# Compensator for $\bigg(\int_0^t f(s)dB(s)\bigg)^2$.

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.

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 $$(M_t^2 - \int_0^t f(s)^2 \, ds)$$.