Standard Brownian Motion Actex Manual Example 2.1.2

Question

Im having difficulty regarding the properties of standard Brownian motion. Can someone briefly describe the 4 properties?

And also, it say $\Bbb E[Z(2)\,Z(4)]$ :-

1. $\Bbb E[Z(2)\,Z(4)\mid Z(2)]$
2. $Z(2)\,\Bbb E[Z(4)\mid Z(2)]$
3. $Z(2) \times Z(2)$
4. $Z^2(2)$

And then he goes calculating the value by double expectation. Can someone please explain the steps 1 and 2?

Thank you!

Answer 1

The properties that model a Wiener process are:

• $Z(0)=0$
• $Z(t)$ is almost everywhere contrinuous.
• $Z(t)$ occurs in independent increments.
• $Z(t)-Z(s)\sim \mathcal N(0, t-s)$ for all $0\leqslant s < t)$

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So you are looking at:

$\begin{align}\Bbb E[Z(2)\,Z(4)] \tag 1 & = \Bbb E[~\Bbb E[Z(2)\,Z(4)\mid Z(2)]~] \\[1ex] \tag 2 & =\Bbb E[~ Z(2)\,\Bbb E[Z(4)\mid Z(2)]~] \\[1ex] \tag 3 & = \Bbb E[~ Z(2) \times Z(2)~] \\[1ex] \tag 4 & = \Bbb E[~Z^2(2)~]\end{align}$

Step $1$ is the Law of Total Epectation. What you have listed is the interior part of the nested expectation: $$\Bbb E[Z(2)\,Z(4)\mid Z(2)]$$

Step $2$ is that $\mathsf E(XY\mid X)=X~\mathsf E(Y\mid X)$, since when measured against $X$, a factor of $X$ itself can be distributed out of the operator.   So: $$Z(2)\,\Bbb E[Z(4)\mid Z(2)]$$

Step $3$ is just evaluating the conditional expectation; $\Bbb E[Z(4)\mid Z(2)]= Z(2)$.

$$\begin{align} &\qquad \Bbb E[Z(4)\mid Z(2)] \\[1ex] &=~ \Bbb E[Z(2)\mid Z(2)]+\Bbb E[Z(4){-}Z(2)\mid Z(2)] &&\text{Linearity of Expectation} \\[1ex] &=~ Z(2)+\Bbb E[Z(4){-}Z(2)\mid Z(2)] && \text{Since }\mathsf E(X\mid X)=X \\[1ex] &=~ Z(2)+ \Bbb E[Z(4){-}Z(2)] && \text{Wiener process occurs in independent increments}\\[1ex] &= ~ Z(2)+0 &&\text{Because } Z(4){-}Z(2)\sim \mathcal N(0, 2) \end{align}$$

Step $4$ is just multiplying the factors.$$Z^2(2)$$

Then take the expectation of that to conclude that $$\Bbb E[Z(2)\;Z(4)] = \mathbb E[Z^2(2)]$$

From here recall again that $Z(2)\sim\mathcal N(0;2)$ so therefore: $\mathbb E[Z^2(2)]= 2$