Derivation of a property of standard Wiener processes

Question

I am reading A Standard Wiener Process and am struggling to piece together how they arrived at their conclusion.

The major properties of any Wiener Process are:

1. $W(t) = 0$
2. $W(t) - W(s) \sim N(0, t-s) $
3. $W(t)$ has continuous paths
4. Increments are independent

Now in the above website I linked the authors indicate that in the Standard Wiener Process the next "step size" is given by:

$$W(t) - W(s) \sim \sqrt{t-s}N(0,1)$$

I am not sure how the authors got $\sqrt{t-s}$. Clearly this is the standard deviation of the normal distribution given in (2). My guess is that with a variance of 1, you can multiply the result of getting a value from $N(0,1)$ by the standard deviation and get the same number you would have got by sampling from $N(0, t-s)$. I have never seen this done before, however. Am I correct in my assumption here?

Answer 1

Yes, that's correct.

If $X \sim N(0,\sigma^2)$ is Gaussian with mean $0$ and variance $\sigma^2$, then the random variable $Y:= X/\sigma$ is also Gaussian and

$$\mathbb{E}(Y) = \frac{1}{\sigma} \mathbb{E}(X)=0$$

and

$$\mathbb{E}(Y) = \frac{1}{\sigma^2} \mathbb{E}(X^2)=1.$$

Hence, $Y \sim N(0,1)$. This shows

$$\underbrace{X}_{\sim N(0,\sigma^2)} = \sigma \underbrace{Y}_{\sim N(0,1)}$$

and this identity can be more compactly written as $N(0,\sigma^2) = \sigma N(0,1)$.