If $X$ and $Y$ are independent $U~(-1,1)$ distributed random varibles. What would be $E(X|Z=aX+Y)$ where $a$ is a constant?
I start to solve this by $$f(x|Z)=\frac{f(Z|x)\cdot f(x)}{f(Z)}$$ where $f(Z|x) =$ \begin{cases} \frac{1}{2a}, & \text{$(-a-1,a+1)$} \\ 0, & \text{otherwise} \end{cases}
But I did not figure out $f(Z)$. And is the expectation still a linear function of $Z$?
The notation is slightly sloppy. Let $Z=a X + Y$
Then $$f_{X|Z}(x|z)=\frac{f_{Z|X}(z|x) f_X(x)}{f_Z(z)} \tag{1}$$
Now $Z$ conditioned on $X=x$ (I don't know why you conditioned on $Y$) is uniform on $[-1+ax,1+ax]$, then
$$f_{Z|X}(z|x)= \frac{1}{2} [-1+ax\le z \le 1+ax] \tag{2}$$
where I'm using Iverson brackets notation.
Then the numerator on $(1)$ is
$$ f_{X,Z}(x,z) = \frac{1}{4} [-1+ax\le z \le 1+ax] [-1 \le x \le 1] \tag{3} $$
Let's assume first $a>0$, and fix $z$. Then the range for $x$ is $\max( \frac{z-1}{a},-1)\le x \le \min( \frac{z+1}{a},1)$. And because, for a given $z$, $f_{X|Z}(x|z)$ is uniform inside its support, $E[X|Z]$ is in the middle of its support interval. Then
$$E[X|Z] = \frac{\max( \frac{Z-1}{a},-1)+\min( \frac{Z+1}{a},1)}{2} \tag{4}$$
I'll leave it for you to get the formula for $a<0$ and $a=0$, and perhaps to express it in a simpler form, partitioning the $z$ range.
In the particular case $a=1$ this gives the (particularly simple) result $E[X|Z]=Z/2$, which is a well known result that can be obtained by simpler methods (eg).
Notice that $E[X|Z]$, in general, is not linear on $Z$, but only piecewise linear.