Suppose $\{\xi_n\}$ are i.i.d uniform random variables on $[0,1]$, and $$\eta_n = 2^n\Pi_{i=1}^n\xi_i$$ then:
Show that $\{\eta_n\}$ is a martingale,
Find the almost sure limit of $\{\eta_n\}$
Determine if $\{\eta_n\}$ is uniformly integrable.
I have no problem with the first part, I also know that $-\log\xi_i$ follows exponential distribution with parameter $1$, so $$\log(2^n\eta_n^{-1})\sim \text{Gamma}(n,1)$$ and consequently the pdf is given by $$f_{\eta_n}(t)=\frac{(-\log(t)+n\log 2)^{n-1}}{(n-1)!2^{-n}}$$ using Stirling's formula, for every fixed $t$ there is \begin{align} f_{\eta_n}(t)&=\frac{(-\log(t)+n\log 2)^{n-1}}{(n-1)!2^{-n}}\\ &=\frac{(n\log 2)^{n-1}(1-\frac{\log t/\log 2}{n})^{n-1}}{(n-1)!2^{-n}}\\ &\approx e^{\log t/\log 2}\sqrt{2\pi n}^{-1}(2^{-1}\cdot e\cdot \log 2)^{n-1}\left(\frac{n}{n-1}\right)^{n-1}\\ &\approx \sqrt{2\pi n}^{-1}e^{\log t/\log 2+1}(0.94)^{n-1} \end{align} which goes to zero as $n$ goes to $\infty$. So what is the almost sure limit? Is the martingale uniformly integrable?
For the second question, taking the logarithm is a good idea. One has $$ \log\eta_n=\sum_{i=1}^n\log\left(2\xi_i\right) $$ hence $$ \frac 1n\log \eta_n=\frac 1n\sum_{i=1}^n\log\left(2\xi_i\right)\to\mathbb E\left[\log(2\xi_1)\right]<0. $$ Consequently, $\log\eta_n\to -\infty$ almost surely and $\eta_n\to 0$ almost surely.