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If $X(t)$ is a martingale, and $X(0) = 0$. $f(t)$ is a left continuous function,

$$ g(t) = \int_0^t f(s) X(s) ds $$

is $g(t)$ also a martingale?

I guess it shall be, but don't know how to prove that?

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up vote 4 down vote accepted

Since $(X(t))_t$ is a martingale in the filtration $(\mathcal F_t)_t$, one knows that $\mathrm E(X(u)\mid\mathcal F_s)$ is $X(u)$ is $u\leqslant s$ and $X(s)$ if $u\geqslant s$. Hence, for every $s\leqslant t$, $$ \mathrm E(g(t)\mid\mathcal F_s)=g(s)+\int_s^tf(u)\mathrm E\left(X(u)\mid F_s\right)\mathrm du=g(s)+X(s)\cdot \int_s^tf(u)\mathrm du. $$ The only cases when $(g(t))_t$ is a martingale are when $f(t)X(s)=0$ almost surely, for almost every $s\leqslant t$.

Edit: Note however that $X(0)=0$ and that $(X(t))_t$ is a martingale hence $\mathrm E(X(t))=0$ for every $t$, and $$ \mathrm E(g(t))=\int_0^tf(s)\mathrm E\left(X(s)\right)\mathrm ds=0. $$

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thank you @Didier, now i understand martingale more... – athos Mar 30 '12 at 18:03
however, can i say that E(g(t))=g(0)? since X(0)=0... – athos Mar 31 '12 at 5:22

No, take $f(s) = s$ so $$ g(t) = \int\limits_0^t s B_s\mathrm ds = \int\limits_0^t B_s\mathrm d\frac{s^2}{2} = \left.\frac{s^2}{2}B_s \right|_0^t - \frac12\int\limits_{0}^ts^2\mathrm dB_s = \frac12 t^2B_t - \frac12\int\limits_{0}^ts^2\mathrm dB_s. $$ Note that $\frac12\int\limits_{0}^ts^2\mathrm dB_s$ is a martingale and if $g$ is a martingale then $\frac12 t^2 B_t$ has to be martingale, but: $$ \mathsf E[t^2 B_t|\mathscr F_s] = t^2 B_s\neq s^2 B_s. $$

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thank you, now I realize my mistake... – athos Mar 30 '12 at 18:03

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