Consider the drift Brownian motion $X_t:=1+bt+W_t$, where $(W_t)_{t\ge 0}$ is a Brownian motion starting at zero. Set $\tau:=\inf\{t\ge 0: X_t=0\}$. Assume $b>0$, then $\mathbb P[\tau=\infty]>0$. What is the conditional law of $X_{\infty}$ knowing $\tau=\infty$?
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By the law of Large numbers, $X_t/t \to b$ almost surely as $t \to +\infty$, hence $X_t \to +\infty$ almost surely as $t \to +\infty$. Therefore $X_\infty = +\infty$ almost surely under $P$ and also under $P[\cdot|\tau = \infty]$.
answered Jun 23, 2022 at 16:18
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$\begingroup$ 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. $\endgroup$– user44143Commented Jun 23, 2022 at 16:22
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2$\begingroup$ @Matt F. You forget the drift! Brownian motion with constant drift is transient. $\endgroup$ Commented Jun 23, 2022 at 16:48
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1$\begingroup$ @GJC20 The density and the distribution function of $X_t$ under $P$ and under $P[\cdot|\tau>t] $converge pointwise to 0 as $t \to +\infty$. $\endgroup$ Commented Jun 23, 2022 at 16:52
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1$\begingroup$ @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 $t\to\infty$. $\endgroup$ Commented Jun 23, 2022 at 20:57
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1$\begingroup$ 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 $X_t$. $\endgroup$ Commented Jun 23, 2022 at 20:58