# How to show that this is a martingale?

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?

• 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 Commented Feb 9, 2016 at 15:50
• 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. $\qquad$ Commented Feb 9, 2016 at 15:55
• @MichaelHardy thanks! Much appreciated :-)
– Ant
Commented Feb 9, 2016 at 16:04
• @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
– Ant
Commented Feb 9, 2016 at 16:06

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.