Example for which the CLT holds but the LLN doesn`t

Question

I am currently thinking about the relationship between the law of large numbers and the central limit theorem and I was wondering whether someone can give me an example of a familiy of random variables $(X_i): (\Omega, \mathscr A, P) \to (\Omega_i, \mathscr A_i) $ such that the central limit theorem holds but the law of large numbers does not.

EDIT: I (believe) I have proved that CLT under these conditions implies WLLN. So i am only interested in the SLLN anymore.

Why would that be the case? Aren't the assumptions of the Lindeberg CLT nested within the SLLN? — AdamO, Jan 18 '18 at 14:58
I Tend to think that as well but when i searched whether CLT-> LLN some people said that this was Not in generally true — Sebastian, Jan 18 '18 at 15:05
Ah i should have specified that. Im interested in both. So whether CLT-> WLLN and whether it implies SLLN — Sebastian, Jan 18 '18 at 15:25
further can you give a prove for your implication? To me it seems that this only implies WLLN — Sebastian, Jan 22 '18 at 12:29
@MartijnWeterings Convergence to a fixed limit such as $N(0,1)$ is one thing, but what does $S_n \to n\mu + \sqrt{n}N(0,\sigma^2)$ mean? — Dilip Sarwate, Jan 12 '19 at 21:38
@MartijnWeterings I don't understand your response. In the statement $S_n \to$ "something", the $n$ is a _variable_, and we are asking what happens as $n$ increases without bound. So $n$ should not showing up in the "something" that is being converged to by the $S_n$'s. — Dilip Sarwate, Jan 13 '19 at 04:23
I must have meant something like $f_n(x) \to g_n(x)$ implicating $f_n(x) - g_n(x) \to 0$, making the $n$ gone on one side. — Sextus Empiricus, Jan 13 '19 at 10:25
@MartijnWeterings Have you seen [this question](https://stats.stackexchange.com/q/353013/6633) and its answers? — Dilip Sarwate, Jan 13 '19 at 18:06
@DilipSarwate that question reminds me I do not like infinity, the degenerate distribution, and the dirac delta function. — Sextus Empiricus, Jan 13 '19 at 18:15

kjetil b halvorsen · Accepted Answer · 2019-01-12T23:04:44.097

Assume a sequence of random variables (independent or not) $X_1, \dotsc, X_n, \dotsc$ with $\DeclareMathOperator{\E}{\mathbb{E}}\DeclareMathOperator{\Var}{\mathbb{V}} \E X_i =\mu$, and which satisfies the conditions for some central limit theorem (CLT) such that $$ c_n (\bar{X}_n -\mu) \stackrel{\mathbb{D}}{\to} \mathcal{N}(0, 1) $$ for some sequence of constants $c_n \to \infty$. For the usual IID case we have $c_n = \sqrt{n}/\sigma$. Then we can show convergence in probability for $\bar{X}_n$ to $\mu$.

So for a counterexample (if it exists) you will have to look for a case where $c_n$ do not grow over any bounds.

And, such examples exist, even if maybe artificial. Less artificial examples would be interesting. Generalize the situation above by replacing $\bar{X}_n$ by $\bar{X}_{wn}$, some weighted mean of the first $n$ variables in the sequence. Then assume that $X_i \sim \mathcal{N}(0,i^2)$, so the variances is increasing fast. Use the usual optimal weighted mean with weights $w_i=i^{-2}$. Then $\mathbb{V} \bar{X}_{wn}=(\sum_{i=1}^n i^{-2})^{-1}$ so we can choose $c_n=(\sum_{i=1}^n i^{-2})^{1/2}$ which do not grow to infinity. So the law of large numbers do not hold, since $$ \sum_{i=1}^n \frac1{i^2}=\pi^2/6. $$

Ha ha, those are the worth typing conditions I have ever heard! — Ben, Jan 12 '19 at 23:06
So the "loophole" is that with the non-classical versions of the CLT you can have growing variances for the different $X_i$ (as long as no single one of the $X_i$ becomes overpowering and the >2 order moments do not grow faster ). Then you have some CLT but the variance of the mean grows (or stabilizes at some constant) instead of shrinks. — Sextus Empiricus, Jan 13 '19 at 01:37

Example for which the CLT holds but the LLN doesn`t

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