Jtdylktuy

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,

$$ E(X mid X < Y) = intlimits_0^y x f_{X mid Y} (x mid y) d x = intlimits_0^y x dx = frac{y^2}{2} $$

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?

Concerning your mistake see the comment of StubbornAtom.

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

$$frac{1}{2}mathbb{E}left(Xmid X<Yright)=mathbb{E}left(Xmid X<Yright)Pleft(X<Yright)=mathbb{E}Xmathbf{1}_{X<Y}=int_{0}^{1}int_{0}^{y}xdxdy=$$$$int_{0}^{1}frac{1}{2}y^{2}dy=left[frac{1}{6}y^{3}right]_{0}^{1}=frac{1}{6}$$

So the correct answer is: $$mathbb{E}left(Xmid X<Yright)=frac{1}{3}$$

We have $$E(Xmid X<Y)=frac{E(Xmathbf1_{X<Y})}{P(X<Y)}=2E(Xmathbf1_{X<Y})$$

And

begin{align}
E(Xmathbf1_{X<Y})&=iint xmathbf1_{x<y}f_{X,Y}(x,y),dx,dy
end{align}