Let $(X_t)$ be a sample continuous stochastic process on $[0,T]$ and a martingale with respect to a given filtration $(\mathcal F_t)$. Let $f:[0,T] \rightarrow \mathbb R$ be a smooth function.

Is $ Y_t= \int_0^t f(s) d X_s $ a martingale with respect to $(\mathcal F_t)$ ? If yes, how to prove it? If not, can we add conditions on $f$ to make the statement true?

  • $\begingroup$ This is true. It can be taken as part of the definition of the stochastic integral $\int_0^t f(s)dX_s$, but otherwise the proof is quite long. You could check Karatzas and Shreve's Brownian Motion and Stochastic Calculus for a reference. $\endgroup$ Commented Mar 29, 2022 at 21:16
  • $\begingroup$ @user6247850 In Karatzas and Shreve, it seems to me that the filtration is not arbitrary, and that extra conditions are needed on $(X_t)$ such as being $L^2$. I do not know if these conditions are just "convenient" or really necessary. You claimed it to be true, could you share a more precise reference? $\endgroup$
    – W. Volante
    Commented Mar 30, 2022 at 11:51
  • $\begingroup$ @W.Volante The filtration cannot be "arbitrary". It must at least be rich enough for $X_t$ to be a martingale. $\endgroup$ Commented Mar 30, 2022 at 11:52
  • $\begingroup$ @JoseAvilez Yes of course, but that's it. No other conditions are imposed. No "usual conditions", not a Brownian filtration, not generated by $(X_t)$... $\endgroup$
    – W. Volante
    Commented Mar 30, 2022 at 12:06
  • $\begingroup$ See this paper for a treatment of stochastic integration against $L^1$ martingales. $\endgroup$ Commented Mar 30, 2022 at 12:13

1 Answer 1


The sketch of just proving the martingale property is the following. Let $(M_t)_{t \in [0,T]}$ be a continuous $L^2$-martingale on $(\mathscr{F}_t)_{t\in [0,T]}$, define $$(f\cdot M)_t:=L^2\textrm{-}\lim_{n \to \infty}(f_n\cdot M)_t \ \ \ \color{red}{(!)}$$ where $(f_n)_{n \in \mathbb{N}}$ is a sequence of simple processes approximating $f$ $\color{red}{(!)}$. Then $(f\cdot M)_t$ is a martingale. To see this, we first use $L^2$ convergence: $$|E[(f_n\cdot M)_t]-E[(f\cdot M)_t]|^2\leq E[|(f_n\cdot M)_t-(f\cdot M)_t|^2]\stackrel{n\to \infty}{\to} 0$$ And the fact that the simple stochastic integrals $(f_n \cdot M)_t$ are martingales $\color{red}{(!)}$: let $F\in \mathscr{F}_s$, $s<t$ $$E[\mathbf{1}_F(f\cdot M)_t]=\lim_{n \to \infty}E[\mathbf{1}_F(f_n\cdot M)_t]=\lim_{n \to \infty}E[\mathbf{1}_F(f_n\cdot M)_s]=E[\mathbf{1}_F(f\cdot M)_s]$$

$\color{red}{(!)}$ Technical details are set aside here. As suggested by @user6247850, look for a complete answer on a standard textbook. In particular, key parts are the existence of $(f\cdot M)_t$ as defined above and how $f_n$ approximates $f$ and how simple stochastic integrals are martingales.

  • $\begingroup$ So the statement is something like: Let $(\mathcal F_t)$ be an arbitrary filtration. $(Y_t)$ is an $(\mathcal F_t)$ martingale iff $(X_t)$ is an $(\mathcal F_t)$ martingale and $E[\int_0^T f^2(s) d[X]_s] < \infty$ ? $\endgroup$
    – W. Volante
    Commented Mar 29, 2022 at 21:54
  • $\begingroup$ No, the stochastic integral is a continuous $L^2$-martingale if $X_t$ is a continuous $L^2$ martingale and $f$ belongs to a specific closure of the space of simple processes in $[0,T]$. @W.Volante $\endgroup$
    – Snoop
    Commented Mar 29, 2022 at 22:07
  • $\begingroup$ here I do not make assumptions about the martingale $(X_t)$ being $L^2$, is that a problem? $\endgroup$
    – W. Volante
    Commented Mar 30, 2022 at 11:30
  • $\begingroup$ @W.Volante the problem with not using $L^2$ martingales seems not conceptual per se, but it can potentially be rough since I haven't seen it in standard literature yet $\endgroup$
    – Snoop
    Commented Mar 30, 2022 at 14:07
  • $\begingroup$ Ok thank you. So let's says it is $L^2$, what are large classes of deterministic functions that are in "the close of the space of simple processes" ? Lebesgue $L^1$ functions? Lebesgue $L^2$ functions? The class I proposed above? $\endgroup$
    – W. Volante
    Commented Apr 1, 2022 at 11:47

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