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Oct 27, 2020 at 23:44 comment added Nate Eldredge Maybe an even worse example is just to let $X_n = 1/n$ independent of both $t$ and $\omega$, with $X=0$. Then $X_n \to X$ uniformly and almost surely, which is about the strongest mode one could imagine, but $\mathbb{P}(\tau_n = \infty)=1$ and $\mathbb{P}(\tau = \infty)=0$. So I don't think you will get your desired result without knowing something more specific about the processes themselves.
Oct 27, 2020 at 23:24 history asked user128095 CC BY-SA 4.0