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Let $\{X_n\}_{n=1}^\infty$ be a sequence of random variables such that,

$$ \forall n,\,\,\mu_n := EX_n < \infty, \quad \mu_n \to \mu < \infty \textrm{ as } n\to\infty. $$

Now for each $n$, let $X_{n,1}, \dots, X_{n,n}$ be i.i.d. copies of $X_n$. Define $$ \bar X_n := \frac{1}{n}\sum_{i=1}^n X_{n,i}. $$

I want to prove that $$ \bar X_n \to \mu \textrm{ as } n\to\infty, $$ almost surely. I'm guessing it should be true, but don't know how to prove it. I tried to utilize existing proofs of law of large numbers, but most of the proofs I could find on Google uses additional assumption like finite variances, which may not hold for my case.

Can anyone give me a hint to prove this? Thanks in advance.

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  • $\begingroup$ It's not true, even if all $X_n$ have the same distribution. See Exercise 24 of Terry Tao's notes $\endgroup$
    – Robert Israel
    Commented Jan 18, 2017 at 19:22
  • $\begingroup$ Thank you for a great reference. I've seen a hint attached to Exercise 24, and it said to prove that $\bar{X}_n$ is unbounded almost surely. But what if $\bar{X}_n$ is always bounded? In my case, $X_n$ is a bounded random variable such that $0 \leq X_n \leq A_n$, with $A_n\to\infty$ as $n\to\infty$. Still not true? $\endgroup$
    – user3141978
    Commented Jan 19, 2017 at 5:40
  • $\begingroup$ E.g. try $X_n = n^2$ with probability $1/n^2$, $0$ otherwise. Then $\mathbb E[X_n] = 1$, but $\mathbb P(\bar{X}_n = 0) \ge 1 - 1/n$. $\endgroup$
    – Robert Israel
    Commented Jan 19, 2017 at 16:34

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