Chapter 12

The Central Limit Problem: The Centered Case

[11/100%]
Show
LaTeX
Problem 12.1

If the independent r.v.s Xj,j≥1X_{j}, j \geq 1, are distributed as U(−j,j)U(-j, j), then show that the Lindeberg condition (see relation (12.24)) holds, so that L(SnSn)⇒n→∞N(0,1)\mathcal{L}\left(\frac{S_{n}}{S_{n}}\right) \underset {n \rightarrow \infty }{\Rightarrow } N(0,1), where Sn=∑j=1nXjS_{n} = \sum_{j = 1}^{n} X_{j} and sn2=∑j=1nσj2,σj2=Var⁡[Xj]s_{n}^{2} = \sum_{j = 1}^{n} \sigma_{j}^{2}, \sigma_{j}^{2} = \operatorname {Var}\left[X_{j}\right].

?
Problem 12.2

If the independent r.v.s Xj,j≥2X_{j}, j \geq 2, are distributed as follows:

P(Xj=−jα)=P(Xj=jα)=1jβ,P(Xj=0)=1−2jβ(α,β>0) \begin{aligned} \mathbb {P}\left(X_{j} = -j^{\alpha }\right) = \mathbb {P}\left(X_{j} = j^{\alpha }\right) = \frac{1}{j^{\beta }}, \quad \mathbb {P}\left(X_{j} = 0\right) \\ & = 1-\frac{2}{j^{\beta }} \quad (\alpha , \beta > 0) \end{aligned}

show that the restriction β<1\beta < 1 ensures that the Lindeberg condition (relation (12.24)) holds.

?
Problem 12.3

Let the r.v.s Xj,j≥1X_{j}, j \geq 1, be distributed as follows:

P(Xj=±jα)=16j2(α−1),P(Xj=0)=1−13j2(α−1),α>1 \begin{aligned} \mathbb {P}\left(X_{j} = \pm j^{\alpha }\right) = \frac{1}{6 j^{2(\alpha -1)}}, \quad \mathbb {P}\left(X_{j} = 0\right) = 1-\frac{1}{3 j^{2(\alpha -1)}}, \quad \alpha > 1 \end{aligned}

Show that the Lindeberg condition (relation (12.24)) holds, if and only if α<32\alpha < \frac{3}{2}. Conclude that L(Snsn)⇒n→∞N(0,1)\mathcal{L}\left(\frac{S_{n}}{s_{n}}\right) \underset {n \rightarrow \infty }{\Rightarrow } N(0,1), where Sn=∑j=1nXjS_{n} = \sum_{j = 1}^{n} X_{j} and sn2=∑j=1nσj2s_{n}^{2} = \sum_{j = 1}^{n} \sigma_{j}^{2}, σj2=Var⁡[Xj]\sigma_{j}^{2} = \operatorname {Var}\left[X_{j}\right].

?
Problem 12.4

For j=1,2,…,nj = 1,2, \ldots , n, let XjX_{j} be independent r.v.s defined as follows:

Xj={±j2 with probability 1/12j2 each ±j with probability 1/12 each 0 with probability 1−1/6−1/6j2 X_{j} = \left\{ \begin{array}{l} \pm j^{2} \text{ with probability } 1 / 12 j^{2} \text{ each } \\ \pm j \text{ with probability } 1 / 12 \text{ each } \\ 0 \quad \text{ with probability } 1-1 / 6-1 / 6 j^{2} \end{array}\right.

Then show that the Lindeberg condition (condition (12.24)) does not hold.

?
Problem 12.5

Let Xn,n≥1X_{n}, n \geq 1, be independent r.v.s such that ∣Xn∣≤Mn\left|X_{n}\right| \leq M_{n} a.s. with Mn=o(sn)M_{n} = o\left(s_{n}\right) where sn2=∑j=1nσj2→∞s_{n}^{2} = \sum_{j = 1}^{n} \sigma_{j}^{2} \rightarrow \infty and σj2=Var⁡[Xj]\sigma_{j}^{2} = \operatorname {Var}\left[X_{j}\right]. Set Sn=∑j=1nXjS_{n} = \sum_{j = 1}^{n} X_{j} and show that

L(Sn−E[Sn]sn)⇒n→∞N(0,1) \mathcal{L}\left(\frac{S_{n}-\mathbb {E}\left[S_{n}\right]}{s_{n}}\right) \underset {n \rightarrow \infty }{\Rightarrow } N(0,1)
?
Problem 12.6

