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?
Best Answer
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.