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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?

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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 $(1)$.

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  • $\begingroup$ So if $T_n < + \infty$ with a positive probability, does this mean that when $T_n = \infty$, $X_{T_n}= X_{\infty}$ is interpreted to be the a.s. limit of $X_t$???? $\endgroup$ – Richard Jan 28 '15 at 19:32
  • $\begingroup$ @Richard Well, I don't know what the author was thinking of when (s)he wrote this... However, note that the limit $$\lim_{t \to \infty} X_t$$ does indeed exist (since $(X_t)_t$ is a non-negative martingale); in fact, one can calculate $X_{\infty}$ explicitly. $\endgroup$ – saz Jan 28 '15 at 19:37

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