma 3.For F-F and Gt-G, define at =(F-F)/t and B=(Gt-G)/t;for Hadamard differentiability we have at-a and f一B with resepct to‖l·lloo for some(bounded)functions o and.Now T(F,G)-T(F,G)_(auB)=t odBt ad(G:-G)+ (at-a)d(G:-G). Since T is continuous,it suffices to show that the right side converges to 0.The second term on the right is bounded by lov-alfdce+dG ≤2l‖la4-a→0. Fix e>0.Since the limit function a in the first term is right-continuous with left limits,there is a step function with a finite number m of jumps,a say,which satisfies lla-alloo<e.Thus the first term may be bounded as follows: Iad(G:-G)I s(a-a)d(G:-G)+1 ad(G:-G) ≤2a-ax+∑1a(g-)l(G-Gxj-1,x川 ≤2e+2 mllalloolGt-Glo→2e. Since e is arbitrary,this completes the proof of A. Here is the proof of B.If T were Frechet-differentiable,it would have to be true that (a)T(Fn;Gn)-T(F,G)-T(Fn-F,Gn-G)=o(llFn-Flloo VIIGn-Glloo) for every sequence of pairs of d.f.'s {(Fn,Gn)}with IlFn-Fllo-0 and IlGn-Gllo0.We now exhibit a sequence ((Fn,Gn)}for which (a)fails. By straightforward algebra using(③)， (b )T(Fn,Gn)-T(F,G)-T(Fn-F,Gn-G)=(Fn-F)d(Gn-G)14 CHAPTER 7. STATISTICAL FUNCTIONALS AND THE DELTA METHOD Example 4.6 For distribution functions F, G on R, define T by T(F, G) = # F dG = P(X ≤ Y ) where X ∼ F, Y ∼ G are independent. Let .F˜ − F.∞ ≡ supx |F˜(x) − F(x)| ≡. F˜ − F.F where F = {1(−∞,t] : t ∈ R}. Proposition 4.1 A. T(F, G) is Hadamard differentiable with respect to . ·. ∞ at every pair of df’s (F, G) with derivative T˙ given by T˙((F, G); α, β) = # αdG − # (3) βdF. B. T(F, G) is not Fr´echet differentiable with respect to . · .∞. Proof. The following proof of A is basically Gill’s (1989), lemma 3. For Ft → F and Gt → G, define αt ≡ (Ft − F)/t and βt ≡ (Gt − G)/t; for Hadamard differentiability we have αt → α and βt → β with resepct to . · .∞ for some (bounded) functions α and β. Now T(Ft, Gt) − T(F, G) t − T˙(αt, βt) = t # αtdβt = # αd(Gt − G) + # (αt − α)d(Gt − G). Since T˙ is continuous, it suffices to show that the right side converges to 0. The second term on the right is bounded by .αt − α.∞ (# dGt + # dG) ≤ 2.αt − α.∞ → 0. Fix ' > 0. Since the limit function α in the first term is right-continuous with left limits, there is a step function with a finite number m of jumps, ˜α say, which satisfies .α − α˜.∞ < '. Thus the first term may be bounded as follows: | # αd(Gt − G)| ≤| # (α − α˜)d(Gt − G)| + | # α˜d(Gt − G)| ≤ 2.α − α˜.∞ +!m j=1 |α˜(xj−1)||(Gt − G)[xj−1, xj )| ≤ 2' + 2m.α˜.∞.Gt − G.∞ → 2'. Since ' is arbitrary, this completes the proof of A. Here is the proof of B. If T were Fr´echet - differentiable, it would have to be true that T(Fn, Gn) − T(F, G) − T˙ (a) (Fn − F, Gn − G) = o(.Fn − F.∞ ∨ .Gn − G.∞) for every sequence of pairs of d.f.’s {(Fn, Gn)} with .Fn − F.∞ → 0 and .Gn − G.∞ → 0. We now exhibit a sequence {(Fn, Gn)} for which (a) fails. By straightforward algebra using (3), T(Fn, Gn) − T(F, G) − T˙(Fn − F, Gn − G) = # (b) (Fn − F)d(Gn − G)