Kolmogorov proved, that, as one considers independent (not necessary equally distributed) Random Variables: $\{X_n\}_{n\ge0}\subseteq \mathcal L^2$

With $\mathrm{Var} (X_n)=\sigma^2_n$ and without loss of generality $E[X_n]=0$.

If $\sum_{n=0}^\infty \frac{\sigma^2_n}{n^2} \lt \infty$ then SLLN holds, that is:

$$\frac1n\sum_{k=0}^n X_k \rightarrow 0 \,\,\,\,\,\,\,\,\,\,\,\,\,\text{a.s.}$$

Something different:

Also it is known, that if $\sup_{n\ge0}\sigma^2_n =: v \lt \infty$ pairwise uncorrelated $(X_n)_{n\ge0}$ will suffice for SLLN to hold.

My question in view of this situation: Is it possible to weaken the condition of $(X_n)_{n\ge0}$ beeing independent and instead merely consider pairwise uncorrelated $(X_n)_{n\ge0}$?

  • $\begingroup$ The following is relevant (almost as weak as pairwise uncorrelated, but not quite), from Y.S. Chow "On a strong law of large numbers for martingales," Annals of Mathematical Statistics, vol. 38, no. 2, article 610, 1967.: If $E[X_i-E[X_i]|X_1, \ldots, X_{i-1}]=0$ for all $i\in\{1, 2, 3, \ldots\}$, and if $\sum_{i=1}^{\infty}\frac{\sigma_i^2}{i^2}<\infty$, then with prob 1 we have $\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^n (X_i-E[X_i])=0$. $\endgroup$ – Michael Jul 26 '15 at 19:57
  • $\begingroup$ Notice that the condition $E[X_i-E[X_i]|X_1, \ldots, X_{i-1}]=0$ for all $i$ implies pairwise uncorrelated, but not vice-versa. $\endgroup$ – Michael Jul 26 '15 at 20:00

If you want to impose just uncorrelatedness between the variables in the sequence rather than independence, then you need to impose a stronger condition than $\sum_{i=1}^{\infty} \operatorname{Var}[X_i]/i^2 < \infty$. To be more precise, a sufficient condition in this case that will guarantee SLLN is the following: $$\sum_{i=1}^{\infty} \operatorname{Var}[X_i] \left(\frac{\log i}{i}\right)^2 < \infty.$$

This follows from Serlfing's SLLN.

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