# 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?

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:

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.

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.

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 $\epsilon=1/2$

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

• Remark: your $X_k$ is exactly the standard example for the tightness of Chebyshev's inequality: en.wikipedia.org/wiki/…
– Ian
Commented Sep 13, 2016 at 13:06
• (+1) Thanks! Does the normalized sum $\frac{S_n}{s_n} \xrightarrow{d} N(0,1)$ for your $X_k$ nonetheless?
– user237392
Commented Sep 13, 2016 at 13:18
• @Bey It is just that $E(X_k^2;|X_k|>\sqrt{n}/2)$ is zero if $\sqrt{k} \leq \sqrt{n}/2$ and is $k \cdot (1/k)=1$ otherwise.
– Ian
Commented Sep 13, 2016 at 13:22
• Actually, now I have a question for @snarfblaat: Does this example satisfy the hypotheses of some other CLT? I ask because I tried it numerically and qualitatively the behavior is consistent with the conclusion of CLT.
– Ian
Commented Sep 13, 2016 at 13:58
• @Bey, Here is how you would know quickly a counterexample exists. For mean-centered $X_i$, Lindeberg implies $\sigma_i^2=E(X_i^2;|X_i|\le\epsilon)+E(X_i^2;|X_i|>\epsilon)\le \epsilon + o(n)$ so the variances of the $X_i,i=1,\ldots,n$ are uniformly small (these are the rows in the triangular array description), whereas the condition you suggested is weaker, saying only that the variances are uniformly small relative to their sum $s_n^2$. Commented Sep 13, 2016 at 16:23