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$X$ is a continuous local martingale with $X_0=0$. I'm trying to show that $\lim_{t\to \infty}X_t$ is finite then $[X]_\infty<\infty$, where $[X]$ denotes quadratic variation.

I want to do this by showing the containments $$\lbrace \lim X_t \text{is finite}\rbrace\subset\cup_{i\ge 1}\lbrace \tau_i=\infty\rbrace\subset\lbrace [X]_\infty<\infty \rbrace,$$ where $\tau_i=\inf \lbrace t\ge 0: |X_t|=i\rbrace$ but I can't quite get this to work:

If the limit is finite then $[X]_{\tau_n}$ is bounded. That's all I have for this direction. The rest of my ideas are only helpful in proving the reverse inclusions. How can I show the containments?

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  • $\begingroup$ what exactly do you mean by "$[X]_{\tau_n}$ is bounded"...? $\endgroup$
    – saz
    Feb 8, 2017 at 8:31

1 Answer 1

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For the first inclusion note that if $\lim_{t \to \infty} X_t(\omega)$ exists, then it follows from the continuity of the sample paths that $t \mapsto X_t(\omega)$ is bounded. This, in turn, implies $\tau_i(\omega)=\infty$ for $i$ sufficiently large.

To prove the second inclusion we use that, by the optional stopping theorem, $$X_{t \wedge \tau_i}^2 - [X]_{t \wedge \tau_i}$$ is a local martingale. If we denote by $(\sigma_k)_k$ a localizing sequence of stopping times, we get

$$\mathbb{E}(X_{t \wedge \tau_i \wedge \sigma_k}^2) = \mathbb{E}([X]_{t \wedge \tau_i \wedge \sigma_k})$$

implying

$$\mathbb{E}([X]_{t \wedge \tau_i \wedge \sigma_k}) \leq i^2.$$

It follows from the monotone convergence theorem that

$$\mathbb{E}([X]_{t \wedge \tau_i}) \leq i^2$$

and

$$\mathbb{E}\left(\sup_{t \geq 0} [X]_{t \wedge \tau_i} \right) \leq i^2.$$

Conclude from the last inequality that $[X]_{\infty}(\omega)<\infty$ almost surely for any $\omega \in \{\tau_i = \infty\}$.

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