I have a Question regarding MCT which I am stuck in, the question goes like this:
Let $X_0 = 1$ and assume that $X_n$ is distributed uniformly on $(0,X_{n-1})$.
and $Y_n = 2^nX_n$.
the questions are:
a) Show that $\left( Y_n\right)$ converges to $0$ a.s.
b) Is $Y_n$ uniformly integrable? (No!)
regarding the first think. I know that $Y_n$ is a non negative Martingale and therefore it is Bounded (Fatou's Lemma). I want to show that $Y_n$ can be written as a product if i.i.d r.v which are Uni(0,2) and then I will finish my proof. how can say this?
after I will solve (a), (b) will be very easy since if it was UI then $E(Y_{\infty})=E(Y_0)=1$ but it is not!
Thanks for the help
As you say, we can write $Y_n = U_1 \cdots U_n$ where $U_i$ are iid $U(0,2)$. That means $\ln Y_n = \sum_{i=1}^n \ln U_i$. Compute $E[\ln U_i]$ and note that it is negative. So by the strong law of large numbers, $\frac{1}{n} \ln Y_n \to E[\ln U_i] < 0$ a.s. This implies $\ln Y_n \to -\infty$ a.s. which is to say $Y_n \to 0$ a.s.
More martingale-y solution:
Since $Y_n$ is a nonnegative martingale, it converges almost surely to some random variable $Y_\infty$. Clearly $Y_\infty \ge 0$ and by Fatou's lemma $E[Y_\infty] \le 1$.
Now let $U$ be a $U(0,2)$ random variable independent of everything in sight. Clearly $U Y_n$ has the same law as $Y_{n+1}$, so passing to the limit, $U Y_\infty$ has the same law as $Y_\infty$.
By Jensen's inequality, $E[\sqrt{Y_\infty}] \le E[Y_\infty] < \infty$. So we may write $E[\sqrt{Y_\infty}] = E[\sqrt{U Y_\infty}] = E[\sqrt{U}] E[\sqrt{Y_\infty}]$. Since $E[\sqrt{U}] \ne 1$ (Jensen or direct computation), we must have $E[\sqrt{Y_\infty}]=0$, which is to say $Y_\infty = 0$ a.s.