I have a Wiener process $W(t)$, which is a normally distributed random variable with mean $\langle W(t)\rangle = \mu = 0$ and variance $\langle W(t)^2\rangle = \sigma^2 = t$. The angled brackets $\langle\ \rangle$ indicate an average over all realisations of the Wiener process.

The Wiener increment $\Delta W$ is defined as:

$\Delta W(t) = W(t + \Delta t) - W(t) $

which corresponds to the time increment $\Delta t$. The mean of $\Delta W$ is zero, since the means of both $W(t + \Delta t)$ and $ W(t) $ are zero.

I am trying to derive the variance of $\Delta W$, which is $\langle (\Delta W)^2\rangle$. So far I have:

$(\Delta W)^2 = W(t + \Delta t)^2 + W(t)^2 -2W(t + \Delta t)W(t) $

$\langle (\Delta W)^2\rangle = \langle W(t + \Delta t)^2\rangle + \langle W(t)^2\rangle -2\langle W(t + \Delta t)W(t)\rangle$

$\langle (\Delta W)^2\rangle = t + \Delta t + t -2\langle W(t + \Delta t)W(t)\rangle$

I am not sure how to evaluate the last term (or if what I have done so far is correct) and would appreciate some help. I have been told that the correct answer is $\Delta t$ and am trying to verify this. Many thanks!


1 Answer 1


So you can take a shortcut if you know that the Wiener process has independent increments. This is stronger than merely being Markov, because it follows that the distribution of the increments starting from a time $t$ do not depend even on $W_t$. Once you know that (which in some definitions of the Wiener process is built into the definition), it follows that the distribution and thus moments of $\Delta W$ do not depend on $W$ at the initial time. Thus in particular moments of $W_t - W_s$ for $t \geq s$ are just the same moments of $W_{t-s}$.

But the alternative is to show that the covariance function of the Wiener process is $E[W_t W_s] = \min \{ t,s \}$. Exactly how to do this will depend on what you already have available to you to use.

  • $\begingroup$ Thanks, that was helpful. I don't really understand the second part, but the first part of your answer is a much clearer way of thinking about the problem and has cleared up my problem. $\endgroup$
    – asph
    Dec 9, 2018 at 15:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .