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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 ?

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2 Answers 2

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$$\mathbb{E}[M_t-M_s|\mathcal{F}_s]=\mathbb{E}\left[\left(\int_0^tf(u)dB_u\right)^2\bigg|\mathcal{F}_s\right]-\mathbb{E}\left[\int_0^tf(u)^2du\bigg|\mathcal{F}_s\right]-$$ $$+\left(\int_0^sf(u)dB_u\right)^2+\int_0^sf(u)^2du$$ 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 $$\mathbb{E}[M_t-M_s|\mathcal{F}_s]=\mathbb{E}\left[\int_s^tf(u)^2du\bigg|\mathcal{F}_s\right]+\left(\int_0^sf(u)dB_u\right)^2-\mathbb{E}\left[\int_0^tf(u)^2du\bigg|\mathcal{F}_s\right]-\left(\int_0^sf(u)dB_u\right)^2+\int_0^sf(u)^2du=0$$

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  • $\begingroup$ 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...) $\endgroup$
    – joshua
    Commented Apr 18, 2021 at 16:52
  • $\begingroup$ 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. $\endgroup$
    – Snoop
    Commented Apr 18, 2021 at 18:01
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Here's another way that uses standard properties of stochastic integrals and local martingales.

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.

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