Suppose I have a continuous-time stochastic process $\{X(t)\}$ defined on a filtered probability space $(\Omega, \mathcal{F},\{\mathcal{F}_t\},\mathbb{P})$ and that I know the distribution of a stopping time $\tau$, can I say something about the distribution (not just the moments) of the random variable $X(\tau)$?

I guess in general the answer is no, but something could be said if the stochastic process $\{X(t)\}$ has some kind of structure (e.g. Geometric Brownian Motion, Lévy processes, martingales, ...). Any help would be kindly appreciated.


1 Answer 1


In general, no, not without knowing the joint distribution of $(\tau, X)$. We can have two processes $X$ and $Y$ with the same distribution, but $X(\tau)$ has a different distribution than $Y(\tau)$.

For example, let $B$ and $W$ be independent Brownian motions and $\tau := \inf\{t : B(t) = 1\}$. Even though $B$ and $W$ have the same structure and distribution, we have $B(\tau) = 1$ but $W(\tau)$ is not identically $1$. One can show that $W(\tau)$ has a Cauchy distribution.

  • $\begingroup$ Right, is it because $\mathbb{P}\left(X(\tau)\in A\right)= \int_A \,dF_{X,\tau}(x,\tau)=\int_A\,f_{X,\tau}(x,\,\tau)\,dx\,d\tau$ (assuming the distribution of $(X,\,\tau)$ is absolutely continuous w.r.t. the Lebesgue measure)? Are there cases in which is possible to retrieve the distribution of $(X,\tau)$ in closed form? $\endgroup$ Oct 26, 2021 at 16:31
  • $\begingroup$ I'm not sure it makes sense to think of the law of $X$ being absolutely continuous w.r.t. Lebesgue measure. $X$ is a process, so its law is a probability measure over paths. For the same reason, I don't know you can really describe the law of $(X,\tau)$ in closed form. $\endgroup$ Oct 26, 2021 at 23:24

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