Would this statistical assertion be true or false? I've tried to understand this, but can't figure it out. Any help would be appreciated! :)
If $Y = 1_{[X ∈ A]}$, $Z = 1_{[X ∈ B]}$, with $A\cap B =\varnothing$, then $\mathbb EY+\mathbb EZ = \mathbb P[X \in A\cup B]$.
 A: If $X$ is an indicator function, then it's expectation is the size of the set where it is positive. Ie, if $X = 1_{Y\in A}$, then
\begin{equation}
E(X) = 1\cdot P(X=1) + 0\cdot P(X=0) = P(X=1) = P(Y\in A).
\end{equation}
Use this to prove that your assertion is true.
A: Realize that: 
$$Y(\omega)=1\iff \omega\in\{X\in A\}\iff X(\omega)\in A$$ and: $$Y(\omega)=0\iff\omega\notin\{X\in A\}\iff X(\omega)\notin A$$ so that: $$\mathbb EY=\mathbb P\{X\in A\}.1+\mathbb P\{X\notin A\}.0=\mathbb P\{X\in A\}$$
Likewise you find: $$\mathbb EZ=\mathbb P\{X\in B\}$$
Next to that we have:
$$A\cap B=\varnothing\implies\{X\in A\}\cap\{X\in B\}=\varnothing$$
So that: $$\mathbb P\{X\in A\}+\mathbb P\{X\in B\}=\mathbb P\{X\in A\cup B\}$$
A: The notation $Y = 1_{[X ∈ A]}$ means
$$ Y = \begin{cases}
 1 & \mbox{if}\quad X \in A \\
 0 & \mbox{if}\quad X \not\in A \\
\end{cases}$$
Therefore $\mathbb EY = \mathbb P(Y=1) = \mathbb P(X\in A).$
In a similar way we can show that $\mathbb EZ = \mathbb P(Z=1) = \mathbb P(X\in B).$
In general for any events $A$ and $B$, we know that
$$
\mathbb P(X\in A \cup B) =
\mathbb P(X \in A) + \mathbb P(X \in B) - \mathbb P(X \in A \cap B).
$$
But in this particular question we know that 
$A\cap B =\varnothing$,
and therefore $\mathbb P(X \in A \cap B) = 0.$
So
$$\mathbb P(X\in A \cup B) = \mathbb P(X \in A) + \mathbb P(X \in B) 
 = \mathbb EY + \mathbb EZ.$$
A: Define events C and D s.t.
$Y = 1_C, Z = 1_D$
E(Y) = P(C)
E(Z) = P(D)
P(C) + P(D) = P(C union D) since C and D are disjoint...I think?
