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Let $X_{1}, X_{2}, X_{3},...$ be a sequence of independent and identically distributed random variables with common (finite) mean $\mu$. Prove that there exists and event A such that $P(A) = 1$ and for all $w \in \Omega$, the quantity $\lim_{n \to \infty}(X_{1}(w)X_{2}(w)...X_{n}(w))^\frac{1}{n}$ converges and determine this limit.

I get the part where these random variables are identically distributed, so the expectation of the sequence is also $\mu$. But how do I proceed further? Will central limit theorem help here?

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    $\begingroup$ $n-th$ roots of negative numbers don't exist in the real line for $n$ even . I think you have missed the hypothesis that $X_n$'s are positive random variables. $\endgroup$ Jun 23 at 5:42
  • $\begingroup$ @geetha290krm It was not mentioned in the question that the $X_{n}'s$ are positive r.v. But, let's assume they are, how do I prove it? $\endgroup$ Jun 23 at 5:46
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    $\begingroup$ Let me also point out that CLT is of no use in proving convergence with probability $1$ since this theorem only gives convergence in distribution. $\endgroup$ Jun 23 at 6:07
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    $\begingroup$ Please check your source and see if there is any other hyptothesis. The result in the present form cannot be proved. $\endgroup$ Jun 23 at 7:37
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    $\begingroup$ Is it "for all $\omega\in \mathbf A$, the quantity $\lim$ [...]" ? $\endgroup$
    – P. Quinton
    Jun 23 at 8:12

1 Answer 1

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Assuming that $X_i$'s are positive and $E\ln X_1 >-\infty$ which implies $E |\ln X_1| <\infty$ note that $\ln [(X_1X_2...X_n)^{1/n}]=\frac 1 n \sum\limits_{k=1}^{n} \ln X_i \to E\ln X_1$ almost surely by SLLN's since $(\ln X_i)$ is also an i.i.d. sequence. Taking exponential we get $ (X_1X_2...X_n)^{1/n} \to e^{E\ln X_1}$.

Proof for the case $E \ln X_1=-\infty$:

Let $0<\epsilon <1$ and $Y_j=\max \{\epsilon, X_j\}$. Then $0 \leq (X_1X_2...X_n)^{1/n} \leq (Y_1Y_2...Y_n)^{1/n} \to e^{E\ln Y_1}$ by the prevous case. I leave it to you to check the fact that $E\ln Y_1 \to -\infty$ as $\epsilon \to 0$. It follows that $(X_1X_2...X_n)^{1/n} \to 0$ almost surely.

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    $\begingroup$ Why $E[\ln(X_i)]=\ln\mu$? $\endgroup$
    – Feng
    Jun 23 at 5:52
  • $\begingroup$ @Feng Thansk for pointing out the error. The limit is not $\mu$. I have corrected the answer. $\endgroup$ Jun 23 at 5:56
  • $\begingroup$ $\ln X_1$ does not have to be integrable, so the case $E\ln X_1 = -\infty$ should be addressed separately. $\endgroup$
    – zhoraster
    Jun 23 at 6:40
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    $\begingroup$ @zhoraster I have now given a proof for that case. $\endgroup$ Jun 23 at 8:44
  • $\begingroup$ Surely $(X_1X_2...X_n)^{1/n} \to e^{E \ln X_1}$ not $\to e^{\ln EX_1}$ $\endgroup$
    – Henry
    Jun 23 at 10:17

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