If $M_t$ is a martingale, is this process a martingale too? 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 $Y_t=M_t$  ?
 A: 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 $\mathcal M^2_0$
A: 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.
