# Conditional probability distribution of a Brownian particle surviving forever

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 $$\tau=\infty$$?

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 $$P[\cdot|\tau = \infty]$$.
• 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 at 16:22
• @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 at 16:52
• @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 at 20:57
• 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 at 20:58