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Suppose $X_i\in\mathcal{L}^2$ with expectation $0$ such that $\sum_{i=1}^\infty \mathbb{E}[X_i^2]/i^2<\infty$ and suppose they are pairwise non correlated. Does then the SLLN still hold?

Liealgebrabach
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  • Maybe(?) related: http://math.stackexchange.com/questions/1375925/when-does-sum-i-1-infty-x-i-exist-for-random-sequences-x-i-i-1#comment2802355_1375925 – PhoemueX Aug 06 '15 at 16:42
  • I don't have any answer ready yet but if $\sum \frac{X_i}{i}$ converges a.s. then SLLN follows from Kronecker's Lemma as $n \uparrow \infty$. – Saty Aug 06 '15 at 18:47

1 Answers1

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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 $\sum_{i=1}^{\infty} Var[X_i]/i^2 < \infty$. To be more precise, a sufficient condition in this case that will guarantee SLLN is the following: $$\sum_{i=1}^{\infty} Var[X_i] \left(\frac{log \, i}{i}\right)^2 < \infty.$$

This follows from Serlfing's SLLN.

MerylStreep
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