I am reading a text on Brownian motion and don't understand the following:
Let $X_t = \exp \{ W_t - \frac{t}{2} \}$, where $W$ is a standard Brownian motion on $\mathbb{R}$. Let $T_n = \inf \{ t \geq 0 : X_t >n \}$ be a sequence of stopping times.
It claims that for any fixed $n$, $\mathbb{E} ( X_{T_n} | \mathcal{F}_t ) = X_{T_n \wedge t}$ a.s, for any $t > 0$.
My reasoning:
By optional stopping theorem, $\forall m \geq t$, $ \mathbb{E} ( X_{T_n \wedge m} | \mathcal{F}_t ) = X_{T_n \wedge t}$.
We know that $(X_{T_n \wedge m} )_{m \in \mathbb{N}}$ is uniformly bounded. If we could show that $T_n < + \infty$ a.s, then we can apply the dominated convergence theorem and claim that $$ \mathbb{E} ( X_{T_n } | \mathcal{F}_t ) = \lim_{m \to \infty} \mathbb{E} ( X_{T_n \wedge m} | \mathcal{F}_t ) = X_{T_n \wedge t}.$$
My question is: Is it true that $T_n < + \infty$ a.s?
No, $T_n<\infty$ does not hold true.
Proof: Suppose that $T_n<\infty$ almost surely. Since $(X_t)_{t \geq 0}$ is a martingale, $X_0 = 1$, the optional stopping theorem yields
$$\mathbb{E}X_{T_n \wedge t} = 1$$
for any $t \geq 0$. Since $|X_{T_n \wedge t}| \leq n$ for all $t$, we get
$$\mathbb{E}X_{T_n}=1 \tag{1}$$
by the dominated convergence theorem. On the other hand it follows from the continuity of Brownian motion that $(X_t)_{t \geq 0}$ has continuous sample paths and therefore $X_{T_n}=n$ almost surely. Obviously, this contradicts $(1)$.