I thought that I understand conditional probability and expectation until I saw this question:
The problem for conditional expectation.
Basically, it is given that: $$(X,Y)\sim f(x,y)=\begin{cases} 2xy &\text{ if $0<x<2y<2$} \\ 0 &\text{ otherwise } \end{cases}$$ And it is asked to find $E[Y|X=aY]$
Background: The good
I understand Method 2, where one would write:
$E[Y|X=aY]=E\left[Y|\frac{X}{Y}=a\right]=E[Y]=\frac{4}{5}$
$\frac{X}{Y}=a$ is dropped from expectation after proving that $Y$ and $\frac{X}{Y}$ are independent using the transformation $(X,Y)\to(X/Y,Y)$
I also (kind of) understand Method 1 in the answer, with conditioning on $Y$ (although my intuition tells me the result is correct only because $Y$ and $X/Y$ are independent):
$$E[Y|X=aY]=\int_0^1E[Y|X=aY,Y=y]f_Y(y)\,dy=\int_0^1yf_Y(y)\,dy=E[Y]=\frac{4}{5}$$
But when I try my own intuitive approaches, I'm getting stuck.
Question 1: The Bad
I interpret the conditional pdf $f_{Y|X}(y|t)=Cf_{Y,X}(y,t)$ as sectioning the joint pdf surface with the plane $y=t$ and scaling the resulting curve to a pdf.
My intuition tells me that the conditional pdf of $Y$ given $X=aY$ should similarly be found by sectioning the joint pdf with the plane $x=ay$ and scaling to a pdf.
The pdf would be $Cf(ay,y)=Cay^2=3y^2, 0<y<1$, and the conditional expectation:
$E[Y|X=aY]=\int_0^13y^3\,dy=\frac{3}{4}$
What am I missing here?
Question 2: The Ugly
I'm trying now to do the same thing as in Method 1, but condition on $X$ rather than on $Y$:
$$E[Y|X=aY]=E[X/a|X=aY]=\frac{1}{a}E[X|X=aY]\\=\frac{1}{a}\int_0^aE[X|X=aY,X=x]f_X(x)\,dx=\frac{1}{a}\int_0^a xf_X(x)\,dx$$
Which is something very ugly depending on $a$, instead of $4/5$.
Again, many thanks for anybody who could point out the mistakes in my thinking.