I am studying Stochastic Integral theory. My professor said that given $M_t$ a continuous martingale and $p\in (1,\infty)$ the stochastic integral
$$\int_0^t M_s^{2p-1} \mathrm{sgn}(M_s) \mathrm{d}M_s, \quad \quad \quad (*)$$
is always a local martingale but not necessarily a martingale.
I cannot understand why $(*)$ is not a martingale. Since $M_s^{2p-1} \mathrm{sgn}(M_s)$ is continuous, then $M_s^{2p-1} \mathrm{sgn}(M_s)$ is predicatable, and therefore $(*)$ should be a martingale. What am I confusing?
Moreover, why is $(*)$ a local martingale?
Best Answer
If a progressively measurable $(H_s)_{s\in[0,t]}$ satisfies $\int_0^t\vert H_s\vert^2\,d\langle M\rangle_s<+\infty$ almost surely, then the stochastic integral is a local martingale. If moreover $\mathbb E\left[\int_0^tH_s^2\,d\langle M\rangle_s\right]<+\infty$, then it is a martingale.
Here you would need that $\mathbb E\left[\int_0^tM_s^{4p-2}\,d\langle M\rangle_s\right]$ to conclude that it is a martingale.