I am given random variables x and y where $f(x,y) = 1$ for $0 <= x, y <= 1$ and 0 elsewhere. I am asked to find $E[X | Y > X]$.
Here's what I have. I know I made a mistake somewhere or am misunderstanding something so any help would be appreciated.
I get $f_{x|y}(x, y > \frac{1}{2}) = \frac{f_{xy}(x, y > x)}{f_{y}(y > x)}$.
I get $f_{xy}(x, y > x) = \int_{0}^{1} \int_{x}^{1} 1 {\,dy} {\,dx} = \int_{0}^{1} (1-x) {\,dx} = 1/2$
I also get ${f_{y}(y > x)} = \int_{x}^{1} 1 {\,dy} = 1-x$
However, that means $f_{x|y}(x, y > \frac{1}{2}) = \frac{1}{2(1-x)}$.
This can't be right since when I plug this into $E[X | Y > X] = \int_{0}^{1} x * \frac{1}{2(1-x)} {\,dx}$, this is an undefined value...so clearly I messed up. What step am I doing wrong? Or am I not even approaching the problem in the right way?
Thanks!
Not sure where $\frac{1}{2}$ is coming from. You can use the formula
$$\mathbb{E}\left(X\mid X < Y\right) = \frac{\mathbb{E}\left(X1_{\left\{X<Y\right\}}\right)}{P\left(X<Y\right)}.$$
Both the numerator and denominator can be computed using
\begin{eqnarray*} P\left(X<Y\right) &=& \int^1_0 \int^y_0 1 \,dx\, dy,\\ \mathbb{E}\left(X1_{\left\{X<Y\right\}}\right) &=& \int^1_0 \int^y_0 x \,dx\, dy. \end{eqnarray*}