Example of why $\lim_{n\to \infty} \max_{1\leq i \leq n} \frac{\sigma_i^2}{s_n^2} = 0 $ does not imply Lindeberg's Condition?

Question

Note that $s_n^2:=\sum_1^n \sigma_i^2$ is the sum of the variances of a sequence of rv's up to $n$.

This question is related to the two below, but I couldn't make sense of the answers:

1. Does this condition imply the Lindeberg condition?
2. https://math.stackexchange.com/questions/318317/conditions-that-imply-lindebergs-condition

On the Wikipedia page, it seems to say that the limit in my title is not sufficient to imply the Lindeberg Condition when applied to a sequence of independent random variables.

Question: I'd like a concrete example of where the limit holds but Lindeberg's Condition does not.

I am assuming that such an example exists, because if this simple limit were sufficient for Lindeberg's Condition, then it would replace the more complex formulation of the actual Lindeberg Condition.

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I tried to demonstrate this to myself using a simple sequence:

$$X_i: E[X_i]=0; Var[X_i] = i; X_i\; \mathrm{independent}$$

In this case:

$$\lim_{n\to \infty}\; \max_{1\leq i \leq n} \frac{\sigma_i^2}{s_n^2} = \lim_{n\to \infty}\; \frac{n}{\frac{n(n+1)}{2}}=0$$

And Lindeberg's Condition becomes (for a given $\epsilon > 0$):

$$\lim_{n\to \infty} \frac{2}{n(n+1)}\sum_{i=1}^nE\left[X_i^2\mathbf{1}_{|X_i|\geq \epsilon\sqrt{\frac{n(n+1)}{2}}}\right]$$

This appears to be equivalent to showing:

$$\sum_{i=1}^nE\left[X_i^2\mathbf{1}_{|X_i|\geq \epsilon\sqrt{\frac{n(n+1)}{2}}}\right] = o(n^{2})$$

However, I know that:

$$E\left[X_i^2\mathbf{1}_{|X_i|\geq \epsilon\sqrt{\frac{n(n+1)}{2}}}\right]$$ $$ = P\left(|X_i|\geq \epsilon\sqrt{\frac{n(n+1)}{2}}\right)E\left [X_i^2 \Bigg| |X_i|\geq \epsilon\sqrt{\frac{n(n+1)}{2}}\right] \leq E\left[X_i^2\right] = n$$

Hence,

$$\sum_{i=1}^nE\left[X_i^2\mathbf{1}_{|X_i|\geq \epsilon\sqrt{\frac{n(n+1)}{2}}}\right] \leq \sum_{i=1}^n n = O(n^{2})$$

Of course, equality would require that

$$|X_i|\geq \epsilon\sqrt{\frac{n(n+1)}{2}}\;\; \mathrm{a.s.}$$

However, from Chebyshev's Inequality, we know:

$$P\left(|X_i|\geq \epsilon\sqrt{\frac{n(n+1)}{2}}\right) \leq \frac{2\sigma_i^2}{n(n+1)\epsilon^2} = \frac{2i}{n(n+1)\epsilon^2}$$

This means that we can show:

$$\forall \epsilon>0\;\exists n: \frac{2i}{n(n+1)\epsilon^2} < 1$$

So we have shown that the $X_i$ will eventually fall below some threshold with a non-zero probability. Thus,

$$\exists k>0: \forall n> k\;\;\sum_{i=1}^nE\left[X_i^2\mathbf{1}_{|X_i|\geq \epsilon\sqrt{\frac{n(n+1)}{2}}}\right] < \sum_{i=1}^n n$$

That's where I'm stuck, since I don't know if I can proceed to show that the LHS sum is $o(n^2)$. I suspect this is why the limit in the title of my post is insufficient, since we'd have to know how fast the "tails" of the $X_i$ distributions fall off to make this assessment. This is something that Lindeberg's Condition forces us to deal with but not the simple limit.

Of course, I have probably made some errors in my attempt. I would be grateful for examples (or corrections/extensions to my attempt) where it is clear that the sum is greater than $o(n^2)$ or at least that Lindeberg's Condition will not hold.

Answer 1

How about $X_k=-\sqrt{k},0,\sqrt{k}$ with probability $1/2k,1-1/k,1/2k$, respectively.

Then $E(X_k)=0,\sigma_k^2=E(X_k^2)=1,s_n^2=n$ and

$$\lim_{n\to\infty}\max_{1\le k\le n}\sigma_k^2/s_n^2=\lim_{n\to\infty}1/n=0$$

But, for $\epsilon=1/2$

$$s_n^{-2}\sum_{k=1}^nE(X_k^2;|X_k|>\epsilon s_n)= n^{-1}\sum_{k=1}^nE(X_k^2;|X_k|>\sqrt{n}/2)=\\ n^{-1}\sum_{k=n/4+1}^nE(X_k^2;|X_k|>\sqrt{n}/2)=n^{-1}(3n/4-1)\to3/4. $$