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Suppose $X$ is 1-d stochastically continuous process with $X(0) = 0$, i.e. $X_s \to X_t$ in probability as $s\to t$ for all $t\ge 0$. Let $\tau = \inf\{t>0: |X_t|>1\}$.

[Q.] Is $\tau>0$ almost surely?

I think the answer shall be yes, because $X$ has a Cadlag version (modification). Thanks.

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  • $\begingroup$ what is wrong with the process $X_t=0$? $\endgroup$
    – ofer zeitouni
    Commented Nov 10, 2015 at 10:53

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This is not true. Consider some random sequence that converges to $0$ a.s., for example, $Y_n=(Z_1+\cdots+ Z_n)^{-1}$, where $Z$'s are i.i.d. (for instance) Exp(1) random variables. Set $$ X_t = \begin{cases} 2, & \text{if $t=Y_k$ for some $k$},\\ 0, &\text{otherwise}. \end{cases} $$ Then $X$ is stochastically continuous (since $X_t=0$ a.s. for any fixed $t$), but $\tau=0$ a.s.

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  • $\begingroup$ Thanks for this cute example. I am considering a Levy jump process, ex. $\alpha$-stable process for $\alpha = 3/2$. Is $\tau>0$ a.s.? Thanks. $\endgroup$
    – kenneth
    Commented Nov 10, 2015 at 15:01
  • $\begingroup$ Sorry, I'm not a specialist in Levy processes :) $\endgroup$
    – Serguei Popov
    Commented Nov 10, 2015 at 15:27
  • $\begingroup$ For $\alpha$-stable process, it seems to me yes. From scaling property, for any $c>0$, we have $\max_{0\le t <c} |X_t| \sim c^{1/\alpha} \max_{0\le t <1} |X_t|$. This implies $P(\max_{0\le t <c} |X_t| <1) = P(\max_{0\le t <1} |X_t| < c^{-1/\alpha}) \to 1$ as $c\to 0$. Thus, $P(\tau>0) \ge P(\tau\ge c) = P(\max_{0\le t <c} |X_t| <1) \to 1$ as $c\to 1$. Thanks for your help. $\endgroup$
    – kenneth
    Commented Nov 10, 2015 at 15:32

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