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Oct 21, 2018 at 11:57 comment added Marc You are right. I guess I stared at this problem for too long, I'll delete this question. Thank you for the reply!
Oct 19, 2018 at 15:27 comment added Nate Eldredge I think perhaps you misunderstood Anthony's comment. When the product converges to zero almost surely, so that $P(\bigcap_{n=1}^\infty \{X_n \in A\}) = 0$, then it certainly converges in $L^1$ as well. This is just the "continuity from above" property of probability measures and follows immediately from countable additivity. (Or you could use the dominated convergence theorem.) I understood Anthony's comment to be saying that you can't always get exponentially fast convergence to zero.
Oct 19, 2018 at 9:35 history asked Marc CC BY-SA 4.0