Show that $M_t = ( \int_0^t b(w,u) dB_u)^2 - \int_0^t b^2 (w,u) du$ is a martingale

Question

$M_t = ( \int_0^t b(w,u) dB_u)^2 - \int_0^t b^2 (w,u) du$ 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.

Answer 1

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 $$\mathbb{E}\left[\left.\mathcal{I}_t^2 -\int_0^t b^2(\omega, u)du \right| \mathcal{F}_s\right] = \mathcal{I}_s^2 - \int_0^s b^2(\omega, u)du. $$