Is the following a Wiener process?

Question

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:

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!

Answer 1

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 $b=\sqrt{s}$.

Answer 2

I know this is a very old question but I recently came upon this very same example and I don't agree with the discussion above, although I agree with the conclusion that W(t) is not a Weiner process.

Assuming two random variables X,Y are independent we have these facts: Var (X + Y) = Var(X) + Var(Y), and also: Var (X - Y) = Var (X) + Var(Y) Also if a is a constant, then Var(aX) = a^2*Var(X)

We also know that a normal random variable Z multiplied by a constant "a" is another normal random variable with variance a^2*Var(Z).

According to these facts, we can determine: Each of W(t + s) and W(s) are normal random variables with variances of (sqrt(t + s))^2 = t + s and (sqrt(s))^2 = s, respectively.

In other words, Var( W(t + s) ) = t + s, and Var ( W(t) ) = t

Next we know that Var ( W(t + s) - W(s) ) = Var (W(t + s)) + Var (W(s)) = t + s + s = 2s + t

So W(t) is not a Weiner process because its variance is "2s + t", not because its variance is (sqrt(t + s) - sqrt(s))^2.