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Jun 23, 2022 at 21:15 comment added Christophe Leuridan @GJC20 Yes. Moreover, the distribution of $X_t$ given $X$ never reached $0$ is above (for the stochastic ordering) the distribution of $X_t$. I think that all these distributions can be computed explicitly.
Jun 23, 2022 at 20:58 comment added Iosif Pinelis Previous comment continues: At least intuitively, this should be clear, as $X_t$ for large $t$ will "forget" whether the process previously reached $0$ or not. So, for large $t$, the conditional distribution of $X_t$ given $\tau>t$ will be close to the unconditional distribution of $X_t$.
Jun 23, 2022 at 20:57 comment added Iosif Pinelis @GJC20 : As Christophe Leuridan noted, the conditional density of $X_t$ given $\tau>t$ will go to $0$ pointwise as $t\to\infty$. However, it is not hard to find a simple expression for the conditional density of the standardized version $Z_t:=(X_t-EX_t)/\sqrt{Var\,X_t}$ given $\tau>t$ and show that it will go to the standard normal density pointwise as $t\to\infty$.
Jun 23, 2022 at 16:52 comment added Christophe Leuridan @GJC20 The density and the distribution function of $X_t$ under $P$ and under $P[\cdot|\tau>t] $converge pointwise to 0 as $t \to +\infty$.
Jun 23, 2022 at 16:48 comment added Christophe Leuridan @Matt F. You forget the drift! Brownian motion with constant drift is transient.
Jun 23, 2022 at 16:43 comment added user44143 To the contrary, $P[\tau=+\infty]=0$, by the recurrence property — see eg the reference at math.stackexchange.com/questions/4460636/…
Jun 23, 2022 at 16:30 vote accept GJC20
Jun 23, 2022 at 16:29 comment added GJC20 @ChristopheLeuridan Thanks a lot for your answer. Let $p_t$ be the conditional density function of $X_t$ knowing $\tau>t$, i.e. $\mathbb P[X_t\in dx|\tau>t]=p_t(x)dx$. What can we conclude for $\lim_{t\to\infty}p_t(x)$?
Jun 23, 2022 at 16:26 comment added Christophe Leuridan @Matt F. You have $P[\tau=\infty]>0$, so there is no problem.
Jun 23, 2022 at 16:22 comment added user44143 I see that $X_\infty = +\infty$ almost surely under $P$ — but I don’t think that implies much about $P[\cdot|\tau = \infty]$ since $\tau=\infty$ is a measure-0 condition.
Jun 23, 2022 at 16:18 history answered Christophe Leuridan CC BY-SA 4.0