## Transcribed Text

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 En
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 11 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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