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Let $H_s$ be a predictable and bounded process. How can I show that $$M_t = \int_0^t H_s \, dW_s$$ is a martingale?

Clearly since $H_s \in L^2_\text{loc} (W)$ we have that $M_t$ is a local martingale, but then how to proceed?

I have tried to use Lebesgue dominated convergence theorem, writing

$$\lim_{k \to \infty} E[M_{t \wedge \tau_k} \mid \mathcal F_s] = M_s$$

where $\tau_k$ is the localising sequence for the local martingale $M_t$

So If I can bring the limit inside I am done. I need to show that exists $Y$ such that

$$|M_{t \wedge \tau_k}| \le Y \in L^1$$

that is

$$\left|\int_0^{t \wedge \tau_k} H_s \, dW_s\right| \le Y \in L^1$$

If this was a normal integral, I would just write

$$\left|\int_0^{t \wedge \tau_k} H_s \, dW_s\right| \le M\left|\int_0^{t \wedge \tau_k} \, dW_s \right|= M |W_{t \wedge \tau_k} | \le M |W_{\max}| \in L^1$$

where $\max(\omega, t) $ is the value where the brownian motion assumes the highest value in the interval $[0,t]$ (of course as a function of $\omega$)

But I am not sure of my argument, especially when I substitute the $M$ there in the inequality seems very suspicious.. Can you provide some help?

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  • $\begingroup$ Check Lemma 3 on George Lowther's website - here almostsure.wordpress.com/2010/03/25/… . NB: I suppose that $W$ is a Brownian motion which is a square integrable martingale. Best regards $\endgroup$
    – TheBridge
    Commented Feb 9, 2016 at 15:50
  • $\begingroup$ I changed $L^2_{loc}(W)$ to $L^2_\text{loc}(W)$, $W_{max}$ to $W_{\max}$, $max(\omega,t)$ to $\max(\omega,t)$, and $\displaystyle\int H_s dW_s$ to $\displaystyle\int H_s\,dW_s$. All standard usage. $\qquad$ $\endgroup$
    – Michael Hardy
    Commented Feb 9, 2016 at 15:55
  • $\begingroup$ @MichaelHardy thanks! Much appreciated :-) $\endgroup$
    – Ant
    Commented Feb 9, 2016 at 16:04
  • $\begingroup$ @TheBridge Great! It seems to be what I was looking for. Indeed $W$ is the brownian motion. If you want to make that an answer (maybe also elaborating a little bit) I'll gladly accept it $\endgroup$
    – Ant
    Commented Feb 9, 2016 at 16:06

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So the thing is that what you are looking for is exactly the content of lemma 3. I report the claim here :

Lemma 3 : Let $X$ be a càdlàg square integrable martingale and ${\xi}$ be a bounded predictable process. Then, $\int\xi\,dX$ is a square integrable martingale.

In your case $W$ is a Brownian motion, so you fit the conditions for the lemma as a Brownian motion is a continuous square integrable martingale.

The content of the proof is a bit too long to report here but G. Lowther's blog is GREAT and really self contained, he provides all the details in a much better way that I would (or more realistically "could"). Just to say a few words on the argument of the proof, it is based in spirit on the approach of "standard machine" used in the classical measure theory as it goes back to elementary process where the result is clear and then extends it thanks to topological stability properties of the limits. Best regards

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  • $\begingroup$ @Ant : Welcome. As a last remark, you see that you get the "square integrable property" for the same price. Best regards. $\endgroup$
    – TheBridge
    Commented Feb 9, 2016 at 16:39

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