Let $M=(M_t)_{t\ge 0}$ be a continuous martingale defined on some filtered probability space taking values in $\mathbb{R}$. Let $H=(H_t)_{t\ge 0}$ be some bounded progressively measurable process, i.e.
$$\sup_{t,\omega}|H_t(\omega)|<+\infty$$
My question is whether we could show the stochastic integral
$$\int_0^{\cdot}H_udM_u$$
is a martingale? I stronly believe that the answer is no, but I can't find a counterexample. Does someone know the result? Many thanks for the reply!
Best Answer
It is a local martingale, by definition, with quadratic variation given by $$ Q_t = \int_0^t H_u^2 d< M >_u. $$ Now, $Q_T$ is upper bounded by an almost surely finite random variable times $<M>_T$. So that the expectation of the quadratic variation is finite, because $M$ is a martingale, and hence the stochastic integral is a martingale.