Assume $Y,X,Z_1,Z_2 $ are random variables, and assume that they all continuous and have density function, such that $Y=Z_1+Z_2$.
Can it be that $ \mathbb{E}[Y|X]=\mathbb{E}[Z_{1}|X]+\mathbb{E}[Z_{2}|X] $?
(Where $Y|X$ is the random variable whom density is $ f_{Y|X}\left(y,x\right):=\frac{f_{Y,X}\left(y,x\right)}{f_{X}\left(x\right)}$)
For me it seems very unlikely to be true, since conditioning is not a linear operation. But I cannot find a counter example nor prove equality.
I'd really appreciate a counter example/proof for this.
Thanks in advance.
The function $U\mapsto \mathbb E[U\mid X],$ is linear. To prove it, you have to prove that $$\mathbb E\big[\mathbb E[\alpha A+\beta B\mid X]V\big]=\mathbb E\big[(\alpha \mathbb E[A\mid X]+\beta \mathbb E[B\mid X])V\big],$$ for all $V$ being bounded and $\sigma (X)-$measurable, all $\alpha ,\beta \in \mathbb R$ and all bounded r.v. $A$ and $B$. But this is clear since if $V$ is bounded and $\sigma (X)-$measurable, $A$ and $B$ are bounded r.v. and $\alpha ,\beta \in \mathbb R$, then
\begin{align*} \mathbb E\big[\mathbb E[\alpha A+\beta B\mid X ]V\big]&=\mathbb E[(\alpha A+\beta B)V]\\ &=\alpha \mathbb E[AV]+\beta \mathbb E[BV]\\ &=\alpha \mathbb E\big[\mathbb E[A\mid X]V \big]+\beta \mathbb E\big[\mathbb E[B\mid X]V\big]\\ &=\mathbb E\big[(\alpha \mathbb E[A\mid X]+\beta \mathbb E[B\mid X])V\big], \end{align*} what prove the claim.