I'm studying an exercise in which we must determine what kinds of convergence holds for the following martingale $$ X_n = \sum_{i=1}^n Y_i $$ where $Y_i$ are i.i.d standard normal distributed r.v.'s.
First, we check convergence in $L^2$, and we can see that $$ E[X_n^2] = \sum_{i=1}^n E[Y_i^2] = n $$ From which we conclude that $\sup_n E[X_n^2] = \infty$. By the $L^2$ martingale convergence theorem, we do not have convergence in $L^2$.
Then, we check convergence in $L^1$ $$ E[|X_n|] = \sqrt{n}\sqrt{\frac{2}{\pi}} $$ from which we again conclude that $\sup_n E[|X_n|] = \infty$.
1) How does $\sup_n E[|X_n|] = \infty$ directly imply that $X_n$ is not uniformly integrable and so we do not have convergence in $L^1$?
Finally, we check almost sure convergence. The exercise hints that we should show that $$ P(\liminf X_n = -\infty) = 1, P(\limsup X_n = \infty) = 1 $$ using Kolmogorov's 0-1 law to show that $X_n$ does not converge a.s.
2) How does one show this using Kolmogorov's 0-1 law?
I know you're not supposed to say this in posts, but... Many thanks for your help.
If $(X_n)$ were uniformly integrable then it would be $L^1$ bounded, in the sense that $\sup_n\|X_n\|_1<\infty$.
Can you show that $P(\limsup_nX_n =+\infty)>0$? (As a start, $P(\sup_{k\ge n}X_k>C)\ge P(X_n>C)$ for each $n$ and each $C>0$. That last probability tends to $1/2$ as $n\to\infty$.)