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I've got the following question: We have random variables $X_1, X_2,...$, which are i.i.d. of uniform distribution on $[-1,1]$. How to check if $\frac{X_1 + X_2^2 + ... + X_n^n}{n}$, $n=1,2,...$ converges with probability equal to $1$?

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There's something called Kolmogorov's version of the strong law of large numbers where the random variables don't need to be identically distributed.

If $Y_i$ is a sequence of independent random variables such that $$\sum_{k=0}^\infty \frac{var[Y_k]}{k^2} < \infty$$ and $$\frac{\sum_{k=0}^n {\mathbb E[Y_k]}}n\to A\in\mathbb R$$ as $n\to\infty$

then the sequence $\frac{\sum_{k=0}^n Y_k}n\to A$ almost surely as $n\to\infty$.

You should be able to check that if $Y_k = X^k_k$ then each variance $var[Y_k] < 1$ and so the series converges.

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  • $\begingroup$ But the thesis of this theorem is that $\frac{\sum Y_k - E(\sum Y_k)}{n}$ converges almost surely to $0$... $\endgroup$ – Anne Jun 2 '13 at 12:38
  • $\begingroup$ Sorry, I was slightly lazy typing that, you also need the average of the means to be convergent (easy in your case). You can of course consider $\tilde Y_k = Y_k - \mathbb E(Y_k)$ to see that the two versions of the theorem are equivalent. $\endgroup$ – Tim Jun 2 '13 at 12:55
  • $\begingroup$ Well, I'm still wondering how to show that $\frac{ E(\sum_{k=1}^n(Y_k))}{n}$ converges. I mean it's equal to: $\frac{E(X_1)}{n} + ... + \frac{E(X_n^n)}{n}$. And I computed that $E(X_i^i) = \frac{1}{2} (\frac{1+(-1)^{i+1}}{i+1})$. So, what can I do next to prove that it converges? $\endgroup$ – Anne Jun 2 '13 at 13:55
  • $\begingroup$ @Anne You have that $E(X^i_i)\to 0$ as $i\to 0$. It's quite an easy exercise to check that if a sequence $x_i\to 0$ then the Cesaro means $S_n\frac 1n\sum_{i=1}^n x_1$ converge to zero. $\endgroup$ – Tim Jun 3 '13 at 9:27

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