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?

New contributor
user1070420 is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
  • 2
    $\begingroup$ $n-th$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Is it "for all $\omega\in \mathbf A$, the quantity $\lim$ [...]" ? $\endgroup$
    – P. Quinton
    Jun 23 at 8:12

1 Answer 1


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.

  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ @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

Your Answer

user1070420 is a new contributor. Be nice, and check out our Code of Conduct.

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.