Distribution of a stochastic process at a stopping time.

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.

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.
• 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? Oct 26, 2021 at 16:31
• 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. Oct 26, 2021 at 23:24