Conditional expectation disjoint events

Question

Let $X$ be a random variable and $A$ and $B$ be two mutually exclusive events, is it true that:

$$E(X \mid A \cup B) = \frac{P(A)E(X|A) + P(B)E(X|B)}{P(A \cup B)} $$ If so, how can I prove it?

Answer 1

$\newcommand{\one}{\mathbf{1}}$

\begin{align*} E(X \,|\, A \cup B) &= \frac{E(X \cdot \one_{A\cup B} )}{P(A\cup B)} & \text{by definition}\\ &=\frac{E(X[\one_A + \one_B])}{P(A \cup B)} & \text{by mutual exclusivity}\\ &=\frac{E(X\cdot\one_A) + E(X \cdot \one_B)}{P(A \cup B)} &\text{by linearity of expectation} \\ &=\frac{P(A)E(X \,|\,A) + P(B)E(X \,|\,B)}{P(A \cup B)} \end{align*}

Answer 2

It is only true when it happen to be that $A\cap X$ and $B\cap (X=x)$ are disjoint (ie $A\cap B\cap (X=x)$ is a null set) for all values $x$ in the support.

To show this, use the definition of conditional probability and the principle of inclusion and exclusion.

ie $\mathsf P(U\cup V)=\mathsf P(U)+\mathsf P(V)-\mathsf P(U\cap V)$

(Where mutual exclusion will make the last term vanish.)