Conditional expectation with uniform variable Let $Y$ be a random variable on $[0,1]$ and let $X\sim U[0,1]$ be a uniformly distributed random variable that is independent of $Y$.
Prove that $$P(X<Y\mid Y)=Y.$$
Actually that looks kind of trivial, only use the independence of $X$ and $Y$ and then use the uniform distribution of $X$. But I am not able to write down an intermediate step in mathematical formulas, something like
$$P(X<Y\mid Y)= \dots=Y.$$
Any help available?
 A: Start with conditioning on the event $Y=y$, which is simpler to think through:
$$
P(X<Y|Y=y) = P(X<y|Y=y) \stackrel{\text{indep.}}{=}P(X<y)
= \int_{0}^{y}1\, \mathrm dx
=y.
$$
Conditioning an event $A$ on a random variable $U$ yields another random variable defined by $P(A|Y)(\omega) := P(A|Y = Y(\omega))$, therefore, by performing the same steps, we obtain:
\begin{align*}
P(X<Y|Y)(\omega)
&= P(X<Y|Y=Y(\omega)) = P(X<Y(\omega)|Y=Y(\omega)) \stackrel{\text{indep.}}{=}P(X<Y(\omega))
\\
&=
\int_{0}^{Y(\omega)}1\, \mathrm dx
=Y(\omega).
\end{align*}
Since this identity holds for every $\omega\in\Omega$, you get the desired result.
A: Consider a probability space on which $X,Y$ are independent and (on $[0,1]$)  uniformly distributed random variables. By the definition of conditional expectation one has to prove that
$$\int_A P(X<Y\mid Y)dP= \int_A YdP,$$
for any event $A$.
Because of independence and the uniform distribution:
$$P(X<Y\mid Y=y)=P(X<y)=y.$$
Also, $$P(X<Y\mid Y=y)=0$$ if $Y\not=y$.
Let $B_y$ the event when $Y=y$.
With all this, we have
$$\int_{A} P(X<Y\mid Y)dP=\int_{A\cap B_y}y dP=\int_A YdP.$$
Edit
Way simpler:
If for an elementary event $\omega  \ Y(\omega)=y$, because of independence and uniformity
$$P(X<Y\mid Y=y)=P(X<y)=y.$$
So 
$$P(X<Y\mid Y)=Y.$$
