Conditional Expectation

As we all know, Conditional probability of event $A$ given $B$:

$$P(A|B) = \frac{P(A \bigcap B)}{P(B)}$$

Conditional distribution function of r.v. $X$ given $B$:

$$F_X(x|B) = \frac{P(X \leq x, B)}{P(B)}, x \in \mathbb{R}$$

Conditional expectation of r.v. X given B:

$$E(X|B) = \frac{E(XI_B)}{P(B)} = \begin{cases} \sum_{k=1}^{\infty}x_{k}P(X=x_{k}|B), X \ is \ discrete\\ \frac{1}{P(B)} \int_B xf(x)dx, X \ is \ continuous \end{cases}$$

where $I_B(\omega)$ is the indicator function of the event $B$.

We could get $E(X|Y)(\omega) = E(X_i|A_i) = E(X|Y=y_i)$ according to the figure in the last chapter, we know the $X$ is the function of $A_i$, so this is easy to understand.

Then we could approve $E(E(X|Y)) = E(X)$:

$$\begin {aligned} E(E(X|Y))&=\sum_{i=1}^{\infty} E(X|Y=y_i) \times P(Y=y_i) \\ &=\sum_{i=1}^{\infty} E(X|A_i) \times P(A_i) \\ &=\sum_{i=1}^{\infty} E(X \times I_{A_i}) \end {aligned}$$

We know the $I$ is :

$$I_{A_i} = \begin{cases} 1, if \ \omega \in A_i\\ 0, other \end{cases}$$

So we get $\sum_{i=1}^{\infty}I_{A_i} = 1$, let's go on the approvement.

$$\begin {aligned} E(E(X|Y))&=\sum_{i=1}^{\infty} E(X \times I_{A_i})\\ &= E(X(\sum_{i=1}^{\infty}I_{A_i}))\\ &=E(X) \end {aligned}$$

• Now we have got $E(E(X|Y)) = E(X)$.