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Let $(X_s)_{s\geq 0}$ be a stochastic process. In which cases is $\int_0^t f(X_s,s) dX_s$ a martingale where f is progressively measurable function?

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    In general, we have $H\dot\ X$ is a martingale, if $H$ is simple and $X$ a martingale, or $H$ bounded with càglàd paths in $\mathbb{R}$ and $X$ a martingale of bounded variation with $sup_{t\in\mathbb{R}_{\geq 0}}E[X_t^2]<\infty$2017-01-12
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    Typically, you have to assume that $(X_t)_t$ is a martingale and that $f$ satisfies a certain integrability condition.2017-01-12

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