Product of Uniform Random Variables Let $X_1, X_2, \cdots, X_n$ be iid random variables such that $X_i \sim U(0,1)$ for $i=1,2,\ldots,n$. Find the value of a constant $k$ such that $P(\prod_{i=1}^n X_i \ge k)=0.05$ if $n=100$.
Attempt:
Let $Y_i=\ln(X_i)$. Then, $Y$ is an exponential r.v. with mean $1$. Then the problem reduces to a CLT problem, with $P(\prod_{i=1}^n X_i \ge k)=0.05 \Longrightarrow P(\sum_{i=1}^n Y_i \ge \ln k)=0.05$. Setting $n=100$, the probability reduces to $P\left(Z \ge \frac{\ln k - 100}{10}\right)=0.05$. Solving this gives me $k=e^{116.45}$, which I believe is wrong since $k$ is too large.
 A: as Did wrote earlier that has little to do with the CLT. The approach is the following: take the logarithm, thus obtaining sum of exponentially distributed RVs. This sum is $\chi^2$ distributed with $2n$ degrees of freedom:
As you wrote earlier using exponentially distributed $Y_i$ with $\lambda = 1$ ($Y_i=-\ln(X_i)$), we obtain
$$
P(\prod_{i=1}^n X_i \ge k)=0.05 \Longrightarrow P(-\sum_{i=1}^n Y_i \ge \ln k)=0.05 \Longrightarrow P(\sum_{i=1}^n Y_i \le -\ln k)=0.05.
$$
As we know from Wikipedia:
$$
\sum_{i=1}^n \mathrm{Exp}(1) \sim \frac{1}{2}\chi_{2n}^2
$$
Thus using rather sloppy notation, we obtain
$$
P(\chi_{2n}^2 \le -2\ln k)=0.05
$$
and followingly
$$
k=\exp\bigl(-0.5\cdot Q_{200}(0.05)\bigr)
$$
with quantile function $Q$ for $\chi^2$ distributed RV with degree of freedom 200. Using language R we obtain

exp(qchisq(p = 0.05, df = 200)*(-0.5))

which equals $k=2.875916e-37$. Looks very small, but if we test (again with R)

quantile(-log10(sapply(1:1000, function(x) prod(runif(100)))),probs = 0.05)

we obtain 36.19577, thus our calculation seems plausible.
