Lindeberg and Lyapunov's condition on symmetric bernoulli random variables

Question

I'm self-studying for my qualifying exam and I cannot solve this problem (from Probability Theory by S. R. S. Varadhan):

Consider the case of mutually independent random variables $\{X_j\}$, where $X_j = \pm a_j$ with probability $1/2$ . What do Lyapunov's and Lindeberg's conditions demand of $\{a_j\}$? Can you find a sequence $\{a_j\}$ that does not satisfy Lyapunov’s condition for any $\delta > 0$ but satisfies Lindeberg's condition? Try to find a sequence $\{a_j\}$ such that the central limit theorem is not valid.

Lindeberg's condition would be: $$ \sum_{k=1}^n E[X_k^2 1_{(|X_k|>\epsilon s_n)}]/\sum_{k=1}^n a_k^2 \to 0 \text{ as }n\to\infty \text{ , } \forall \epsilon>0 $$ where $s_n^2 = \sum_{k=1}^n a_k^2$.

Lyapunov's condition would be: $$ \sum_{k=1}^n a_k^{2+\delta}/s_n^{2+\delta}\to 0 \text{ as }n\to\infty \text{ , for some } \delta>0 $$

However, I cannot figure out the sequence $\{a_j\}$ such that $\{a_j\}$ that does not satisfy Lyapunov's condition for any $\delta > 0$ but satisfies Lindeberg's condition.

I have tried sequence $\{k\}, \{k^2\}, \{2^k\}, \{1/k\}, \{1/\sqrt{k}\}, \{1/k^2\}$... Can someone give any hints or ideas?

Answer 1

For the last example, check that if $a_j=2^{-j}$, then $\sum_{i=1}^n X_i/s_n$ converges in distribution to a uniform random variable on -1 to 1, and further in this case both the Lindeberg and Lyapounov conditions do not hold (since basically the limits of all terms involved are nonzero constants).

One class of examples that can be ruled out, thanks to the work of PhoemueX in this post Class of functions whose integral behaves as an exponential function., is $a_j$ increasing. In general if Lindebergs condition holds, then you must have that $a_n/s_n\to 0$, since if not, there exists $\epsilon > 0$ and a subsequence $n'$ so that $|a_{n'}|/s_{n'} > \epsilon$, and on this subsequence

$(1/s_{n'}^2) \sum_{j=1}^n a_j^2 1_{|a_j|>\epsilon s_n} \ge a_{n'}^2/s_{n'}^2 > \epsilon^2.$

But when $a_n$ is increasing, $a_n/s_n \to 0$,

$ \frac{\sum_{j=1}^n a_j^{2+\delta}}{s_n^{2+\delta}} \le \left(\frac{\sum_{j=1}^n a_j^{2}}{s_n^{2}}\right)\left(\frac{a_n^\delta}{s_n^{\delta}}\right) \to 0,$

and so Lyapounovs condition is satisfied. Decreasing sequences can be ruled out, just replace $a_n$ with $a_1$.

Another class of examples that will not work is $a_j^2$ summable, since in this case Lindebergs conditions will evidently not be satisfied. $a_j$ bounded above and below can also be ruled out. It seems whatever sequence would work would be quite special! Perhaps we should focus our efforts on showing that in this limited example, the conditions are equivalent.

This leads then to the following proof that Lindeberg implies Lyapounov in this case: In order for Lindebergs conditoin to be satisfied, $\max_{1\le j \le n}a_j/s_n \to 0$. If not you arrive at the above contradiction. However, when $\max_{1\le j \le n}a_j/s_n \to 0$

$ \frac{\sum_{j=1}^n a_j^{2+\delta}}{s_n^{2+\delta}} \le \left(\frac{\sum_{j=1}^n a_j^{2}}{s_n^{2}}\right)\left(\frac{\max_{1\le j \le n}a_j^\delta}{s_n^{\delta}}\right) \to 0,$ and Lyapounov's condition holds.