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

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
• Are we talking strong law or weak law? – Greenparker Jan 18 '18 at 15:23
• 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
• @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

## 1 Answer

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 $$\sum_{i=1}^n \frac1{i^2}=\pi^2/6.$$

• Typo, fixed (I'm on my mobile on board a ship in sea) – kjetil b halvorsen Jan 12 '19 at 23:06
• Ha ha, those are the worth typing conditions I have ever heard! – Ben - Reinstate Monica 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 1:37