5. HIGH ORDER DERIVATIVES 17 5 High Order Derivatives The following example illustrates the phenomena which we want to consider here in the simplest possible setting: Example5.1 Suppose that Xi,X2,,Xn are i..id.Bernoulli(pl.Then√元(n-p)）→aZ~ N(0,p(1-p)),and,if g(p)=p(1-p), Vm(g(xn)-g(p)→dg(p)Z=(1-2p)Z~N(0,p(1-p)1-2p)2) by the delta-method (or g-prime theorem).But if p =1/2,since g'(1/2)=0 this yields only (g(区n)-1/4→a0. Thus we need to study the higher derivatives of g at 1/2.since g is,in fact,a quadratic,we have 9p)=9(1/2)+0(p-1/2)+(-2)0p-1/2)2=1/4-(p-1/2)2. Thus ngX)-91/2)=-n区。-1/22-a-z2w This is a very simple example of a more general limit theorem which we will develop below. Now consider a functional T:FR as in sections 1-4. Definition 5.1 T is k-th order Gateaux differentiable at F if,with F=F+t(G-F), dT(F;(G-F)=T(F)0 exists Note that T(F;G-F)=d(T;G-F)if it exists.It is usually the case that dT(F:G-F)=. k(x1,...,xk)d(G-F)(z1)...d(G-F)(xk) =… Vk.F(x1;...,Ik)dG(x1)...dG(k); here the function F is determined from v by a straightforward centering recipe: 1,F()=1()- 2(c1,x2)dF(c1)dF(x2),5. HIGH ORDER DERIVATIVES 17 5 High Order Derivatives The following example illustrates the phenomena which we want to consider here in the simplest possible setting: Example 5.1 Suppose that X1, X2, . . . , Xn are i.i.d. Bernoulli(p). Then √n(Xn − p) →d Z ∼ N(0, p(1 − p)), and, if g(p) = p(1 − p), √n(g(Xn) − g(p)) →d g% (p)Z = (1 − 2p)Z ∼ N(0, p(1 − p)(1 − 2p) 2) by the delta-method (or g-prime theorem). But if p = 1/2, since g% (1/2) = 0 this yields only √n(g(Xn) − 1/4) →d 0. Thus we need to study the higher derivatives of g at 1/2. since g is, in fact, a quadratic, we have g(p) = g(1/2) + 0 · (p − 1/2) + 1 2!(−2)(p − 1/2)2 = 1/4 − (p − 1/2)2. Thus n(g(Xn) − g(1/2)) = −n(Xn − 1/2)2 →d −Z2 ∼ −1 4 χ2 1. This is a very simple example of a more general limit theorem which we will develop below. Now consider a functional T : F → R as in sections 1 - 4. Definition 5.1 T is k−th order Gateaux differentiable at F if, with Ft = F + t(G − F), dkT(F; (G − F)) = dk dtk T(Ft) ' ' ' t=0 exists. Note that T˙(F; G − F) = d1(T; G − F) if it exists. It is usually the case that dkT(F; G − F) = # ··· # ψk(x1, . . . , xk)d(G − F)(x1)··· d(G − F)(xk) = # ··· # ψk,F (x1, . . . , xk)dG(x1)··· dG(xk); here the function ψk,F is determined from ψk by a straightforward centering recipe: ψ1,F (x) ≡ ψ1(x) − # ψ1dF, ψ2,F (x) ≡ ψ2(x1, x2) − # ψ2(x1, x2)dF(x2) − # ψ2(x1, x2)dF(x1) + # # ψ2(x1, x2)dF(x1)dF(x2)