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If $E\int_0^T X_t^2 dt < \infty$ then the Ito integral can be defined for every $0 \leq t \leq T$ such that it is a continuous process in $[0,T]$. The convention that I am working with for a simple process is that $X_t = \sum e_j \chi_{[t_j,t_{j+1})}$ (some books tend to use $X_t = \sum e_j \chi_{(t_j,t_{j+1}]}$) with the jump times $0=t_0<t_1<t_2<\cdots < t_N = T$.

If $\tau \leq T$ is a stopping time, $$\int_0^{t\wedge \tau} X_s dW_s= \int_0^t X_sI_{s < \tau} dW_s$$ The notes (see Lemma 4.2.9) outlines a proof for this. The steps in the proof are

  • Assume that there exist simple processes $X^n$ which converge to $X$ fast enough.
  • For every $n$, define the stopping time $\tau^n$ to be equal to the smallest jump time of $X^n$ greater than or equal to $\tau$
  • Prove that $$\int_0^{\sigma} X_s^n dW_s= \int_0^T X_s^nI_{s < \sigma} dW_s$$ for every stopping time $\sigma \leq T$
  • Using the previous result prove that $$\int_0^{\tau} X_s dW_s= \int_0^T X_sI_{s < \tau} dW_s$$
  • Extend for general case by using the stopping time $\tau \wedge T$.

I could prove the third point above (but it was not as simple as suggested in the notes). The notes say that step four can be proved using continuity of paths.

As $n \to \infty$, $\int_0^T X_s^nI_{s < \tau^n}dW_s \to \int_0^T X_sI_{s < \tau}dW_s$, if it can be shown that $X_s^nI_{s < \tau^n} \to X_sI_{s < \tau}$ in $\mathcal{L}^2$. It also remains to show that $\int_0^{\tau^n} X_s^n dW_s \to \int_0^\tau X dW_s$. I do not understand how to show both these claims. Is there any reference that proves this particular form of the theorem. Any help is appreciated.

Thanks,

jpv
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  • You can find a proof of this result e.g. in Schilling's book on Brownian motion; however, he uses a different idea to prove the result (namely, approximation of the stopping time $\tau$ by discrete stopping times). – saz Oct 04 '15 at 06:57
  • saz: Thanks for the reference. – jpv Oct 04 '15 at 07:14
  • You are welcome. If you are still interested in proving it the way suggested in the notes, I can have a look at it later today. – saz Oct 04 '15 at 08:14
  • saz: Thanks. It would be great if you can give a proof. The book you suggested seems to be very nice. – jpv Oct 04 '15 at 09:40
  • Yeah, I like it very much :). Do you know the maximal inequality for stochastic integrals; that is $$\mathbb{E} \left( \sup_{s \leq t} \left| \int_0^s f(r) , dW_r \right|^2 \right) \leq 4 \mathbb{E} \int_0^t f(s)^2 , ds$$....? – saz Oct 04 '15 at 17:07
  • No, I am not aware of that. Is that useful for the proof? I shall try to read about it. – jpv Oct 05 '15 at 09:14

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