Counterexample for Omitting UAN Condition in Feller's Proof

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Recall the Lindeberg-Feller Central Limit Theorem.

Let $\{X_{nj}\}$, $n=1,2,...$, $j=1,2,...,k_n$, be a double array of random variables and for each $n$, $X_{n1},\ldots,X_{nk_n}$ are independent.  Define $S_n=\sum_{j=1}^{k_n}X_{nj}$ and
$$\begin{array}{ll}
\mathscr{E}(X_{nj})=\alpha_{nj}, & \mathscr{E}(S_n)=\sum_{j=1}^{k_n}\alpha_{nj}=\alpha_n; \\
\sigma^2(X_{nj})=\sigma^2_{nj}, & \sigma^2(S_n)=\sum_{j=1}^{k_n}\sigma^2_{nj}=s^2_n. \\
\end{array}$$Suppose $\alpha_{nj}=0$ for all $n$ and $j$, and $s^2_n=1$.  In order that as $n\rightarrow\infty$ the two conclusions below both hold:

(1) $S_n$ converges in distribution to $\Phi$.
(2) $\{X_{nj}\}$ is uniformly asymptotically negligible (UAN);

it is necessary and sufficient that for each $\eta>0$, we have $$\underset{n\rightarrow\infty}{\lim}\sum_{j=1}^{k_n}\mathscr{E}\left[X_{nj}^2\,I\left(|X_{nj}|>\eta\right)\right]=0$$
It is important that the sufficient conditions for the Lindeberg's criterion should hold simultaneously.  Here is a counterexample that we omit the requirement of UAN.

$\bullet$ Counterexample.
Let $\{X_n\}$ be a sequence of independent random variables with densities $X_n\sim N\left(0,\sigma^2_n\right)$, where $\sigma^2_1=1$ and $\sigma^2_k=2^{k-1}$ for $k\geq2$.  Then $$B_n^2=\sigma^2(S_n)=\sum_{k=1}^n\sigma^2_k=1+1+2+4+\cdots+2^{n-2}=2^{n-1}.$$First, we find the limit distribution of $S_n$.  Since for each $k$, $$\frac{X_k}{B_n}\sim N\left(0,\sigma^2_k/2^{n-1}\right),$$ and $$\sum_{k=1}^n\frac{\sigma^2_k}{2^{n-1}}=\frac{1}{2^{n-1}}+\frac{1}{2^{n-1}}\sum_{k=2}^n2^{k-2}=\frac{1}{2^{n-1}}+\frac{2^{n-1}-1}{2^{n-1}}=1,$$we have $$\frac{S_n}{B_n}\sim\mathscr{N}\left(0,1\right).$$
The sequence $\{X_n\}$ is not UAN, since for $\epsilon>0$,
$$\begin{array}{rl}\underset{n\rightarrow\infty}{\lim}\underset{1\leq k\leq n}{\max}\mathscr{P}\{|X_k|>\epsilon\}&=\underset{n\rightarrow\infty}{\lim}\underset{2\leq k\leq n}{\max}\mathscr{P}\left\{\frac{|X_k|}{2^{(k-1)/2}}>\frac{\epsilon}{2^{(k-1)/2}}\right\}\\
&=\underset{n\rightarrow\infty}{\lim}\underset{2\leq k\leq n}{\max}2\Phi\left(-\frac{\epsilon}{2^{(k-1)/2}}\right)\\
&=\underset{n\rightarrow\infty}{\lim}2\Phi\left(-\frac{\epsilon}{2^{(n-1)/2}}\right)\\
&=2\Phi(0)=1\neq0. \end{array}$$
Note that, the Lindeberg's condition implies no significant large variance among $\{X_n\}$.  In this case, we have $$\underset{n\rightarrow\infty}{\lim}\underset{1\leq k\leq n}{\max}\frac{\sigma^2_k}{B_n^2}=\underset{n\rightarrow\infty}{\lim}\frac{2^{n-2}}{2^{n-1}}=\frac{1}{2}\neq0.$$Which implies the Lindeberg's condition does not hold.