Conditional Expectation with respect to a Set is Integral divided by Probability

Last updated Nov 1, 2022

Let $(\Omega, \mathcal{B}, \mathbb{P})$ be a Probability Space and let $X$ be a Random Variable in $\bar{L^{1}}(\mathcal{B})$. Suppose $B \in \mathcal{B}$ so that $\mathbb{P}(B) \neq 0$. Then

$$\mathbb{E}(X|B) = \frac{1}{\mathbb{P}(B)} \int\limits_{B} X d \mathbb{P}(\omega) = \frac{\mathbb{E}(1_{B} X)}{\mathbb{P}(B)}$$

Recall that $\mathbb{E}(X|B) = \mathbb{E}(X|1_{B})(\omega)$ for any $\omega \in B$. So we are really checking that $$\mathbb{E}(X|1_{B})(\omega) = \begin{cases} \frac{\mathbb{E}(1_{B}X)}{\mathbb{P}(B)} &\text{if } \omega \in B\\ \frac{\mathbb{E}(1_{B^{C}}X)}{\mathbb{P}(B^{C})} &\text{if } \omega \in B^{C} \end{cases} = \frac{\mathbb{E}(1_{B}X)}{\mathbb{P}(B)} 1_{B} + \frac{\mathbb{E}(1_{B^{C}}X)}{\mathbb{P}(B^{C})} 1_{B^{C}}$$ All we need to do is check our candidate for $\mathbb{E}(X|1_{B})$ satisfies the definition of Conditional Expectation. That is, we must check that for all Sets in $\sigma(1_{B}) = {\emptyset, B, B^{C}, \Omega}$ we satisfy the integral criterion of Conditional Expectation. To see this is indeed the case observe that

1. $$\int\limits_{\emptyset} \Big( \frac{\mathbb{E}(1_{B}X)}{\mathbb{P}(B)} 1_{B} + \frac{\mathbb{E}(1_{B^{C}}X)}{\mathbb{P}(B^{C})} 1_{B^{C}} \Big) d \mathbb{P}(\omega) = 0 = \int\limits_{\emptyset} X d \mathbb{P}(\omega) \text{ }\checkmark$$
2. $$\begin{align*} \int\limits_{B} \Big( \frac{\mathbb{E}(1_{B}X)}{\mathbb{P}(B)} 1_{B} + \frac{\mathbb{E}(1_{B^{C}}X)}{\mathbb{P}(B^{C})} 1_{B^{C}} \Big) d \mathbb{P}(\omega) &= \int\limits \Big( \frac{\mathbb{E}(1_{B}X)}{\mathbb{P}(B)} 1_{B} 1_{B} + \frac{\mathbb{E}(1_{B^{C}}X)}{\mathbb{P}(B^{C})} 1_{B^{C}} 1_{B} \Big) d \mathbb{P}(\omega)\\\ &= \int\limits \frac{\mathbb{E}(1_{B}X)}{\mathbb{P}(B)} 1_{B} d \mathbb{P}(\omega)\\ &= \frac{\mathbb{E}(1_{B}X)}{\mathbb{P}(B)} \mathbb{P}(B)\\ &=\mathbb{E}(1_{B}X)\\ &=\int\limits_{B} X d \mathbb{P}(\omega) \checkmark \end{align*}$$
3. $B^{C}$ works almost exactly the same as $B$.
4. $$\begin{align*} \int\limits_{\Omega} \Big( \frac{\mathbb{E}(1_{B}X)}{\mathbb{P}(B)} 1_{B} + \frac{\mathbb{E}(1_{B^{C}}X)}{\mathbb{P}(B^{C})} 1_{B^{C}} \Big) d \mathbb{P}(\omega) &= \int\limits_{B} \frac{\mathbb{E}(1_{B}X)}{\mathbb{P}(B)} d \mathbb{P}(\omega) + \int\limits_{B^{C}} \frac{\mathbb{E}(1_{B^{C}}X)}{\mathbb{P}(B^{C})} d \mathbb{P}(\omega) \\ &=\int\limits_{B} X d \mathbb{P}(\omega) + \int\limits_{B^{C}} X d \mathbb{P}(\omega) &\text{by (2) and (3)}\\ &= \int\limits_{\Omega} X d \mathbb{P}(\omega) \checkmark \end{align*}$$

Therefore our candidate for $\mathbb{E}(X|1_{B})$ is indeed the Conditional Expectation and we have (with $\omega \in B$) $$\mathbb{E}(X|B) = \mathbb{E}(X|1_{B})(\omega) = \frac{\mathbb{E}(1_{B}X)}{\mathbb{P}(B)}$$ $\blacksquare$

• Sigma Algebra induced by an Indicator Function
• Integration of Simple Functions
• Integration is Linear