Chapter 13

The Central Limit Problem: The Noncentered Case

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Problem 13.1

Under the assumption of Theorem 1, the convergence in (13.29) holds if and only if (13.30) holds for every $\varepsilon > 0$ . With the notation $\bar{X}_{n j} = X_{n j}-\alpha_{n j}$ and $\bar{X}_{n j}$ having d.f. $\bar{F}_{n j}$ , and $\bar{S}_{n} = \sum_{j = 1}^{n} \bar{X}_{n j}$ , show that the convergence in (13.29) follows from the condition

\sum _{j = 1}^{k_{n}} \int _{\left(\left|x\right| \geq \varepsilon s_{n}\right)} x^{2} d \bar{F}_{n j}(x) \underset {n \rightarrow \infty }{\rightarrow } 0 \text{ for every } \varepsilon > 0

Problem 13.2

In Theorem 1 of this chapter, as well as Theorem 1 of Chapter 12, under some basic assumptions on the r.v.s $X_{n j}, j = 1, \ldots , k_{n}$ , given in the theorems cited, a necessary and sufficient condition is given (namely, $g_{n}(\varepsilon ) \underset {n \rightarrow \infty }{\longrightarrow } 0$ for every $\varepsilon > 0$ ) so that

\mathcal{L}\left(S_{n}\right) \underset {n \rightarrow \infty }{\Rightarrow } N\left(\mu , \sigma ^{2}\right), \quad \mu \in \mathbb {R}, \quad \sigma ^{2} \in (0, \infty )

where $S_{n} = \sum_{j = 1}^{k_{n}} X_{n j}$ . Thus,

\mathbb {P}\left(S_{n} \leq x\right) = F_{S_{n}}(x) \underset {n \rightarrow \infty }{\longrightarrow } \Phi \left(x ; \mu , \sigma ^{2}\right), \quad x \in \mathbb {R},

where $\Phi \left(x ; \mu , \sigma^{2}\right)$ is the d.f. of the $N\left(\mu , \sigma^{2}\right)$ distribution. Actually, the preceding convergence is uniform in $x \in \mathbb {R}$ . That is, for every $\varepsilon > 0$ there exists $N(\varepsilon ) > 0$ , independent of $x \in \mathbb {R}$ , such that $n \geq N(\varepsilon )$ implies

\left|F_{S_{n}}(x)-\Phi \left(x ; \mu , \sigma ^{2}\right)\right| < \varepsilon \text{ for all } x \in \mathbb {R} . \tag {*}

This result is a consequence of the following fact (known as Pólya's Lemma). Namely, if $F$ and $\left\{ F_{n}\right\}$ are d.f.s of r.v.s such that $F_{n}(x) \underset {n \rightarrow \infty }{\longrightarrow } F(x), x \in \mathbb {R}$ , and $F$ is continuous, then the convergence is uniform in $x \in \mathbb {R}$ . Establish this fact.

Remark.

The uniformity asserted in $\left({ }^{*}\right)$ is what legitimizes as simple an approximation as that in the Binomial case. Thus, if $X_{1}, \ldots , X_{n}$ are independent, distributed as $B(1, p)$ , and if $a$ and $b$ are integers with $0 \leq a < b \leq n$ , then, for sufficiently large $n, \mathbb {P}\left(a < S_{n} \leq b\right) \simeq \Phi \left(\frac{b-n p}{\sqrt{n p q}}\right)-\Phi \left(\frac{a-n p}{\sqrt{n p q}}\right)$ , where $S_{n} = \sum_{j = 1}^{n} X_{j}, q = 1-p$ , and $\Phi (x) = \Phi (x ; 0,1)$ . This is so, because $\mathbb {P}\left(S_{n} \leq b\right) = \mathbb {P}\left(\sum_{j = 1}^{n} Z_{j} \leq \frac{b-n p}{\sqrt{n p q}}\right)$ , where $Z_{j} = \frac{1}{\sqrt{n}} \frac{X_{j}-p}{\sqrt{p q}}, j = 1, \ldots , n$ , and $\left|\mathbb {P}\left(\sum_{j = 1}^{n} Z_{j} \leq \frac{b-n p}{\sqrt{n p q}}\right)-\Phi \left(\frac{b-n p}{\sqrt{n p q}}\right)\right| < \varepsilon$ for $n \geq N(\varepsilon )$ independently of $\frac{b-n p}{\sqrt{n p q}}$ ; and likewise for $\mathbb {P}\left(S_{n} \leq a\right)$ .