This is a worked example on Wiener processes.

Question: Pick a normally distributed random variable $Z \sim N(0,1)$, then define $W(t) = Z\sqrt{t}$. Is $W(t)$ a Wiener process?


  1. It is continuous.
  2. $W(0) = 0$.

Therefore two required properties are satisfied.

However, $W(t+s) - W(s) = Z(\sqrt{t+s}-\sqrt{s})$; which has variance $(\sqrt{t+s}-\sqrt{s})^2$ so it is not a Wiener process as the incremental change in such a process should be $Z \sim N(0,t)$ also.

I don't understand how the example arrived at $(\sqrt{t+s}-\sqrt{s})$ as the number evaluated within the normal distribution (why the square root?) nor how the variance of the increment is $(\sqrt{t+s}-\sqrt{s})^2$. I think I am missing some knowledge about manipulating $t + s$ and $t$. Can someone kindly break down the manipulation into simple steps, I would be so grateful! Or otherwise explain how to arrive at the conclusion based on the third property? Yeah, I know I am not the best at maths!


Since $W(t+s)=aZ$ and $W(s)=bZ$ for some $a$ and $b$, $W(t+s)-W(s)=(a-b)Z$. Since the variance of $Z$ is $1$, the variance of $(a-b)Z$ is $(a-b)^2$.

Now, set $a=\sqrt{t+s}$ and $b=\sqrt{s}$.

  • $\begingroup$ Oh we are just using the formula $W(t) = Z\sqrt{t}$. $Z$ is a random variable so we are multiplying by the bracketed expression. Would it be correct to state $W(t+s) - W(s) = Z(\sqrt{t+s}) -Z(\sqrt{s}) = Z(\sqrt{t+s}-\sqrt{s})$ $\endgroup$ – user1905552 Jan 26 '13 at 11:46
  • $\begingroup$ With regard to the variance would it also be correct to state $var((a+b)Z) = (a+b)^2var(z)$? My error was not to realise in the notation that $Z$ is just like any other variable! $\endgroup$ – user1905552 Jan 26 '13 at 11:54
  • $\begingroup$ Yes, that would be correct (twice), and in fact this is what is in my answer, ain't it? $\endgroup$ – Did Jan 26 '13 at 11:56
  • $\begingroup$ Yes it is. Thanks so much for your help. I just wanted as much confirmation as possible due to my sketchy maths skills. $\endgroup$ – user1905552 Jan 26 '13 at 11:57
  • $\begingroup$ To turn the function $W(t)$ into a Wiener process could we state $W(t) = Z\sqrt{h} $ where $t_j = jh$ $\endgroup$ – user1905552 Jan 26 '13 at 12:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.