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If $M_t$ is a martingale, is this process a martingale too ?

$X_t=\int_0^tY_sdM_s$ where $Y_t$ is some process that makes $\int_0^tY_sdM_s$ defined

If not, what about the case $Y_t=M_t$ ?

mdrlol
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  • What exactly do you mean by "defined"? – saz Feb 07 '16 at 19:11
  • @saz Good point.. I took it for "when the integral exists" but unlike riemann integrals we only defined stochastic integral for elementary bounded functions, which we then extended to $L^2_{loc}(M)$. It doesn't really make sense to try to apply the definition to functions which are not in $L^2_{loc}(M)$ – Ant Feb 07 '16 at 19:24
  • @Ant I agree with you. However, it might well be that the OP doesn't know about $L_{\text{loc}}^2(M)$ (but only about $L^2(M)$). – saz Feb 07 '16 at 19:34

2 Answers2

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No, in general $X_t$ is not a martingale; if $Y \in L^2_{loc}(M)$ then you can say that $X_t$ is a local martingale though, and then try to see if it's indeed a true martingale using a number of different criteria. In general it's not true though.

If $Y_t = M_t$ then it's not guaranteed, either;

since $$d(M_t^2/2) = M_t dM_t + \frac 12d[M]_t$$ you get that

$$\int_0^t M_sdM_s = \frac 12M^2_t - \frac 12[M]_t$$

So $\int_0^t M_sdM_s$ is a martingale if and only if $M_t^2 - [M]_t$ is. But the latter in general is not a martingale, only a local martingale. It is a martingale if some conditions are met, for example $\sup_{0 \le s \le t} |M_s| \in L^2$.

If $Y \in L^2(M)$ though, then $\int YdM$ is a martingale (it is even in $\mathcal M^2_0$). Or if $M$ is a martingale in $\mathcal M^2_0$ and $Y$ is predictable and bounded, then $\int YdM$ is again a martingale is $\mathcal M^2_0$

Ant
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  • Thank you for your answer, it happens that I know that $M_t=\int_0^t f(s)dW_s$ where $W$ is a standard brownian process and $f \in L^2(0,T)$ can I say that $\int_0^t M_sdM_s$ is a martingale ? – mdrlol Feb 07 '16 at 20:16
  • @mdrlol You mean $f \in L^2(0, T)$ for every $T$? – Ant Feb 07 '16 at 20:18
  • Yes $ f \in L^2(0, T)$ for every T – mdrlol Feb 07 '16 at 20:20
  • @mdrlol I still don't think you can say that. That implies that $f \in L^2_{loc}(W)$ so the integral (that is, $M$) would be in $\mathcal M^2_{0, loc}$, and you know that it's a martingale. But from this I don't know if you can conclude that $M \in \mathcal M^2_0$, which would imply that $\int M dM$ is indeed a martingale (and also in $\mathcal M^2_0$) – Ant Feb 07 '16 at 20:23
  • @mdrlol If you instead know that $\int_0^\infty f^2(s)ds < \infty$, then $f \in L^2(W)$, then $M = \int f dW$ is in $\mathcal M^2_0$, then $\int MdM \in \mathcal M^2_0$ and in particular it's a martingale – Ant Feb 07 '16 at 20:40
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For $M$ of the form $M_t=\int_0^t f(s)\,dW_s$ with $f$ locally square integrable, the (local martingale) $K_t:=\int_0^t M_s\,dM_s$ satisfies $$ E[K_t^2]=\int_0^t E[M_s^2f(s)^2]\,ds\le E[M_t^2]\int_0^t f(s)^2\,ds=\left(E[M_t^2]\right)^2<\infty, $$ and $K$ is indeed a martingale.

John Dawkins
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