Let Xnj,j=1,…,kn→∞X_{n j}, j = 1, \ldots , k_{n} \rightarrow \infty be row-wise independent r.v.s with E[Xnj]=0\mathbb {E}\left[X_{n j}\right] = 0, σ2(Xnj)=σnj2<∞,sn2=∑j=1knσnj2=1\sigma^{2}\left(X_{n j}\right) = \sigma_{n j}^{2} < \infty , s_{n}^{2} = \sum_{j = 1}^{k_{n}} \sigma_{n j}^{2} = 1, and max⁡1≤j≤knσnj2→n→∞0\max_{1 \leq j \leq k_{n}} \sigma_{n j}^{2} \underset {n \rightarrow \infty }{\rightarrow } 0. Then, with Sn=∑j=1knXnjS_{n} = \sum_{j = 1}^{k_{n}} X_{n j}, and under the assumption that L(Sn)⇒N(0,1)\mathcal{L}\left(S_{n}\right) \Rightarrow N(0,1), show that ∑j=1knXnj2→→P1\sum_{j = 1}^{k_{n}} X_{n j}^{2} \xrightarrow [\rightarrow ]{P} 1.

?
Problem 12.7

For a function g:R→Cg: \mathbb {R} \rightarrow \mathbb {C}, set Δg(u)=g(u+h)−g(u−h),h∈R\Delta g(u) = g(u+h)-g(u-h), h \in \mathbb {R}, and define Δ(n)g(u)\Delta^{(n)} g(u) recursively. Then

?
(i)

Show that

Δ(m)g(u)=∑r=0m(−1)r(mr)g(u+(m−2r)h). \Delta ^{(m)} g(u) = \sum _{r = 0}^{m}(-1)^{r}\binom {m}{r} g(u+(m-2 r) h) .
(ii)

From part (i), and by expanding f(t)f(t) around 0 up to terms of order 2n2 n, obtain

Δ(2n)f(0)(2h)2n=f(2n)(0)+o(h2n)h2n( where o(t)t→0 as t→0), \begin{aligned} \frac{\Delta ^{(2 n)} f(0)}{(2 h)^{2 n}} = & f^{(2 n)}(0)+\frac{o\left(h^{2 n}\right)}{h^{2 n}} \\ & \left(\text{ where } \frac{o(t)}{t} \rightarrow 0 \text{ as } t \rightarrow 0\right), \end{aligned}

so that f(2n)(0)=lim⁡Δ(2n)f(0)(2h)2nf^{(2 n)}(0) = \lim \frac{\Delta^{(2 n)} f(0)}{(2 h)^{2 n}} as h→0h \rightarrow 0.

Problem 12.8

For m=1,2,…m = 1,2, \ldots and kk integer, show that

∑r=0m(−1)r(mr)(m−2r)k={0 for (0≤)k<m2m×m!k=m \begin{align} & \sum _{r = 0}^{m}(-1)^{r}\binom {m}{r}(m-2 r)^{k} \\ & \quad = \begin{cases} 0 & \text{ for }(0 \leq ) k < m \\ 2^{m} \times m! & k = m \end{cases} \end{align}
?
Problem 12.9

For j=1,…,kn→∞j = 1, \ldots , k_{n} \rightarrow \infty, let XnjX_{n j} be row-wise independent r.v.s with respective d.f.s FnjF_{n j}. Then show that, for every ε>0\varepsilon > 0, the convergence

P(max⁡1≤j≤kn∣Xnj∣≥ε)→n→∞0 \mathbb {P}\left(\max _{1 \leq j \leq k_{n}}\left|X_{n j}\right| \geq \varepsilon \right) \underset {n \rightarrow \infty }{\rightarrow } 0

is equivalent to

∑j=1kn∫(∣x∣≥ε)dFnj(x)→n→∞0 \sum _{j = 1}^{k_{n}} \int _{(\left|x\right| \geq \varepsilon )} d F_{n j}(x) \underset {n \rightarrow \infty }{\rightarrow } 0

and implies the convergence

max⁡1≤j≤knP(∣Xnj∣≥ε)→n→∞0 \max _{1 \leq j \leq k_{n}} \mathbb {P}\left(\left|X_{n j}\right| \geq \varepsilon \right) \underset {n \rightarrow \infty }{\rightarrow } 0
?
Problem 12.10

Show that, for every complex number zz, it holds ∣z2∣=∣z∣2\left|z^{2}\right| = \left|z\right|^{2}.

?
Problem 12.11

In the proof of Proposition 2 (see discussion following relation (12.36)), Lemma 6 is used for n=1n = 1 to ensure finiteness of E[X2]\mathbb {E}\left[X^{2}\right], which, according to the lemma, follows from the existence and finiteness of f(2)(0)f^{(2)}(0). Prove this fact directly; i.e., show that the existence and finiteness of f(2)(0)f^{(2)}(0) implies finiteness of E[X2]\mathbb {E}\left[X^{2}\right].

?