Can the result be 1? Finding the conditional PDF: X given Y=y. I am struggling on a concept. If given a PDF:
f(xy)=\begin{cases}y & 0<y<min{[1/x,1/(1-x)], 0<x<1}\\ 0 & otherwise \\ \end{cases}
looking a the support and finding
P(X > Y) I get the answer of being 2. I use a double integral to find it and have to do two double integrals.
Then finding fx(X) and fy(Y) where we integrate out the variables. fx(X) =from 0 to 1/x fxy(x,y)dy
$\mathbf{f}_{x}(x) = \int\limits_{x=0}^1 f_{X,Y}(x,y)  \,dy$
and then similarly for fy(y):
$\mathbf{f}_{Y}(y) = \int\limits_{x=0}^{1/x} f_{X,Y}(x,y)  \,dx+ \int\limits_{x=0}^{1/(1-x)}  \,dx$
Getting to find the conditional pdf of f(X|Y) I am not sure if I did this right. I got....
$\mathbf{f}_{x|y}(x|y)=\frac {{f}_{xy}(x,y)} {{f}_{y}(y)}= y/y = 1 $
is that possible for a conditional PDF? It is a constant just not sure if it works or not.
Sorry I am not sure how to do all the formatting for the math code. Anything helps!! Thank YOU!
 A: you are totally lost.

*

*looking at the joint support, the region where $X>Y$ is just a triangle... thus

$$\mathbb{P}[X>Y]=\int_{X>Y} f(x,y)dxdy=\int_0^1\left[\int_0^x y dy   \right]dx=\frac{1}{6}$$


*to get $f_X(x)$ by integration w.r.t. $dy$ you get

$$f_X(x)=\int_0^{1/(1-x)}y dy=\frac{1}{2(1-x)^2}\cdot\mathbb{1}_{(0;0.5)}(x)$$
$$f_X(x)=\int_0^{1/x}y dy=\frac{1}{2x^2}\cdot\mathbb{1}_{[0.5;1)}(x)$$
that is
$$f_X(x)=\frac{1}{2(1-x)^2}\cdot\mathbb{1}_{(0;0.5)}(x)+\frac{1}{2x^2}\cdot\mathbb{1}_{[0.5;1)}(x)$$


*similarly,

$$f_Y(y)=y\int_0^1 dx=y\cdot\mathbb{1}_{(0;1)}(y)$$
$$f_Y(y)=y\int_{1-1/y}^{1/y}dx=(2-y)\cdot\mathbb{1}_{[1;2)}(y)$$
that is
$$f_Y(y)=[1-|1-y|]\cdot\mathbb{1}_{(0;2)}(y)$$
(a triangular density)
4)
$$f_{X|Y}(x|y)=\frac{f_{XY}(x,y)}{f_Y(y)}=\begin{cases}
\mathbb{1}_{(0;1)}(x),  & \text{if $0<y<1$ } \\
\frac{y}{2-y}\cdot\mathbb{1}_{(1-1/y;1/y)}(x), & \text{if $1\le y<2$ }
\end{cases}$$
thus the conditional distribution is uniform.
Example: given $y=3/2$ we have $X\sim U\left(\frac{1}{3};\frac{2}{3}   \right)$
