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Let $B: \Omega \to \mathbb{R}$ be a standard $1$-dim. Brownian motion (started at $0$). Then, the event $\{ \lim_{t \to 1} B_t = 0 \}$ has probability zero, but we can study the Brownian bridge, a stochastic process which can be interpreted as a Brownian motion conditioned to hit $0$ at time $1$, in a sense which is made rigorous e.g. here.

Is there a stochastic process $X$ which can be interpreted as a Brownian motion conditioned to vanish at $\infty$?

I thought about setting $X_t = \beta_{\frac{t}{t+1}}$ where $\beta$ is a standard Brownian bridge, but after rescaling to get $[X]_t = t$, I don’t think this vanishes at $\infty$ any more.

In general, the methods used to construct the Brownian bridge do not seem like they work here. For example, it is difficult to use weak convergence of conditioned processes because $\mathbb{P}(\limsup_t B_t - \liminf_t B_t \le M) = 0$ for all $M > 0$.

The only other thought I had is to concatenate infinitely many rescaled Brownian bridges, but it’s not clear what the right rescaling is.

J. S.
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    Maybe I missed the context, but what do you mean by interpreted? – openspace Jan 10 '25 at 10:33
  • @openspace Given that you can’t directly condition on an event of probability zero, I think it’s more precise to say that e.g. a Brownian bridge can be interpreted as a BM conditioned to hit $0$ at time $T$. Ideally there would be some compelling reasons to support the interpretation, as there are for Brownian bridges. – J. S. Jan 10 '25 at 10:39
  • So it's like they equal in $d$ for any $t$ ? – openspace Jan 10 '25 at 10:42
  • @openspace Maybe, if such a process exists and that uniquely characterises the process. – J. S. Jan 10 '25 at 10:51
  • When $\beta$ is Brownian bridge that starts at zero and ends at zero for $t=1$ you have in $X_t=\beta_{\frac t{t+1}}$ undoubtedly a process that ends at zero for $t=\infty,.$ From $[X]_t=t$ (which is true) you cannot conclude that $X$ does not vanish at $t=\infty$ anymore because that quadratic variation is the one of $X's$ martingale part only. Recall that $X$ satisfies the SDE $$dX_t=-\frac{X_t,dt}{1-t}+dB_t$$ where $B$ is a Brownian motion. – Kurt G. Jan 10 '25 at 15:41
  • @KurtG. Indeed, the process $X$ converges to $0$ at infinity, but that is not what I was talking about. I do not agree that $[X]_t = t$. Here is a reference which would prove that it is $\frac{t}{t+1}$, which was also my belief. I said after rescaling to get $[X]_t = t$, i.e. multiplying $X$ by a deterministic function so that its quadratic variation at time $t$ really is $t$, then it is no longer vanishing at infinity. I still believe this to be true. – J. S. Jan 10 '25 at 20:12
  • For clarification if it is needed, noting that for a Brownian motion $B$, the quadratic variation at time $t$ is $t$ independent of the realisation, it logically follows that a quadratic variation of $t$ would be a sensible pre-condition for any stochastic process intended to be interpreted as conditioned Brownian motion. – J. S. Jan 10 '25 at 20:16
  • Yes I agree (stumbled about my own notation). We definitely have $[X]_t=\frac t{t+1},.$ What you want seems hard for formulate rigorously. – Kurt G. Jan 11 '25 at 09:02
  • @KurtG. I think the concatenation of Brownian bridges idea seems promising. For that approach, it may be useful to have an answer to the following question. For all $x \in \mathbb{R}$, let $\tau_x = \inf {t \ge x : B_t = 0}$. What is the distribution of $\tau_x$ given that $\tau_x > 0$? I think this one is more amenable and that I may be able to solve it, but I may ask it at some point. Do you know the answer? – J. S. Jan 11 '25 at 14:11

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