Let $(X_n : n = 1,2, \ldots)$ be an i.i.d. sequence of random variables uniformly distributed on $(0,1)$. For $n = 0, 1,2, \ldots$ , set $$ S_n = \prod_{k=1}^{n} X_k$$. Calculate $E(S_n)$. Compare $S_n$ and $E(S_n)$ for large n.
I tried this:
First $E(S_n)=E(\prod_{k=1}^{n} X_k) = \prod_{k=1}^{n} E(X_k)=_{uniformly} 2^{-n}$ then $E(S_n)=2^{-n}$ Is it correct?
Second I'm trying to apply the Strong Law of Large Numbers to the sequence dened by $$\log S_n = \log ( \prod_{k=1}^{n}X_k) = \sum_{k=1}^{n} \log X_k$$ then $$ \lim_{n\rightarrow \infty} \frac{\log S_n}{n} =\lim_{n\rightarrow \infty} \frac{\sum_{k=1}^{n} \log X_k}{n} $$
but here I stuck I don't know How to continue this exercise....
Could someone help to compare this $S_n$ and $E(S_n)$ for large n? Pls.
Thanks for your time and help.
Your computation for $E[S_n]$ appears to be correct. So let's discuss $\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \log X_k$
What is $E[\log X_1]$? It's $\int_{0}^{1} \log x dx$, which visually looks like this:
It's obvious from the picture that we can think of the integral in terms of the exponential function: that is, $\int_{0}^{1} \log x dx = -\int_{-\infty}^0 e^x dx = -\int_{0}^{\infty} e^{-x} dx = -1$, so by the SLLN, $\frac{1}{n}\log S_N = \frac{1}{n} \sum_{k=1}^n \log X_k \to E[\log X_1] = -1$ almost surely, and $\log S_n = n \left(\frac{1}{n}\log S_n\right) \to -\infty$ almost surely (this is true since $\lim_{n \to \infty} n \left(\frac{1}{n}\log S_n\right) = \left(\lim_{n \to \infty} n\right) \left(\lim_{n \to \infty} \left(\frac{1}{n}\log S_n\right)\right) = -1 \times \left(\lim_{n \to \infty} n\right)$ almost surely), so $S_n \to e^{-\infty} = 0$ almost surely, which agrees with the limit of the expected value of $S_n$.