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$M_t = ( \int_0^t b(w,u) dB_u)^2 - \int_0^t b^2 (w,u) du$ where $B_u$ is a standard brownian motion and $b$ is such that $E[\int_0^t b^2(w,u) du]$ is finite.

I want to show that $M_t$ is a martingale with respect to the standard brownian filtration.

Showing that $M_t$ is $L^1$ is ok. But I can't prove that $E[M_t | F_s]=M_s$ for $t>s$. I'm pretty sure that one has to use Ito isometry, but even with that, I can't manage to arrive at $M_s$.

Thank you for your help.

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For simplicity denote $\mathcal{I}_t := \int_0^t b(\omega, u)dB_u$ for each $t \geq 0$.

The generalised Ito's isometry provides us with the following equality $$ \mathbb{E}[\left.(\mathcal{I}_t - \mathcal{I}_s)^2 \right| \mathcal{F}_s] =\mathbb{E}\left[\left. \int_s^t b^2(\omega, u)du\right|\mathcal{F}_s\right] $$ for all $0\leq s < t <\infty$.

To prove this formula start first with the case when $b$ is a simple process, as it is done in the standard approach for proving Ito's isometry (the proof of the generalised Ito's isometry can be found in Brownian motion and stochastic calculus by Karatzas and Shreve in section 3.2.B. ) Then for the general case you take a sequence of simple processes $(b_n)$ to approximate $b$, that is, $$\mathbb{E} \left[ \int_0^t (b(\omega, s) - b_n(\omega, s))^2 ds\right] \to 0 \mbox{ as } n \to \infty $$ and use Cauchy-Schwartz inequality.

Then by applying the generalised Ito's isometry we can show that $\left( \mathcal{I}_t^2 -\int_0^t b^2(\omega, u)du\right)_{t \geq 0}$ is a martingale w.r.t $(\mathcal{F}_t)_{t \geq 0}$.

We obtain \begin{align*} \mathbb{E}\left[\left. \int_s^t b^2(\omega, u)du\right|\mathcal{F}_s\right]=& \mathbb{E}[\left.(\mathcal{I}_t - \mathcal{I}_s)^2 \right| \mathcal{F}_s] \\=& \mathbb{E}[\left.\mathcal{I}_t^2 \right| \mathcal{F}_s]-2\mathcal{I}_s\mathbb{E}[\mathcal{I}_t | \mathcal{F}_s] + \mathbb{E}[\left.\mathcal{I}_s^2|\mathcal{F}_s \right| \\=& \mathbb{E}[\left.\mathcal{I}_t^2 \right| \mathcal{F}_s]-2\mathcal{I}_s^2 + \mathcal{I}_s^2\\ =&\mathbb{E}[\left.\mathcal{I}_t^2 \right| \mathcal{F}_s]- \mathcal{I}_s^2 \end{align*} on the other hand \begin{align*} \mathbb{E}\left[\left. \int_s^t b^2(\omega, u)du\right|\mathcal{F}_s\right] =& \mathbb{E} \left[\left. \int_0^t b^2(\omega, u)du - \int_0^s b^2(\omega, u)du \right| \mathcal{F}_s\right]\\ =& \mathbb{E} \left[\left. \int_0^t b^2(\omega, u)du\right| \mathcal{F}_s\right] - \mathbb{E} \left[\left. \int_0^s b^2(\omega, u)du \right| \mathcal{F}_s\right]\\=& \mathbb{E} \left[\left. \int_0^t b^2(\omega, u)du\right| \mathcal{F}_s\right] - \int_0^s b^2(\omega, u)du .\end{align*} Hence, $$\mathbb{E}[\left.\mathcal{I}_t^2 \right| \mathcal{F}_s]- \mathcal{I}_s^2 = \mathbb{E} \left[\left. \int_0^t b^2(\omega, u)du\right| \mathcal{F}_s\right] - \int_0^s b^2(\omega, u)du $$ and so $$\mathbb{E}\left[\left.\mathcal{I}_t^2 -\int_0^t b^2(\omega, u)du \right| \mathcal{F}_s\right] = \mathcal{I}_s^2 - \int_0^s b^2(\omega, u)du. $$

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  • $\begingroup$ Can you explain why $E[\int_0^s b^2(w,u) du | F_S] = \int_0^s b^2(w,u) du$ ? I tend to say $E[\int_0^s b^2(w,u) du | F_S]= E[\int_0^s b^2(w,u) du]$ $\endgroup$
    – W. Volante
    May 11, 2017 at 17:23
  • $\begingroup$ Sure. Let $\mathcal{G}$ be a $\sigma$-algebra if a random variable $X$ is $\mathcal{G}$-measurable then $\mathbb{E}[X| \mathcal{G}] = X$. In order to obtain your claim, $X$ has to be independent from $\mathcal{G}$ and then $\mathbb{E}[X|\mathcal{G}] = \mathbb{E}[X]$. Here we have that the random variable $\int_0^s b^2(\omega, u)du$ is $\mathcal{F}_s$-measurable for all $s \geq 0$. $\endgroup$
    – m_gnacik
    May 11, 2017 at 17:48
  • $\begingroup$ Why is it $\mathcal{F}_s$-mesurable ? This integral has nothing to do with Brownian motion right ? I'd say that it's independent from $\mathcal{F}_s$ $\endgroup$
    – W. Volante
    May 11, 2017 at 21:24
  • $\begingroup$ It depends on your assumptions on $b(\omega, t)$. In order to have $(M_t)$ well defined (we take an Ito integral of $b(\omega, t)$) so it is assumed that $(b(\cdot, t))_t$ is adapted with respect to $(\mathcal{F}_t)$. Now because the time integral of it exists and is finite, then this integral say $\int_0^t b^2(\omega, u)$ is $\mathcal{F}_t$-measurable for all $t \geq 0$. Check this as well math.stackexchange.com/questions/206344/… $\endgroup$
    – m_gnacik
    May 11, 2017 at 21:36
  • $\begingroup$ I got it, thanks ! $\endgroup$
    – W. Volante
    May 11, 2017 at 21:40

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