Why does $ds$ integral have zero quadratic variation? Even if I have a integral of the form

$$\int X_s ds$$

where $X$ is a stochastic process? I know that a continuous process of finite variation has zero quadratic variation, but I do not see why this should be the case here. thanks for your help

  • $\begingroup$ Which part of the question is not covered by this page? $\endgroup$
    – Did
    Nov 15, 2012 at 17:33

1 Answer 1


In general if $X$ is a semimartingale and $H$ is a locally bounded predictable process, then $$ \Delta \left(\int_0^{\cdot} H_s\,\mathrm dX_s\right)_t=H_t \Delta X_t,\quad t\geq 0, $$ so if $X$ is continuous, then so is any integral with respect to $X$. Now, integration with respect to the Lebesgue measure is just integration with respect to the semimartingale $X_t=t$ (which actually is of finite variation). Since $\Delta X_t=t-t=0$ we have that this $X$ is continuous and hence the integral is as well.

The integral is also of finite variation because the following holds for the stochastic integral:

If $X$ is a semimartingale and $H$ is locally bounded and predictable, then $(\int_0^\cdot H_s\,\mathrm d X_s)_{t\geq 0}$ is also a semimartingale. If $X$ is a local martingale then so is $(\int_0^\cdot H_s\,\mathrm d X_s)_{t\geq 0}$, and lastly, if $X$ is of finite variation, then so is $(\int_0^\cdot H_s\,\mathrm d X_s)_{t\geq 0}$.

This can be seen in Jacod and Shiryaev's Limit Theorems for Stochastic Processes for example.


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