I was reading on conditional expectation online when I came to this curious passage:
I can easily understand that $\mathbb E[X|Y]$ can be seen as a function of $Y$: for any $\omega\in\Omega$ in the underlying sample space, the value of $\mathbb E[X|Y](\omega)$ is obtained from $Y$ through $$\mathbb E[X|Y](\omega)=\mathbb E[X|Y(\omega)].$$
What baffles me though is the claim that if $Y$ is a simple random variable, i.e., can be written as $$Y=\sum_{i=1}^n a_i\chi_{A_i},$$ where $a_1,\ldots,a_n$ are constants and $A_1,\ldots,A_n$ are measurable and disjoint, then $\mathbb E[X|Y]$ is a polynomial function of $Y$. In the simplest case when $Y=x1_A$ for some measurable set $A$ and constant $X$, one easily shows that $$\mathbb E[X|Y]=\mathbb E[X|Y=0]+\left(\frac{\mathbb E[X|Y=x]-\mathbb E[X|Y=0]}x\right)Y.$$
However, as soon as $Y$ can take up to three distinct values (including $0$), I have absolutely no idea how to proceed. Any hint would be appreciated.
I suppose in the case of $n=3$ you could write it as \begin{align*} &\mathbb{E}[X \mid Y]\\ &= \mathbb{E}[X \mid Y=a_1] \frac{(Y-a_2)(Y-a_3)}{(a_1-a_2)(a_1-a_3)}+\mathbb{E}[X \mid Y=a_2] \frac{(Y-a_1)(Y-a_3)}{(a_2-a_1)(a_2-a_3)}\\ &\qquad+\mathbb{E}[X \mid Y=a_3] \frac{(Y-a_1)(Y-a_2)}{(a_3-a_1)(a_3-a_2)}, \end{align*} and you can use the same pattern for higher $n$.
I don't really see any advantage in writing it in this way, though.