# Let $X_{1}, X_{2}, X_{3},...$ be a sequence of independent and identically distributed random variables, prove the following

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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• $n-th$ 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. Jun 23 at 5:42
• @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? Jun 23 at 5:46
• 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. Jun 23 at 6:07
• Please check your source and see if there is any other hyptothesis. The result in the present form cannot be proved. Jun 23 at 7:37
• Is it "for all $\omega\in \mathbf A$, the quantity $\lim$ [...]" ? Jun 23 at 8:12

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.

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