Suppose I have $f(x,y)=1$ , $-x<y<x$, $0<x<1$. Show that $E(X|y)$ is not a straight line. My approach is, I find $h(x|y)$ (in fact i got $h(x|y)$ a piecewise function). Then I find $E(X|y)$ in usual way. But I get $E(X|y)=1$, which is a straight line. But the question asked me to prove it is not a straight line. Am I misunderstanding the question?
*If possible, let me know some hint if you have successfully proved that it is not a straight line.
Just pay attention to the support.
The marginal distribution of $Y$ is $$f_Y(y) = \int_{|y|}^1 1 \mathrm{d}x = 1 - |y|,$$ for $y \in (-1, 1)$. So the conditional density of $X$ given $Y$ is $$ f_{X|Y}(x, y) = \frac{ 1} {1 - |y|},$$ for $-x < y < x$ and $0 < x < 1$. Therefore, $$E(X|y) = \int_{|y|}^1 \frac{x}{1 - |y|} \mathrm{d}x = (1 + |y|) / 2$$ which is obvious not a straight line.