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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.

1 Answers 1

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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

$$Var\{\Delta z\} = (\sqrt{\Delta t})^{2} Var\{ \varepsilon\}=\Delta t\cdot 1 = \Delta t$$.