# How to prove convergence of process and stopping time

We consider the random function $X^n=(X^n_t)_{t\geq 0}$ with values in the Skorokhod space $\mathcal{D}$ of càdlàg paths, and suppose that it weakly converges (i.e in distribution) to $X=(X_t)_{t\geq0}$ as $n\rightarrow \infty$. If we define the first passage time to level $x\in\mathbb{R}$, by: $$\tau_x(Y)=\inf\{t\geq 0 : Y_t\geq x\},$$ for any random function $Y$ with values in $\mathcal{D}$, I would like to prove that $(X^n,\tau_x(X^n))$ weakly converges to $(X,\tau_x(X))$ as $n\rightarrow \infty$. I am wondering if there is already an existing result in the literature of weak convergence of stochastic processes, or if there is any result (possibly under more restrictive assumptions) which enables one to prove joint weak convergence? Any ideas or references to the literature would be greatly appreciated.

update: My general idea about how to proceed is to use something along the lines of: if $X^n$ weakly converges to $X$ as $n\rightarrow\infty$, then we should expect $(X^n,f(X^n))$ to weakly converge to $(X,f(X))$ as $n\rightarrow \infty$ for any measurable $f$, and apply this to the case of the first passage time, but I am not sure whether this is entirely rigorous?

You have to be more concrete. In general, the distribution of $\tau(X^n)$ will not converge to the distribution of $\tau(X)$, because this is not a continuous function of the path.
$$X^n(t) = \begin {cases} 0 & t \in [0,\frac 12) \\ 1-\frac 1n & t \in [\frac 12,1) \\ 1 & t \in [1,\infty)\end{cases}$$
Now $X^n$ converges uniformly to $X$, the indicator of $[\frac 12,\infty)$.
Let $\tau$ denote the hitting time of $1$. We have that $\tau(X)=\frac 12$, however, $\tau(X^n) = 1$ for all $n$.
• Thanks for your input. If I add the assumption that $X^n$ is continuous, does the result hold? Mar 22, 2016 at 19:05
• @user223935, no. Let $X^n(t) = 1-(1-t)/n$. Then $\tau_1(X^n) = 1$, $\tau_1(X) = 0$. You should also add some condition of uniform variability of $X^n$. Mar 23, 2016 at 14:33