I am new to stopping time and is unsure of the following problem. Any hints/ suggestions?
Let $X_t = W_t + 1$, where $W_t$ is a Brownian motion and denote $$T = \mathrm{inf}\{t > 0 : X_t \leq0\}$$
show as-$\lim_{t \to \infty} Y_t = 0$, where $Y_t = X_{T \land t}$
This is equivalent to showing $$P(\omega : \lim_{t \to \infty}Y_t(\omega)=0) = 1$$
My thought process is as such:
I want to show $\lim_{t \to \infty}Y_t(\omega)=0$ thus,
consider $\lim_{t \to \infty}Y_t(\omega)$
That is equivalent to $\lim_{t \to \infty}X_{T\land t}(\omega) = X_T$.
But by definition of $T$, $X_T \leq 0$.
Since $X_T = W_T +1$, then $W_T \leq -1$.
And I am stuck. I do not think this is the right approach…
Thanks!
Best Answer
It is well known that $T$ is finite with probability $1$. If $T(\omega) <\infty$ then $Y_t (\omega) \to X_{T(\omega) }(\omega)$ which is $0$ by continuity of paths: you only need the fact that if $f$ is a continuous function with $f(0)=1$ and you know that $f(t) =0$ for some $t$ then $f(T)=0$ where $T =\inf \{t: f(t)=0\}$.