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?
Answer:
- It is continuous.
- $W(0) = 0$.
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!
Best Answer
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 $b=\sqrt{s}$.