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7.4 Exercises 2. Let f and g be continuous functions on [0,1] satisfying 0 ≤ f(x) ≤ Cg(x) for all x and some constant C > 0. Show that 􏰐 1 􏰐 1 􏰍n f(xi) 􏰎1f(x)dx = 􏰎0 􏰎1 0 􏰍i=1 0 ni=1g(xi) 1 n g(x)dx ̸= 0. , In fact, a more general result is: for i.i.d. r.v. X, X1, X2, ..., we have lim ... n→∞ 0 dx ...dx 1g(x)dx 0 where 􏰍ni=1 f(Xi) lim E􏰍n Ef(X) = , whereEg(X)̸=0. n→∞ i=1 g(Xi) (Hint: Apply the WLLN.) Eg(X) 11.9 Exercises 1. Show that if {Xn,n ≥ 1} are independent r.v.s with EXn = 0 and |Xn| ≤ Cn a.s., n ≥ 1, and 2 􏰍n 2 Cn = o(Bn), where Bn = i=1 EXn → ∞, then Sn/Bn =⇒ N(0,1). 2. if {Xn, n ≥ 1} are independent r.v.s with α11 2nβ , P(Xn =0)=1− , 2α>β−1, nβ P(Xn =±n )= the Lindeberg condition holds when and only when 0 ≤ β < 1. 3. Failure of the Lindeberg condition does not preclude asymptotic normality. Let {Yn, n ≥ 1} be i.i.d. with EYn = 0, EYn2 = 1. Let {Zn,n ≥ 1} be independent with P(Zn = ±n) = 1/(2n2) andP(Zn =0)=1−1/n2 and{Zn,n≥1}areindependetof{Yn,n≥1}. DefineXn =Yn+Zn, 􏰍n √ Sn = i=1 Xn. Show that the Lindeberg condition can not hold, but Sn/ n =⇒ N(0,1). Explain why this does not contravene Lindeberg-Feller Theorem. 4. Let {Xn, n ≥ 1} be independent with 1 1􏰅 1􏰆1 􏰅 1􏰆􏰅 1􏰆 P(Xn =±1)= , P(Xn =±n)= 1− , P(Xn =0)= 1− 1− , 2a2an2 an2 where n ≥ 1, a > 1. Again, Sn/√n =⇒ N(0,a) despite the Lindeberg condition being violated. 5. Let {Yn,n ≥ 1} be i.i.d. r.v.s with finite variance σ2 (say σ2 = 1), and let {σn2,n ≥ 1} be nonzero 2 􏰍n 2 constants with Bn = i=1 σi ↗ ∞. Show that the weighted i.i.d. r.v.s {σnYn,n ≥ 1} obey the CLT, i.e., 1 􏰏n σjYj =⇒N(0,1) B 6. LetX1,X2,...beindependentandSn =X1+...+Xn. Showthatif|Xi|≤M and􏰍∞i=1Var(Xi)= ifEY1 =0andσn =o(Bn). ∞, then Sn −ESn 􏰑 =⇒ N(0,1). P(Xi =±i)= (i) Find limn→∞ V ar(Sn)/n3; Prove that Tn =⇒ N (0, 1). n j=1 V ar(Sn) 7. Let X1, X2, ... be i.i.d. r.v.s such that P (Xi = ±1) = 1/2. Show that 􏰒 3 􏰏n n3 8. Let X1, X2, ... be independent r.v.s with 2 (ii) What is the limiting distribution of Sn/n3/2? 9. LetX1,X2,...bei.i.d. r.v.swithEXi =μandVar(Xi)=σ2 <∞. Put k=1 kXk =⇒N(0,1). 1121 − , P(Xi =±i )= , fori=1,2,.... 2i2 2i2 1 􏰏n 1 􏰏n √ X ̄= X, S2= (X−X ̄)2, T= n(X ̄−μ)/S. ninninn i=1 i=1

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