How do I find this conditional expectation?

Question

$X$ and $Y$ are independent $U(0,1)$ random variables.

I have to find $E[X|X>Y]$.

I found it out by integrating $\int_0^1 \int_y^1 x dxdy$ to get $\frac{1}{3}$ as the answer but this is wrong according to the answer key. What is the mistake that I have made?

Answer 1

You have computed $EXI_{X>Y}$ instead of $(EX|X>Y)$. You should divide your answer by $P(X>Y) =\int_0^{1}\int_y^{1}dxdy$.

Answer 2

$$\mathbb{E}\left[X\mid X>Y\right]P\left(X>Y\right)=\mathbb{E}X\mathbf{1}_{X>Y}=\int_{0}^{1}\int_{0}^{x}xdydx=\int_{0}^{1}x\int_{0}^{x}dydx=\int_{0}^{1}x^{2}dx=\frac{1}{3}$$

By symmetry we find that $P\left(X>Y\right)=\frac{1}{2}$ so we end up with $\mathbb{E}\left[X\mid X>Y\right]=\frac{2}{3}$.