I have a exercise, which is given as Let the random variables $X$ and $Y$ have $f(x,y)=1$, where $-x<y<x$ and $0<x<1$. I need to show that the $E(X|y)$ is not linear, where I was told that the limits of $X$ would be $|y|<x<1$, but I cannot figure out, why?
Can someone help me?
The reason that for any given fixed $y$ the range of $x$ is $|y| < x < 1$ is that if $|y| \geq x$, then either $y$ is positive and $y>x$, or $y$ is negative and $y < --x$; either of those contradicts $-x<y<x$. (The $x<1$ part of the condition is more obvious: it is needed to ensure that $0<x<1$.)
Now fix $Y$ at some value $y$. Then since $f(x,y)$ is constant on the entire $+x$ facing right isosceles triangle, $X$ is uniformly distributed on $(|y|,1)$. Thus $$ E(X|y) = \frac{|y|+1}{2} $$ That is in fact a linear function of $y$ in the domain $y > 0$ or the domain $y < 0$, but for positive $y$ the slope is positive $\frac{1}{2}$ while for negative $y$ the slope is negative $\frac{1}{2}$. So the overall function $E(X|y)$ is not linear.