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I have a question about local martingale. I am reading Karatzas and Shreve's Brownian Motion and Stochastic Calculus(Second Edition). On page 146, it wrote ($M$ is a local continuous martingale $M_{0}=0$ a.s. $P$ )

" For $X \in \mathcal{P}^{\ast}$,one construct... \begin{eqnarray*} R_{n}(\omega)=n \wedge \inf \{0 \leq t<\infty \,; \int_{0}^{t}X_{s}^{2}(\omega)d \left< M\right>_{s}(\omega) \geq n \} \end{eqnarray*} This is also a nondecreasing sequence and because of (2.31), $\lim_{n \to \infty}R_{n}=\infty$ a.s. $P$"

(2.31) : $P\left[\int_{0}^{T}X_{t}^{2}d \left<M\right>_{t} <\infty \right]=1 $ for every $T\in [0,\infty)$

$\mathcal{P}^{\ast}$ is the collection of equivalence classes of all progressively measurable processes satisfying the condition (2.31).

I see "$R_{n} $ is a nondecreasing sequence", but I can't understand "$\lim_{n \to \infty} R_{n}= \infty$ a.s. $P$".

I think that it is sufficient to prove for all $\omega \in \cap_{N=1}^{\infty}A_{N}, A_{N}:=\{\int_{0}^{N}X_{t}^{2}d \left<M\right>_{t}< \infty\}$ , $R_{n}(\omega)\to \infty$.

I'm never certain about the use of (2.31). How do you use (2.31)?

s34
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1 Answers1

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Since

$$\mathbb{P} \left( \int_0^T X_t^2 \, d\langle M \rangle_t < \infty \right)=1$$

for all $T \geq 0$, we have

$$\mathbb{P} \left( \forall n: \int_0^n X_t^2 \, d\langle M \rangle_t < \infty \right)=1.$$

Consequently, it suffices to prove $\lim_{n \to \infty} R_n(\omega)=\infty$ for all $$\omega \in \Omega_0 :=\left\{\forall n: \int_0^n X_t^2 \, d\langle M \rangle_t < \infty\right\}.$$ Pick $\omega \in \Omega_0$ and $n \in \mathbb{N}$. Then we can choose $m=m(n,\omega) \in \mathbb{N}$, $m \geq n$, such that

$$\int_0^n X_t^2(\omega) \, d\langle M \rangle_t(\omega) \leq m.$$

It follows from the definition of the stopping time and the fact that $m \geq n$ that $R_m(\omega) \geq n$. Since the sequence is increasing, we have shown that for any $n \in \mathbb{N}$ there exists $m_0 \in \mathbb{N}$ such that for all $m \geq m_0$:

$$R_m(\omega) \geq n.$$

Hence, $\lim_{m \to \infty} R_m(\omega)=\infty$.

saz
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