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I've seen two types of conditions for strong law of large numbers: one requires i.i.d and first order moment condition: $X_n$ i.i.d with $E|X_1| < \infty$; the other requires second order moment condition but may not be identical: $X_n$ independent (may not identical) with $\sum_{n} Var(X_n)/n^2 < \infty$. I have known the proof under these two types of conditions.

However, I cannot find a proof for not identical case requiring only first order moment. I am wondering whether it is true that: if $X_n$ independent (may not identical) with $\sup_{n} E|X_n| < \infty$, then $\frac{1}{n}\sum_{i=1}^n (X_i - E(X_i)) \xrightarrow{a.s.}0$.

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Example: Independent $X_k\sim\mathcal N(0,k)$? In this case the average $n^{-1}\sum_{k=1}^n X_k$ is normally distributed with mean $0$ and variance $(n+1)/2n$, and so is asymptotically normal with variance $1/2$. In view of the Kolmogorov zero-one law, the average cannot converge a.s.

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  • $\begingroup$ Thanks a lot! Now I realized that the case I thought cannot be true. $\endgroup$ – Titan_Tong Mar 9 '16 at 0:12
  • $\begingroup$ However, there is a problem in your example that $E|X_k| = k \sqrt{2/\pi} \to \infty$. $\endgroup$ – Titan_Tong Mar 9 '16 at 0:25
  • $\begingroup$ $E|X_k|=\sqrt{2k/\pi}$. $\endgroup$ – John Dawkins Mar 9 '16 at 3:00

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