Product of n uniformly distributed Let ${Y_n}$ be independent random variables uniformly distributed in [0, 1]. For $ N \geq 1 $, let $ X_n = Y_1...Y_n$, the product of the $Y_i$. What is the limit, as $n \rightarrow \infty$, of $Prob(X_n > (0.4)^n)$?
 A: This follows from the law of large numbers (the weak version, to be precise). 
To see this, note that, for every $x$, $[X_n\gt x^n]=[S_n\gt n\log x]$ where $S_n=\sum\limits_{k=1}^n\log Y_k$. By the weak law of large numbers, $\frac1nS_n\to z=E[\log Y_1]$ in probability. 
This means that if $\log x\gt z$ then $P[S_n\gt n\log x]\to0$ and that if $\log x\lt z$ then $P[S_n\gt n\log x]\to1$.
In the case at hand, $x=.4$ and $z=\int_0^1\log y\,\mathrm dy=-1$ hence $x\mathrm e^{-z}=.4\cdot\mathrm e\gt1$ and $\log x\gt z$, which implies that $P[X_n\gt (.4)^n]\to0$.
Note finally that $x=.4$ is a somewhat close call since $P[X_n\gt (.36)^n]\to1$.
A: Hint: We describe a more or less mechanical method of finding the answer by using the Central Limit Theorem. One can also solve the problem using less machinery. 
(i) Let $W_i=-\ln Y_i$. Then $W_i$ has exponential distribution (not really needed) with mean and variance equal to $1$ (this part is needed).
It follows that $-\ln X_n$ has mean and variance equal to $n$.
(ii) For large $n$, the random variable $-\ln X_n$ has nearly normal distribution, with mean $n$ and variance $n$. This can be made more precise by quoting the Central Limit Theorem.
(iii) We have $X_n \gt (0.4)^n$ if and only if $\ln X_n \ge n\ln(0.4)$, that is, iff the nearly normal random variable $-\ln X_n$ is $\le n\ln(2.5)$. For large $n$, this is many "standard deviation units" away from $n$. 
