For random variables $X$ and $Y$, I would like to simplify expression $E[X \, \vert \, Y = y_1 \, \lor \, Y = y_2]$. Idea is getting rid of "or" statement from condition of expectation.
Is this statement valid? \begin{aligned} E[X \, \vert \, Y = y_1 \, \lor \, Y = y_2] &= E[X \, \vert \, Y = y_1] \cdot P[Y = y_1 \, \vert \, Y = y_1 \, \lor \, Y = y_2] \\ &+ E[X \, \vert \, Y = y_2] \cdot P[Y = y_2 \, \vert \, Y = y_1 \, \lor \, Y = y_2] \end{aligned}
Yes. Basically, when $U$ and $V$ are disjoint events, with non-zero measure: $$\begin{align}\mathsf E(X\mid U\cup V)&=\dfrac{\mathsf E(X\mathbf 1_{U\cup V})}{\mathsf P(U\cup V)}\\[1ex]&=\dfrac{\mathsf E(X\mathbf 1_U)+\mathsf E(X\mathbf 1_V)}{\mathsf P(U\cup V)}\\[1ex]&=\dfrac{\mathsf E(X\mid U)\,\mathsf P(U)+\mathsf E(X\mid V)\,\mathsf P(V)}{\mathsf P(U\cup V)}\\[1ex]&=\mathsf E(X\mid U)\,\mathsf P(U\mid U\cup V)+\mathsf E(X\mid V)\,\mathsf P(V\mid U\cup V)\end{align}$$
It is just the Law of Total Expectation applied over a conditioning event. So when $y_1\neq y_2$.
$$\mathsf E(X\mid Y{\in}\{y_1,y_2\})= {\mathsf E(X\mid Y{=}y_1)\mathsf P(Y{=}y_1\mid Y{\in}\{y_1,y_2\})+\mathsf E(X\mid Y{=}y_1)\mathsf P(Y{=}y_1\mid Y{\in}\{y_1,y_2\})}$$