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Is the following correct and/or a (simple) known result?

Let $X$ be a local martingale and $H$ an integrand for $X$, such that the stochastic integral $\int H\cdot dX\ge x$ for some random variable. Then $\int H\cdot dX$ is also a local martingale.


I would prove it in analogy to the proof in Corollaire 3.5 of this paper:

Define the process $V_t=\int^t_0 H\cdot dX-x\ge 0$ and the sequence of stopping times $T_n=\inf\{t:V_t\ge n\}$. Thus, $(V^{T_n})_{t-}\le n$ which together with $V_t\ge 0$ yields $\Delta V^{T_n}=\Delta\left(\int H\cdot d(X^{T_n})\right)\ge -n$. The assertion then follows from Proposition 3.3 of the same paper.


Proposition 3.3: The stochastic integral $\int H\cdot dX$ is a local martingale if and only if there exists a sequence of stopping times with limit $\infty$ and a series of integrable negative random variables $\theta_n$, such that $ H\cdot\Delta X^{T_n}\ge\theta_n$.

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    $\begingroup$ Hi as soon as H is locally integrable, the stochastic integral $int H\cdot dX$ makes sense is a local martingale itself almostsure.wordpress.com/2010/03/25/… $\endgroup$
    – The Bridge
    Feb 13, 2014 at 13:11
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    $\begingroup$ @TheBridge How can I see that H is locally integrable? Which theorem/corollary of your link are you referring to? $\endgroup$
    – JSG
    Feb 13, 2014 at 14:24
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    $\begingroup$ @TheBridge and upvoters of his comment. Could you provide more insight? Thanks! $\endgroup$
    – JSG
    Feb 14, 2014 at 9:26

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