# Prove that $M_t=\left(\int_0^t f(s)dB_s\right)^2-\int_0^t f(s)^2ds$ is a martingale.

Let $$f:\Omega \times [0,\infty )\to \mathbb R$$ progressively measurable and s.t. $$\mathbb E\int_0^t f(s)^2ds<\infty$$ for all $$t$$. I would like to prove that $$(M_t)_{t\geq 0}$$ is a martingale where $$M_t=\left(\int_0^t f(s)dB_s\right)^2-\int_0^t f(s)^2ds,$$ iwhere $$(B_t)$$ is a Brownian motion. If $$f$$ is predictable, I proved that $$f$$ is measurable. Let $$(f_n)$$ a sequence of predictable sequence s.t. $$f_n\to f$$ in $$L^2(\Omega \times [0,t])$$. So, if I can prove that $$M^n_t:=\left(\int_0^t f_n(s)dB_s\right)^2-\int_0^t f_n(s)^2ds$$ converges to $$M_t$$ for all $$t$$ in $$L^1(\Omega )$$, then $$(M_t)$$ will be a martingale. We have that \begin{align*} \mathbb E|M_t^n-M_t|&=\mathbb E\left[\left|\left(\int_0^t f(s)dB_s\right)^2-\int_0^t f(s)^2ds-\left(\int_0^t f_n(s)dB_s\right)^2+\int_0^t f_n(s)^2ds\right|\right]\\ &\leq \mathbb E\left[\left|\left(\int_0^t f(s)dB_s\right)^2-\left(\int_0^t f_n(s)dB_s\right)^2\right|\right]+\mathbb E\left[\left|\int_0^t f(s)^2ds-\int_0^t f_n(s)^2ds\right|\right]\\ &\leq \mathbb E\left[\left|\left(\int_0^t f(s)dB_s\right)^2-\left(\int_0^t f_n(s)dB_s\right)^2\right|\right]+\mathbb E\left[\int_0^t |f(s)^2 -f_n(s)^2|ds\right] \end{align*} but I don't see how to continue. Any idea ?

$$\mathbb{E}[M_t-M_s|\mathcal{F}_s]=\mathbb{E}\left[\left(\int_0^tf(u)dB_u\right)^2\bigg|\mathcal{F}_s\right]-\mathbb{E}\left[\int_0^tf(u)^2du\bigg|\mathcal{F}_s\right]-$$ $$+\left(\int_0^sf(u)dB_u\right)^2+\int_0^sf(u)^2du$$ By Ito isometry, for $$t > s$$: $$\mathbb{E}\left[\left(\int_0^tf(u)dB_u\right)^2\bigg|\mathcal{F}_s\right]=$$ $$=\mathbb{E}\left[\left(\int_s^tf(u)dB_u\right)^2\bigg|\mathcal{F}_s\right]+\left(\int_0^sf(u)dB_u\right)^2+2\mathbb{E}\left[\left(\int_s^tf(u)dB_u\right)\left(\int_0^sf(u)dB_u\right)\bigg|\mathcal{F}_s\right]=$$ $$=\mathbb{E}\left[\int_s^tf(u)^2du\bigg|\mathcal{F}_s\right]+\left(\int_0^sf(u)dB_u\right)^2$$ Therefore $$\mathbb{E}[M_t-M_s|\mathcal{F}_s]=\mathbb{E}\left[\int_s^tf(u)^2du\bigg|\mathcal{F}_s\right]+\left(\int_0^sf(u)dB_u\right)^2-\mathbb{E}\left[\int_0^tf(u)^2du\bigg|\mathcal{F}_s\right]-\left(\int_0^sf(u)dB_u\right)^2+\int_0^sf(u)^2du=0$$
• Notice that $f$ is not deterministic. However, your proof is easily adaptable. Thanks. Nevertheless, is there a way to prove M_t^n\to M_t$in$L^1$? (it's a hint of my exercise, so I would like to use it...) Apr 18, 2021 at 16:52 • True. I think I fixed it. I am thinking about the$L^1$convergence and tried something to prove that$\lim_{n \to \infty}\mathbb{E}[|M^n_t|]=\mathbb{E}[|M_t|]\$ but I'm not entirely convinced. Will post if I come up with a solid one. Apr 18, 2021 at 18:01
The process $$N_t=\int_0^tf(s)dB_s$$ is a local martingale with quadratic variation $$\langle N, N\rangle_t=\int_0^tf(s)^2ds$$. Better even, it is a square-integrable martingale since $$\mathbb{E}\int_0^\infty f(s)ds<\infty$$. Therefore, $$M_t=N_t^2-\langle N,N\rangle_t$$ is a true martingale.