Let $X$ denote a real-valued bounded random variable and $Y$ denote a discrete-valued random variable with $K$ states. Given
$E[X^m]= \sum_{k} c_{km} P(Y=k)$ with constants $c_{km} \in\mathbb{R}$ and $k=1,...,K$ and $m\in\mathbb{R}\cup\{0\}$.
Can we infer from $E[X^m]= \sum_k E[X^m|Y=k] P(Y=k)$ that $c_{km}=E[X^m|Y=k]$?
What confuses me is that the equality $\sum_{k} c_{km} P(Y=k)=\sum_k E[X^m|Y=k] P(Y=k)$ holds only for the sum. Does it also hold for each element in the sum $c_{km}=E[X^m|Y=k]$ and, if yes, why?
Thanks for your comments / suggestions / counter examples?
Either the conjecture is obviously false, or I am missing something obvious...
Each $c_{km}$ only appears in one equation, that of $E[X^m]$, right? So, for every $m$, you have $K$ unknowns (the $K$ coefficients) and one equation. How can the coefficients be determined?
E.g. lets say $P(Y=1) > 0$. One possible set of coefficients would be:
$$c_{1m} = {E[X^m] \over P(Y=1)}; ~~~~~~~~~~~~~~~ c_{km} = 0 ~~ \forall k \neq 1$$
Does that work? Or am I missing something really obvious...