$\sigma$-field and Measurable

We have understood the $\sigma$-field in the last chapter, then we could connect the conditional expectation with it.

The conditional expectation $E(X|Y)$ can actually be understood as a r.v. constructed from a collection $\sigma(Y)$ of subsets of $\Omega$, which provides us the information about the structure of random variable $Y(\omega), \omega \in \Omega$. Write $E(X|Y) = E(X|\sigma(Y))$.

Then we could define the Borel sets:

Take $\Omega = \mathbb{R}$, and

$\mathscr{C}^{(1)} =\{(a,b]:-\infty < a < b < \infty \}$. The $\sigma$-field $\mathscr{B}_1 =\sigma(\mathscr{C}^{(1)})$

contains every general subsets of $\mathbb{R}$. It is called the Borel $\sigma$-field.

We could understand this by defining the $\sigma$-field contains all the information about $Y$, so we could use $\sigma$-field as the condition.

For example.

$$\left. \begin{aligned} &E(X|Y=y_i \ or \ y_j) \\ =& E(X|A_i \bigcup A_j)\\ =&E(X|Y \neq y_i)\\ =&E(X|A_i^c) \end{aligned} \right\} \sigma-field \Longrightarrow all \ information$$

We know $A_i = \{\omega: \mathscr{F}(\omega) \in B \}$, so $A_i \in \sigma(\mathscr{F})$.

Note:$E(X|Y) = E(X|\sigma(Y))$, because $A_i \in \sigma(Y)$

$\mathscr{F}$-measurable

The measurable means.

$$\{\omega: \xi(\omega) \in B \} \in \mathscr{F} \quad (\xi^{-1} \in \mathscr{F})$$

For example, we could define $X$. $$X = I_{\omega_i,\omega_{\lambda \geq 3 }} = \begin{cases} 1, if \ \omega_i \in \{3,4,5\}\\ 0, if \ \omega_i \in \{1,2\} \end{cases}$$

$$\sigma(X) = \{\emptyset , \{3,4,5\},\{1,2\},\Omega \} = \mathscr{F}_2$$

We all know $X-\mathscr{F}_2$ measurable, so $X-\sigma(X)$ measurable too.

We can see the $\sigma(X)$ contains all the information. So $E(X|Y) = E(X|\sigma(Y))$.

So here are some rules:

• $E(c_1X_1+c_2X_2|\mathscr{F})=c_1E(X_1|\mathscr{F})+c_2E(X_2|\mathscr(F))$;

• $E(E(X|\mathscr{F})) = E(X)$;

• If $X$ and the $\sigma$-field are independent, then $E(X|\mathscr{F})=E(X)$;

• If $\mathscr{F}$ and $\mathscr{G}$ are two $\sigma$-field with $\mathscr{G} \subset \mathscr{F}$, then

$$E(E(X|\mathscr{F})|\mathscr{G}) = E(X|\mathscr{G})$$

$$E(E(X|\mathscr{G})|\mathscr{F}) = E(X|\mathscr{G})$$