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I'm reading Probability Theory, and ran into the following exercise:

Prove that if $X_n \in L^2$ are uncorrelated and identically distributed random variables, then $$\frac{1}{n} \sum_{k=1}^n X_k \to \mathbb{E}[X_1]\text{ almost surely.}$$

This seems a bit weird, since it is given as a theorem that the convergence happens in probability. (Which implies that I can find a subsequence that converges almost surely.) They don't mention this strengthening of the theorem in the text, so I'm wondering if what the exercise claims is true.

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When you say "uncorrelated" do you mean "independent"? (To me "uncorrelated" could mean $\text{Cov}(X_i, X_j) = 0, i \neq j$ which is a weaker statement than "independent" in general.) – Qiaochu Yuan May 23 '11 at 12:17
@Qiaochu Yuan: No. I mean zero covariance like you said. – J. J. May 23 '11 at 12:23
up vote 4 down vote accepted

Yes, what the exercise claims is true. See Theorem 21 here, and consider the paragraph above it for a proof.

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For general interest, it is interesting to consider – Shai Covo May 23 '11 at 13:19

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