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Let $X_t$ and $Y_t$ (for $t\geq 0)$ be stochastic processes such that $P(X_t=Y_t)=1$ for every $t \geq 0$. Assume that $X_t$ has a.s. continuous trajectories. If $t_n\rightarrow t$ then clearly $Y_{t_n}\rightarrow Y_t$ a.s. I have a question concerning a generalization of this observation. Let $\tau_n,\tau$ be bounded random variables such that $\tau_n\rightarrow \tau$ a.s. (or even on whole $\Omega$). Is it true that $Y_{\tau_n}\rightarrow Y_{\tau}$ a.s.?

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  • $\begingroup$ What does "...such that $P(X_t=Y_t)$ for every $t\geq 0.$" mean? Did you mean $P(X_t=Y_1) = 1$ for every $t \geq 0$? $\endgroup$ Jun 5, 2013 at 17:33

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No.

Let $\delta_n$ be independent exponential random variables with mean $\mathbb E(\delta_n) = \dfrac 1{n^2}$ and set $\tau_n = \sum_{i=1}^n \delta_i$.

Let $Y_t$ be the process that agrees with $X_t$ at every point except the $\tau_n$ but $Y_{\tau_n}=0$ for every $n$.

Then $\mathbb P(X_t=Y_t)=1$ for every $t$, but $Y_{\tau_n}\to 0$ with probability one.

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