I have a simple question for the wiener process.
In the book: options, futures and other derivatives by Hull there is a chapter about wiener processes.
There it states, that a wiener process has 2 properties:
1) When ϵ is a standard normal distribution ϕ(0 , 1) (mean zero and variance of 1) and a small change in z during a small period of time is given by:
∆z=ϵ√∆t
2) The values of Δz for two different short time intervals are independent.
My question then is:
How do I get the variance for ∆z?
I know the answer from the book is ∆t, but how is this calculated?
My problem is, that I really don't know how to calculate it, when there is a distribution function in the equation.
We have the following result:
If $a$ is a constant, and $X$ is a random variable, then $$E\{aX\}=aE\{X\}$$ and $$Var\{aX\}=a^{2}Var\{X\}.$$
Given that $\Delta z = \varepsilon \sqrt{\Delta t}$, where $\varepsilon \sim N(0,1)$.
That is, $E\{\varepsilon\}=0$ and $Var\{\varepsilon\}=1$.
The distribution of $\Delta z$ also becomes normal with
$$E\{\Delta z\} = \sqrt{\Delta t}\cdot E\{\varepsilon\}=\sqrt{\Delta t}\cdot 0 = 0,$$ and
$$Var\{\Delta z\} = (\sqrt{\Delta t})^{2} Var\{ \varepsilon\}=\Delta t\cdot 1 = \Delta t$$.