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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!

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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.

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    $\begingroup$ Why should the expectation of the quadratic variation be finite? I thought this is true if and only if the martingale is square integrable? $\endgroup$
    – Stephan Sturm
    Commented Feb 3, 2015 at 0:58
  • $\begingroup$ I have the same question, why the expectation of the quadratic variation is finite? $\endgroup$
    – CodeGolf
    Commented Feb 7, 2015 at 21:33

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