# Is the following a Wiener process?

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 $b=\sqrt{s}$.
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 $W(t+s) - W(s) = Z(\sqrt{t+s}) -Z(\sqrt{s}) = Z(\sqrt{t+s}-\sqrt{s})$ – user1905552 Jan 26 '13 at 11:46
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! – user1905552 Jan 26 '13 at 11:54
To turn the function $W(t)$ into a Wiener process could we state $W(t) = Z\sqrt{h}$ where $t_j = jh$ – user1905552 Jan 26 '13 at 12:09