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$(X_n)$ is a sequence of $L^2$ random variables with $EX_n=0$ for all $n$ and suppose there is a constant $c$ s.t. $\operatorname{Var}(X_{n+k}−X_n)\leq ck$, for all $n,k\geq0$. Show that $X_n/n$ converges to $0$ a.s. (Hint: First prove along a suitable subsequence).

I can see that we are trying to make the probabilities summable along a subsequence but may I know how to choose a subsequence so that the upper bound we use for variance makes the probabilities summable?

$$P(|X_{n_k}|>nϵ)≤\operatorname{Var}(X_{n_k})/{n_k}^2ϵ^2\leq?$$

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  • $\begingroup$ Are you allowed to use the Borel Cantelli lemma ? $\endgroup$ – P. Quinton Dec 25 '19 at 19:21
  • $\begingroup$ Yes, I was trying to use that with summability $\endgroup$ – manifolded Dec 25 '19 at 19:22
  • $\begingroup$ You first need to complete your inequality with an upper bound on the variance as given in the problem. $\endgroup$ – Michael Dec 25 '19 at 21:16
  • $\begingroup$ If it helps you can use $(a+b)^2\leq 2a^2+2b^2$. $\endgroup$ – Michael Dec 25 '19 at 21:54
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Let $ \| X \|_2 $ denote the $ L^2 $ norm of $ X $. Note that if $ EX = 0 $, then $ \| X \|_2^2 = \operatorname{Var}(X) $.

Consider the subsequence given by $ n_k = 2^k $. Then, $$ |\| X_{2^{k + 1}} \|_2 - \| X_{2^k} \|_2| \leq \| X_{2^{k + 1}} - X_{2^k} \|_2 $$ by the reverse triangle inequality. Squaring and using the given condition yields $$ |\| X_{2^{k + 1}} \|_2 - \| X_{2^k} \|_2|^2 \leq c2^k $$ Since $ \| X_{2^k} \|_2 \geq 0 $, it follows that $$ \| X_{2^{k + 1}} \|_2 \leq \| X_{2^k} \|_2 + \sqrt{c}2^{k / 2} $$ By induction, it follows that $$ \| X_{2^{k + 1}} \|_2 \leq \| X_1 \|_2 + \sqrt{2c}\frac{2^{k / 2} - 1}{\sqrt{2} - 1} $$ Finally, $$ \operatorname{Var}(X_{n_k}) / n_k^2 \leq \left( \| X_1 \|_2 2^{-k} + \sqrt{2c}2^{-k}\frac{2^{(k - 1) / 2} - 1}{\sqrt{2} - 1} \right)^2 = O(2^{-k}) $$ so $ \operatorname{Var}(X_{n_k}) / n_k^2 $ is summable.

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