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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 $Z$?

Did
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Quint Z
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  • Welcome to MSE, Quint. I don't understand why you write $f(ax+y|y) $ instead of $f(ax+y|x)$, and further, the expression is wrong, because it does not depend on $y$. Also, it makes little sense to ask if the expectation is "a linear function of $X$ and $Y$": a conditional expectation is funcion of the conditioning variable (here $Z=aX+Y$). Apart from this, if you are just interested in the expectation, it might be easier to find it graphically - it should be easy to see that it's not linear (on $Z$) but piecewise linear. – leonbloy Sep 29 '18 at 03:54

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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.

leonbloy
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  • Thanks! I've changed the questions and I hope it looks better now. I was stuck on $f_Z(z)$ and forgot that $f_X|Z(x|z)$ is also uniform distributed. But could you also illustrate a bit more on how to get $f_Z(z)$ in this case? – Quint Z Sep 29 '18 at 14:27
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    $f_Z(z)$ can be obtained simply by integrating $(3)$. But it's not necessary. $f_Z(z)$ is (for a given $z$) a normalization constant. Once you got (in $(3)$) that the numerator is constant inside some interval (say $[b,c]$, where both extremes depend on $z$), then you already know $f_{X|Z}(x|z) =\frac{1}{c-b} [b \le x \le c] $ – leonbloy Sep 29 '18 at 14:45