# 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? Commented Jan 18, 2018 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 Commented Jan 18, 2018 at 15:05
• Are we talking strong law or weak law? Commented Jan 18, 2018 at 15:23
• Ah i should have specified that. Im interested in both. So whether CLT-> WLLN and whether it implies SLLN Commented Jan 18, 2018 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? Commented Jan 12, 2019 at 21:38

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 $$\sum_{i=1}^n \frac1{i^2}=\pi^2/6.$$
• 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. Commented Jan 13, 2019 at 1:37