# Does this example fail the CLT?

Let $X_i, i \in \mathbb{N}$ be independent random variables with $E[X_i] = \mu_i$ and $\mathrm{var}(X_i) = \sigma_i^2 < \infty$.
Define $S_n := \sum_{i=1}^{n} X_{i}$, and $s_n := \sqrt{\sum_{i=1}^{n} \sigma_{i}^2}$ . Under the above assumptions, following are two different groups of sufficient conditions for $$\frac{ S_n - \sum_{i=1}^n\mu_i}{s_n} \to N(0,1) \text{ in distribution}.$$

1. Lindeberg CLT from Wikipedia

$$\text{Lindeberg condition: }\forall \epsilon >0, \quad \lim_{n \to \infty} \frac{\sum_{i=1}^{n} \mathrm{E} [(X_{i} - \mu_{i})^2 I_{\{|X_{i} - \mu_{i}| > \epsilon s_n\}} ] }{s_n^2} = 0.$$

2. Theorem D.19 of William Greene's Econometric Analysis (p112 of his appendix D file or Theorem 11 on p14 of this note )

$$\text{Feller-Levy condition: } \lim_{n\to\infty} \frac{\max_{i=1,\dots,n}\sigma_i^2}{s_n^2} = 0$$ $$\text{name-unknown condition: }\lim_{n\to\infty} \frac{s_n^2}{n} < \infty$$

Note:

• I have changed the notation a bit.

• In the book, instead of $\frac{\max_{i=1,\dots,n}\sigma_i^2}{s_n^2}$, it writes $\frac{\max_{i=1,\dots,n}\sigma_i}{\sqrt{n} s_n}$. This is said to be a typo, and it should be $\frac{\max_{i=1,\dots,n}\sigma_i}{ s_n}$. And $\frac{\max_{i=1,\dots,n}\sigma_i}{ s_n}$ converges to finite, if and only if $\frac{\max_{i=1,\dots,n}\sigma_i^2}{s_n^2}$ converges to finite. (Note: I have changed the notation a bit.)

• The author said Feller's (1968) Introduction to Probability Theory and Its Applications was the original source for that result. I know it has two volumes, and after some search, I didn't find this version of CLT in either volume. I wonder if I am missing something? (As a side note, in Feller's 1968 Introduction to Probability Theory and Its Applications Volume 1, although the "name-unknown condition" doesn't appear for CLT, it appears for Law of Large Number (see page 254 formula (5.5)). Right below it is Lindeberg condition for Lindeberg CLT, which makes me wonder if the "name-unknown condition" in Greene's Theorem D.19 is a typo?)

Here is an example from Robert Isreal's earlier reply, which seems to indicate something might be wrong in the second one, i.e. the Greene's. Let $X_i$ be independent with $$P(X_i = 2^i) = P(X_i = -2^i) = 2^{-2i-1}, P(X_i = 0) = 1 - 2^{-2i}.$$ Thus $\mu_i = 0$ and $\sigma_i^2 = 1$.

1. Since $$P(S_n = 0) > P(X_i = 0 \text{ for all }i) > 1 - \sum_{i=1}^\infty 2^{-2i} = 2/3,$$ it shows directly that $S_n/s_n$ does not go to $N(0,1)$ in distribution. A normal distribution has $P(Z=0)=0$.
2. The Lindeberg condition in Wikipedia's is not satisfied by the example (take $ϵ=1$ for example). So it is not conclusive that $S_n/s_n$ should or should not converge to $N(0,1)$ in distribution.
3. The two conditions in Greene's are satisfied, since $s_n^2 = n$. So $S_n/s_n$ should converges to $N(0,1)$ in distribution.

So the result from Greene's is not consistent with the results from other ways. I wonder what has gone wrong? Thanks!

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The example mentioned by @Robert Israel goes back (at least) to Turing in his unpublished dissertation written in 1934 (which got him elected Fellow of King's College the next year at the age of 22). See there for a concise description of the relationships between Lindeberg's condition, Feller-Lévy's condition and the CLT. –  Did Jul 6 '12 at 10:15
@did: Thanks, Didier, for the reference! –  Tim Jul 7 '12 at 22:14

A bit too long for a comment. This feels nothing short of a conspiracy. First off, I picked up a copy of Greene's Econometrics and the result there is $\lim_{n\rightarrow\infty}\max_i\frac{\sigma_i}{n\bar{\sigma}_n}=0$, where he defines $\bar{\sigma}_n^2=\frac{1}{n}(\sigma_1^2+\ldots+\sigma_n^2)$ in addition to $\bar{\sigma}^2=\lim_{n\rightarrow\infty}\bar{\sigma}_n^2$, which does not match your lecture notes, where I see the condition to be $\max_{i}\frac{\sigma_i^2}{n\bar{\sigma}^2_n}\rightarrow 0$. However, I still believe this is completely false, regardless of whose condition is correct, Greene or your lecture notes.
Let me call Lindberg Feller Condition '(L)', call $\lim_{n\rightarrow\infty} \sigma_k^2/s_n^2=0$ '(F)' and the statement "the central limit theorem holds" as '(CLT)'. We have that $(L)\Leftrightarrow (CLT)\ and\ (F)$. (F) here is the so called "Feller Condition." Normally, one has $(L)\Rightarrow (CLT)\ and \ (F)$ but the point is that (CLT) does NOT imply (L), unless we also have (F), or vice versa that (F) does not alone imply (L).
I do know that there's something called the Hajek Sirlak Theorem which says that if $X_i$ are i.i.d. with mean zero and variance $\sigma^2$ then $S_n=a_1X_1+\ldots+a_nX_n$ converges upon normalization to a normal distribution if the condition $\max_{i\leq n} a_i/\sqrt{\sum_i^n a_i^2}\rightarrow 0$ holds. The issue is I think you need the i.i.d. condition to hold.
Thanks! (1) Your $\lim_{n\rightarrow\infty}\max_i\frac{\sigma_i}{n\bar{\sigma}_n}=0$ and my $\lim_{n\to\infty} \frac{\max_{i=1,\dots,n}\sigma_i^2}{s_n^2} = 0$ are equivalent, and your $\bar{\sigma}^2=\lim_{n\rightarrow\infty}\bar{\sigma}_n^2$ and my $\lim_{n\to\infty} \frac{s_n^2}{n} < \infty$ are the same, so I don't see there is anything different in the book and in the note. Am I wrong? (2) What do you mean by "even barring this the result in Greene implies the same, that CLT holds and I believe this is completely false"? –  Tim Jul 4 '12 at 16:57
Correct me if I'm wrong, but in the asymptoticsprimer.pdf linked, the condition as written is $\max_{i}\frac{\sigma_i^2}{n\bar{\sigma}^2_n}$. Check out: en.wikipedia.org/wiki/Lindeberg%27s_condition for a partial reference to (F). I believe it's sometimes referred also as the Feller-Levy Condition. –  Alex R. Jul 4 '12 at 17:36