Conditional expectation with proof of linearity

Solution 1:

independence is not stated in the question and not needed to prove what you are requested to do.

$$\mathbb{E}[X+Y]=\mathbb{E}[X]+\mathbb{E}[Y]$$

is valid also if the rv's are not independent. Your proof is quite correct (I amended your $E(X,Y)$ in $E(X+Y)$) but it is not exactly what you are asked to do.

To prove your statement, including conditional probability and constants, simply use the definition finding

$$\begin{align} \mathbb{E}[aY+bZ|X] & = \sum_y\sum_z(ay+bz)p(y,z|x)\\ & = a\sum_y\sum_zyp(y,z|x)+b\sum_y\sum_zzp(y,z|x)\\ & = a\sum_y yp(y|x)+b\sum_z zp(z|x)\\ &=a\mathbb{E}[Y|X]+b\mathbb{E}[Z|X] \end{align}$$