Does $E[X] = E[X|X < a]P\{X < a\} + E[X|X ≥ a]P\{X ≥ a\}$?

Question

Question. Does $E[X] = E[X|X < a]P\{X < a\} + E[X|X ≥ a]P\{X ≥ a\}$?

I would like some hint on how to start this, and mostly if this affirmation is true, whichever hint would help me a lot, I don't know how to start.

Answer 1

Yes. The partitioning of the outcome space and the linearity of expectation, along with the following definition, leads to that result.

By definition, $\mathsf E(X\,\mathbf 1_{\{X\in A\}}) = \mathsf E(X\mid X\in A)~\mathsf P\{X\in A\}$

Where $\mathbf 1_{\{X\in A\}}$ is the indicator function for the event $\{X\in A\}$. It equals $1$ when the event holds, and $0$ where it does not.

So $$\begin{align}\mathsf E(X) &=\mathsf E(X~\mathbf 1_{\{X <a\}\cup\{X\geq a\}}) \\[1ex]&= \mathsf E(X~(\mathbf 1_{\{X< a\}}+\mathbf 1_{\{X\geq a\}}))\\[1ex]&=\mathsf E(X~\mathbf 1_{\{X<a\}})+\mathsf E(X~\mathbf 1_{\{X\geq a\}})\\[1ex]&=\mathsf E(X\mid X<a)~\mathsf P\{X<a\}+\mathsf E(X\mid X\geq a)~\mathsf P\{X\geq a\}\end{align}$$