(t-1/T<r<t/T, we have r1∑Wr(-1)=∑ Wr(r t=1 Wr(r)dr The continuous mapping theorem applies to h(Wr)=Jo Wr(r)dr. It follows that h(WT)= h(W), so that T-3/2 Y-1=oo w(r)dr, as claimed (d). For a random walk, Y=(Yt-1+ut)2=Y_1+2Yt-1ut +ut, imply- ing that Yt-1Wt=(1/2)(Y-Y21-u2 and then 2t-Yi-1ut=(1 /2)Y Y)-(1/2)>h uf. Recall that Yo=0, and thus it is convenient to write 1Y1-1t=是1-∑12. From items(a) we know that T-1( (T-12∑1u)2a2(1)amnd∑1e→o2 by Lln( Kolmogorov);then, ∑=1-1l4=>lW(12-1 (e).Wefirstobservethat∑=1Y-1=[1+(u1+u2)+(1+u2+u3)+…+(u1+ u2+3+.+-1)=T-1)u1+(T-2)u2+(T-3)u3+…+{T-(T-1)ur-= (T-t) T Therefore, T-2 ti tut=T-i2tiut-T-i2t1Yt-1. By applying the con- tinuous mapping theorem to the joint convergence of items(a) and(c), we have W(1)-/w)dr t=1 The proofs of item(f)and(g)is analogous to those of(c) and(b). First rewrite 1 tYi- in term of Wr(rt-1)=T-1/Y-1/o=T-122-lus/o,where Tt-1=(t-1)/T, so that T-5/2>, t Yt-1= oT- E tWr(rt-1).Because Wr(r)is constant for(t-1/T<r<t/T, we have r∑mWr(-1)=∑/rr( r)dr forr=tm rWr(r)dr The continuous mapping theorem applies to h(wr)=o rWr(r)dr. It follows that h(Wr)=h(W), so that T-5/2 2t, tY-1=>0 rw(r)dr,as claimed(t − 1)/T ≤ r < t/T, we have T −1X T t=1 WT (rt−1) = X T t=1 Z t/T (t−1)/T WT (r)dr = Z 1 0 WT (r)dr. The continuous mapping theorem applies to h(WT ) = R 1 0 WT (r)dr. It follows that h(WT ) =⇒ h(W), so that T −3/2 PT t=1 Yt−1 =⇒ σ R 1 0 W(r)dr, as claimed. (d). For a random walk, Y 2 t = (Yt−1 + ut) 2 = Y 2 t−1 + 2Yt−1ut + u 2 t , imply￾ing that Yt−1ut = (1/2){Y 2 t − Y 2 t−1 − u 2 t } and then PT t=1 Yt−1ut = (1/2){Y 2 T − Y 2 0 } − (1/2)PT t=1 u 2 t . Recall that Y0 = 0, and thus it is convenient to write PT t=1 Yt−1ut = 1 2 Y 2 T − 1 2 PT t=1 u 2 t . From items (a)) we know that T −1Y 2 T (= (T −1/2 PT t=1 us) 2 L−→ σ 2W2 (1) and PT t=1 u 2 t p −→ σ 2 by LLN (Kolmogorov); then, PT t=1 Yt−1ut =⇒ 1 2 σ 2 [W(1)2 − 1]. (e). We first observe that PT t=1 Yt−1 = [u1 + (u1 +u2) + (u1 +u2 +u3) +...+ (u1 + u2 +u3 +...+uT −1)] = [T −1)u1 + (T −2)u2 + (T −3)u3 +...+[T −(T −1)uT −1] = PT t=1(T − t)ut = PT t=1 Tut − PT t=1 tut , or PT t=1 tut = T PT t=1 ut − PT t=1 Yt−1. Therefore, T − 3 2 PT t=1 tut = T − 1 2 PT t=1 ut − T − 3 2 PT t=1 Yt−1. By applying the con￾tinuous mapping theorem to the joint convergence of items (a) and (c), we have T − 3 2 X T t=1 tut ⇒ σ[W(1) − Z 1 0 W(r)dr]. The proofs of item (f) and (g) is analogous to those of (c) and (b). First rewrite T −5/2 PT t=1 tYt−1 in term of WT (rt−1) ≡ T −1/2Yt−1/σ = T −1/2 Pt−1 s=1 us/σ, where rt−1 = (t − 1)/T, so that T −5/2 PT t=1 tYt−1 = σT −2 PT t=1 tWT (rt−1). Because WT (r) is constant for (t − 1)/T ≤ r < t/T, we have T −1X T t=1 (t/T)WT (rt−1) = X T t=1 Z t/T (t−1)/T rWT (r)dr for r = t/T = Z 1 0 rWT (r)dr. The continuous mapping theorem applies to h(WT ) = R 1 0 rWT (r)dr. It follows that h(WT ) =⇒ h(W), so that T −5/2 PT t=1 tYt−1 =⇒ σ R 1 0 rW(r)dr, as claimed. 11