Q:

Is $Y_n := \prod_1^n \xi_i$ for $\xi_i$ i.i.d. $\text{Unif}(0,2)$ a sequence of uniformly integrable random variables?

Is $Y_n := \prod_1^n \xi_i$ for $\xi_i$ i.i.d. $\text{Unif}(0,2)$ a sequence of uniformly integrable random variables?

I know that $\mathbb E[Y_n]=1$ for all $n$, and looking at the martingale $\{Y_n, \cal A_n\}$ (for $\mathcal A_n := \sigma[Y_0, \ldots, Y_n], Y_0 := 1$), I can use the martingale convergence theorem to get that $Y_n\to_\text{a.s.} Y_\infty$ for some $Y_\infty \in \mathcal L_1$. Furthermore, from more work with martingales I got that $\mathbb E[Y_n^2]\to \infty$, and so the variance is going off to $\infty$. The $Y_n$ are not bounded because if the $\xi_i$ happen to be above $1$ then we're going to shoot up, but at the same time such events are "rare", and so my intuition is useless. I'm not sure where to go from here -- this example has truly baffled me.

A:

Note that since $(Y_n,\mathcal F_n^Y)$ is a martingale, then uniform integrability is equivalent with convergence in $L_1$. Moreover, if the limit exists (call it $Y$), then it must be the case that $\mathbb E[Y]=1$ since $\mathbb E[Y_n] = 1$ for every $n\in \mathbb N$. However, as you noticed, we know (since it is non-negative martingale) that $Y_n \to Y$ almost surely, for some $Y \in [0,\infty)$. We'll show that $Y = 0$ almost surely.
Note that there exist $\delta,\epsilon > 0$ such that $\mathbb P(|\xi_k-1| > \epsilon) =\delta >0 $
By that we have $\sum_{k=1}^\infty \mathbb P(|\xi_k - 1| > \epsilon) = \infty$, and by Borel Cantelli, there is set $\Omega_0$ of measure $1$ such that for every $\omega \in \Omega_0$ there exist subsequence $(k_m(\omega))_{m \in \mathbb N}$ such that $|\xi_{k_m(\omega)}(\omega) - 1 | > \epsilon$ for every $m \in \mathbb N$.
Now, we need to use some tool from analysis, note that if infinite product $\prod_{k} a_k$ of nonnegative numbers $a_k$ converges to some $a \in (0,\infty)$ then $\lim_{k \to \infty} a_k = 1$ (it is due to taking logarithm of that product and noticing that $\lim_{k \to \infty} \ln(a_k) = 0$ iff $\lim_{k \to \infty} a_k = 1)$ 
But we showed that in our case for $\omega \in \Omega_0$ (so almost surely) $\xi_k(\omega) \not \to 1 $, which means that $Y$ must be $0$ or $\infty$ almost surely (the product must diverge). However, by martingale convergence theorem, we know that limit is finite almost surely, so it must be that $Y=0$ a.s. Hence it cannot converge in $L_1$, since $\mathbb E[|Y_n - Y|] = \mathbb E[|Y_n|] = 1 \not \to 0$

