Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
Which part of the question is not covered by this page? – Did Nov 15 '12 at 17:33
up vote 0 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.