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The question I'm interested in has been asked here https://math.stackexchange.com/questions/2132702/law-of-large-numbers-with-random-index#=

But the proof is lacking in details. Can someone give a more rigorous argument or provide a citation for the steps taken?

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    $\begingroup$ Consider two events $A=\{S_n/n\to\mathbb{E}[X]\}$ and $B=\{I_n\to\infty\}$. Then $A\cap B$ has full probability, and for each $\omega\in A\cap B$ the sequence $(S_{I_n(\omega)}(\omega)/I_n(\omega))_{n\geq1}$ is a subsequence of $(S_n(\omega)/n)_{n\geq1}$ which converges to $\mathbb{E}[X]$, hence itself converges to th same value. $\endgroup$ – Sangchul Lee Aug 3 at 1:17
  • $\begingroup$ So the fact that $S_{I_n}$ and $I_n$ are not independent doesn't effect anything? $\endgroup$ – TPaul Aug 3 at 1:33
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    $\begingroup$ No, it does not affect anything because the argument is no different from the deterministic case. Inceed, once you fix a sample $\omega$ from $A\cap B$, the argument is now completely deterministic because both $(S_n(\omega)/n)_{n\geq1}$ and $(I_N(\omega))_{n\geq 1}$ are just sequences in $\mathbb{R}$. $\endgroup$ – Sangchul Lee Aug 3 at 1:36
  • $\begingroup$ What does this notation $I_n(w)$ mean is it $I_n$ conditioned on $w$? $\endgroup$ – TPaul Aug 3 at 15:18
  • $\begingroup$ Recall that a random variable is a function from the sample space $\Omega$ (which we often sweep under the rug) to $\mathbb{R}$. So $I_N(\omega)$ is simply the function notation indicating the value of $I_N$ evaluated at $\omega$. $\endgroup$ – Sangchul Lee Aug 3 at 15:21

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