Find conditional distribution and expectation if $(X,Y)$ is uniform on the unit square ${0<y<1}$, ${0<x<1}$
SUPPOSE f(x,y) = 1 given that, what is $E(X | X<Y)$?
SO FAR,what I got, 
I draw (in the x-y plane) the region where f(x,y) is nonzero, then the region being conditioned on. So, the density must be 2x, for $0<x<1$ and 0 elsewhere.
$ f_{x|x<y} = \frac{f_{xy}}{f_{x<y}} = \frac{1}{1/x} = x, x \in [0,1]. $
I want to treat $(X|X<Y)$ itself as a random variable.
I'm stuck here, could please give a hint? or another way.(still don't get the point...)
 A: The region being conditioned on is the upper left half of the unit square, above the diagonal $y = x$.  There are different ways to the solution; one is to observe that $f_{X \mid X < Y}(x) = 2(1-x)$ and compute
$$
E(X \mid X < Y) = \int_{x=0}^1 xf_{X \mid X < Y}(x) \, dx
                = \int_{x=0}^1 2x(1-x) \, dx
                = \left. x^2-\frac{2x^3}{3} \right|_{x=0}^1 = \frac{1}{3}
$$
Another is to recognize that $F_{X \mid X < Y}(x) \equiv P(X < x \mid X < Y) = 2x-x^2$, and then
$$
E(X \mid X < Y) = \int_{x=0}^1 [1-F_{X \mid X < Y}(x)] \, dx
                = \int_{x=0}^1 (1-2x+x^2) \, dx
                = \left. x-x^2+\frac{x^3}{3} \right|_{x=0}^1 = \frac{1}{3}
$$
Another is to see that the centroid of the conditioned area lies at $(1/3, 2/3)$, so the average value of $X$ for that condition is $1/3$ (and the average value of $Y$ for that same condition is $2/3$).
ETA: Both Did and the OP have observed that this answer lacks a derivation of the conditional PDF. From the observation in the main post comment, above, we may obtain this PDF as
\begin{align}
f_{X \mid X<Y}(x)
   & = \frac{\int_{y=0}^1 f(x,y) \mathbb{1}_{x<y}\,dy}
            {\int_{z=0}^1 \int_{y=0}^1 f(z,y) \mathbb{1}_{z<y}\,dy\,dz} \\
   & = \frac{\int_{y=x}^1 f(x,y) \,dy}
            {\int_{z=0}^1 \int_{y=z}^1 f(z,y) \,dy\,dz} \\
   & = \frac{\int_{y=x}^1 \,dy}
            {\int_{z=0}^1 \int_{y=z}^1 \,dy\,dz} \\
   & = \frac{1-x}{\int_{z=0}^1 (1-z) \,dz} \\
   & = 2(1-x)
\end{align}
