Find conditional expectation

Question

Let $X$ and $Y$ be two independent uniform on $(0,1)$ . Compute $E(X \mid X < Y )$.

I was asked this question on an exam and here is how I solved it:

We have $f_X(x)=f_Y(y)=1$ and by independence $f_{X,Y}(x,y)=1$. Hence,

$$ f_{X \mid Y} (x \mid y) = \frac{ f_{X,Y}(x,y)}{f_{Y}(y) } = 1 $$

So,

$$ E(X \mid X < Y) = \int\limits_0^y x f_{X \mid Y} (x \mid y) d x = \int\limits_0^y x dx = \frac{y^2}{2} $$

I think my answer is reasonable. I only got 2 points out of 10 on this question which means the entire procedure is flawed? what is my mistake here?

Answer 1

Concerning your mistake see the comment of StubbornAtom.

Note that $P(X<Y)=\frac12$ by symmetry, so that:

$$\frac{1}{2}\mathbb{E}\left(X\mid X<Y\right)=\mathbb{E}\left(X\mid X<Y\right)P\left(X<Y\right)=\mathbb{E}X\mathbf{1}_{X<Y}=\int_{0}^{1}\int_{0}^{y}xdxdy=$$$$\int_{0}^{1}\frac{1}{2}y^{2}dy=\left[\frac{1}{6}y^{3}\right]_{0}^{1}=\frac{1}{6}$$

So the correct answer is: $$\mathbb{E}\left(X\mid X<Y\right)=\frac{1}{3}$$

Answer 2

We have $$E(X\mid X<Y)=\frac{E(X\mathbf1_{X<Y})}{P(X<Y)}=2E(X\mathbf1_{X<Y})$$

And

\begin{align} E(X\mathbf1_{X<Y})&=\iint x\mathbf1_{x<y}f_{X,Y}(x,y)\,dx\,dy \end{align}