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Let $f \in C_b^2(\mathbb{R}_{+} \times \mathbb{R})$. Then my lecturer said that for a Brownian motion $B_t$ \begin{align} X_t = \int_0^t f(s, B_s) \, dB_s \end{align} is a martingale, which should follow from the integral being square integrable, but I can't how to derive this from Ito's formula.

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  • $\begingroup$ A sufficient condition for $X_t$ to be a martingale is that $\mathsf{E}\int_0^t [f(s,B_s)]^2\, ds<\infty$ for all $t$, which holds if $f$ is bounded. $\endgroup$
    – user140541
    Commented Dec 17, 2020 at 9:03
  • $\begingroup$ @d.k.o. Do you know the lemma where this follows from? $\endgroup$
    – Hamilton
    Commented Dec 17, 2020 at 9:07
  • $\begingroup$ Sure. Look at Corollary 3.2.6 here. The definition of $\mathcal{V}$ is on page 25. $\endgroup$
    – user140541
    Commented Dec 17, 2020 at 9:16

1 Answer 1

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The condition @d.k.o. provided in the comments implies that $E[\int_s^t f(u,B_u)dB_u | \mathscr F_s]=0.$ Then, using the linearity property of the integral, we have $$E[X_t |\mathscr F_s]=E[X_s |\mathscr F_s]+E[\int_s^t f(u,B_u)dB_u | \mathscr F_s]=X_s.$$

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