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In the context of stochastic integration, we showed how it's possible to define the stochastic integral $\int H dM$ for $H \in L^2(M)$ and $M \in \mathcal M^2_0$ (martingales null at $0$ such that $\sup_{t} E[|M_t|^2] < \infty$) Then we want to expand this, and we introduce the "local" version of the spaces, and we say that $M \in \mathcal M^2_{0,loc}$ if exists a sequence of stopping times $\tau_n \nearrow \infty$ such that the stopped process $M^{\tau_n} = M_{t \wedge \tau_n} \in \mathcal M^2_0$ for every $n$.

Similarly we say that a predictable process $H \in L^2_{loc}(M)$ if there exists a sequence of stopping times $\tau_n \nearrow \infty$ such that $HI_{]0, \tau_n]} \in L^2(M)$ for each $n$, where $]0, \tau_n] = \{(\omega, t) \in \bar \Omega: 0 < t \le \tau_n(\omega)\}$

Now the professor's notes say that for

$M \in \mathcal M^2_{0,loc}$ and $H \in L^2_{loc}(M)$, defining the stochastic integral is straightforward: set $$\int H dM = \int HI_{]0, \tau_n]} dM^{\tau_n}$$ on $]0, \tau_n]$.

Question (finally):

It seems to assume that it's possible to find a sequence of stopping times $\tau_n$ that localise simultaneously both $H$ and $M$, which seems surprising to me. Is it true that it's always possible to find such a sequence or am I misunderstanding?

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  • $\begingroup$ Should the sequence of stopping times $\tau_n$ converge to $\infty$ almost surely? $\endgroup$
    – Olorun
    Jan 28, 2016 at 12:54
  • $\begingroup$ @Olorun Indeed. I was a bit lazy and didn't include that :) $\endgroup$
    – Ant
    Jan 28, 2016 at 13:06
  • $\begingroup$ Haha, all righty then! It's just that I've seen that issue several times by others, so just wanted to make sure :) $\endgroup$
    – Olorun
    Jan 28, 2016 at 13:11

1 Answer 1

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Yes, it is possible to find a common localizing sequence.

By the very definition, there exist localizing stopping times $(\sigma_n)_{n \in \mathbb{N}}$ and $(\tau_n)_{n \in \mathbb{N}}$ for $M$ and $H$, respectively. If we set

$$\varrho_n := \min\{\tau_n, \sigma_n\}$$

then $\varrho_n$ is a stopping time satisfying $\varrho_n \uparrow \infty$. Since $\varrho_n \leq \tau_n$, it is obvious that $H 1_{]0,\varrho_n]} \in L^2(M)$. Moreover, the optional stopping theorem (applied for the martingale $M^{\sigma_n}$) gives that

$$M^{\varrho_n}_t = M^{\sigma_n}_{t \wedge \tau_n}$$

is a martingale. Hence, $M^{\varrho_n} \in \mathcal{M}_0^2$.

This shows that $(\varrho_n)_{n \in \mathbb{N}}$ is a localizing sequence for both $M$ and $H$.

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  • $\begingroup$ Great answer, as usual.. thank you! It is clear now. Only thing, you mean $H 1_{]0,\varrho_n]} \in L^2(M)$ right? $\endgroup$
    – Ant
    Jan 27, 2016 at 19:53
  • $\begingroup$ @A Yeah, sure, thanks. You are welcome :) $\endgroup$
    – saz
    Jan 27, 2016 at 20:14

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