# Law of large numbers with random index more detailed proof

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?

• 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. – Sangchul Lee Aug 3 at 1:17
• So the fact that $S_{I_n}$ and $I_n$ are not independent doesn't effect anything? – TPaul Aug 3 at 1:33
• 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 $\mathbb{R}$. – Sangchul Lee Aug 3 at 1:36
• What does this notation $I_n(w)$ mean is it $I_n$ conditioned on $w$? – TPaul Aug 3 at 15:18
• 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 $\omega$. – Sangchul Lee Aug 3 at 15:21