My Math Forum Limit of Product of iid Random Variables

 January 26th, 2018, 10:25 PM #1 Senior Member   Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 Limit of Product of iid Random Variables Let $X_1, X_2, ... , X_n$ be i.i.d random variables with common density function: $f(x)=\frac{\alpha}{x^{\alpha+1}}I(x\geq 1)$ where $I(x\geq 1)$ is the indicator function and $\alpha >0$. Now define: $$S_n=\left[ \prod_{i=1}^{n}X_i \right ]^{n^{-1}}$$ Prove that $\lim_{n\rightarrow \infty}S_n=e^{\alpha^{-1}}$, and use the law of large numbers and the continuous mapping theorem to conclude about the convergence in probability of $ln(S_n)$. I honestly don't have the slightest inkling on how to go about evaluating that limit. However, since the value is given in the problem statment, my intuition tells me that $ln(S_n)$ converges in probability to $\alpha^{-1}$, but I'm not certain.
 January 26th, 2018, 10:47 PM #2 Senior Member   Joined: Oct 2009 Posts: 402 Thanks: 139 What is $ln(S_n)$. Compute it.
 January 26th, 2018, 11:06 PM #3 Senior Member   Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 $\ln(S_n) =\frac{1}{n}\sum_{k=1}^n\ln(X_k)$, but I still don't know how to compute the limit of $S_n$ as $n$ goes to infinity. Apparently I've forgotten part of calculus.
 January 26th, 2018, 11:10 PM #4 Senior Member   Joined: Oct 2009 Posts: 402 Thanks: 139 This is where the law of large numbers comes in, which says something about limits like these.
 January 26th, 2018, 11:13 PM #5 Senior Member     Joined: Sep 2015 From: USA Posts: 1,931 Thanks: 999 you'll probably need $\lim \limits_{n \to \infty} ~\left(1+\dfrac x n\right)^n = e^x$
 January 27th, 2018, 07:11 AM #6 Senior Member   Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 I understand how to do the problem using the strong law of large numbers, but I wasn't sure if I was supposed to prove the value of that limit some other way initially (the setup of the problem suggests so since it mentions so first).

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