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Let $\{B(t)\}_{t\geq 0}$ be a brownian motion, $ \{\mathcal F_t\}_{t\geq 0}$ an admissible filtration for the BM.

Let $f$ be an $\mathcal F_t$-adapted function satisfying $\mathbb E \int_a^b f(s)^2ds <\infty$

Now I define $M_t$ to be the following stochastic process $$M_t=\int_0^t f(s) dB(s), \\ a\leq t\leq b$$

It's clear that under our assumptions $M_t$ is a martingale and by Jensen's inequality for conditional expectations we have that $M_t^2$ is a sub-martingale.

Then using Ito's formula we obtain the following

$$M^2_t=\color{red}{2\int_0^t M_s f(s)dB(s)}+\color{green}{\int_0^t f(s)^2 ds}$$

now, clearly the second term is an increasing process, so I am tempted to say that it's the compensator for the sub-martingale $M^2_t$, but nonetheless to claim that I would need the first term to be a martingale, namely the above expression needs to be the Doob-Meyer decomposition.

But to be honest it's not clear to me whether or not $2\int_0^t M_s f(s)dB(s)$ is a martingale.

I know that it would suffice to prove that $$\mathbb E \int_a^b M_s^2 f(s)^2 ds<\infty$$

But so far I haven't been able to do so.

I would appreciate any help.

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If the function $f$ is bounded, then it follows from Itô's formula that $$\mathbb{E} \int_a^b M_s^2 f(s)^2 \, ds < \infty, \tag{1}$$ and therefore the stochastic integral $\int_0^t M_s f(s) \, dB_s$ is a martingale. If $f$ is unbounded, then the situation is more difficult. Somehow, $(1)$ is a too strong condition since $(1)$ actually implies that the stochastic integral is an $L^2$-martingale (which we cannot expect in general; square integrability may fail).

For a general function $f$ (satisfying $\mathbb{E}\int_0^t f(s)^2 \, ds < \infty$) define $$f_n := (-n) \vee f \wedge n.$$ It follows from the dominated convergence theorem that $$\mathbb{E} \int_0^t |f_n(s)-f(s)|^2 \, ds \xrightarrow[]{n \to \infty} 0,$$ and so $$\lim_{n \to \infty} \underbrace{\int_0^t f_n(s) \, dB_s}_{=:M_t^{(n)}} = \underbrace{\int_0^t f(s) \, dB_s}_{=:M_t}\quad \text{in $L^2$}.$$ This entails that

$$(M_t^{(n)})^2 - \int_0^t f_n(s)^2 \, ds \to M_t^2 - \int_0^t f(s)^2 \, ds \quad \text{in $L^1$}$$

for each $t >0$. Since $f_n$ is bounded, we know from our earlier consideration that $(M_t^{(n)})^2 - \int_0^t f_n(s)^2 \, ds$ is a martingale (w.r.t the canonical filtration of Brownian motion), and so is the $L^1$-limit $(M_t^2 - \int_0^t f(s)^2 \, ds)$.

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  • $\begingroup$ Thanks again Saz!, I'm happy to see that it wasn't such a trivial result. In my textbook it was stated that " it's easy to prove that ...", not so easy to be honest. $\endgroup$
    – Chaos
    Feb 7, 2020 at 16:43

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