Conditional Expectation involving Uniform Random Variables. [duplicate]

$X,Y \stackrel{i.i.d}{\sim} U[0,1]$. I am supposed to find $E[X|X>Y]$.

My approach was:

Let $X>Y$ be denoted as event $A$ for the time being.

$f(A)=\int_{0}^{1}\int_{0}^{x}1 dydx = 0.5$

Now, $E(X|A) = \frac{\int_{0}^{1} x f (X\cap A)}{f(A)} dx$ = $\frac{2}{3}$

How do I Approach this problem using the Indicator function?

Solution 1:

You... basically did so.$$\begin{align}\mathsf E(X\mid X>Y)&=\dfrac{\mathsf E(X\,\mathbf 1_{X>Y})}{\mathsf E(\mathbf 1_{X>Y})}\\[1ex] &=\dfrac{\displaystyle\iint_{[0,1]^2} x\,\mathbf 1_{x>y}\, \mathrm d y\,\mathrm d x}{\displaystyle\iint_{[0,1]^2}\mathbf 1_{x>y}\,\mathrm d y\,\mathrm d x}\\[1ex] &=\dfrac{\displaystyle\int_0^1\int_0^x x\,\mathrm d y\,\mathrm d x}{\displaystyle\int_0^1\int_0^x \,\mathrm d y\,\mathrm d x}\\[1ex]&=\dfrac{1^3/3}{1^2/2}\\[1ex]&=\dfrac{2}{3}\end{align}$$