Cumulative distribution function of $W(T)/\sqrt{T}$, where $W$ is a standard Wiener process.

Question

$W(T)$ is the standard Wiener process and $\phi(\cdot)$ is the CDF of $\mathcal N(0,1)$.

Can anyone explain how I can go from (7.9) to the last step; i.e. how does $W(T)$$/$$\sqrt(T)$ disappear? I am aware of the all the properties of the Wiener process, but can't see how those will work here.

Answer 1

By definition of Brownian motion, $W(T) \sim N(0, T)$. Hence,

$$\frac{W(T)}{\sqrt{T}} \sim N(0,1)$$

and hence the CDF of your random variable is just the CDF of the standard normal distribution $Z$, i.e.

$$\Phi (z) = P(Z\leq z) = \int_{-\infty}^z \frac{1}{\sqrt{2\pi}} \exp(-1/2 t^2) dt$$