Condtional probability of uniform random variable

Question

Let $X$ and $Y$ be iid. $U(0,1)$ random variable .Then $ E(X|X>Y) $ equal:

I know that $ E(Y|X=x) = \int_{0}^{1} y *f(y|x) *dy $

Should we find the joint distribution X and Y to solve this question

How do we solve this question?..

Answer 1

$$\begin{align}\mathsf E(X\mid X>Y)&=\dfrac {\mathsf E(X\,I_{X>Y})} {\mathsf P(X>Y)}\\[2ex]&=\dfrac {\displaystyle\int_0^{1}\int_y^{1} x\,\mathrm dx\,\mathrm dy} {1/2}\\[2ex]&=\dfrac 2 3\end{align}$$

Answer 2

Should we find the joint distribution X and Y to solve this question

As $X>Y$ the joint distribution you are looking for is uniform on the purple triangle

Say $f(x,y)=2$ in the triangle and zero otherwise.

Now you can derive the marginal distribution you want and calcualate the mean finding

$$\mathbb{E}[X|X>Y]=\frac{2}{3}$$

$$\mathbb{E}[Y|X>Y]=\frac{1}{3}$$