I need some help with the following exercise:
Let $X$ and $Y$ be two independent $\exp(1)$ random variables. Let $U= \max \{X,Y\}$. Compute the conditional law of $X$ given $U = u$.
I have tried to solve the exercise above, however it's wrong since the obtained conditional density does not sum $1$.
Thanks in advance. ${{{{}}}}$
EDIT:
My try is here (in light of one of the solutions (for uniform r.v.'s) in here Conditional Distribution of uniform random variable given Order statistic):
First, I computed
\begin{align*} P(U=u\mid X=x) &= P(Y\leq x)\mathbb{1}_{\{U=x\}} + P(Y > x)\mathbb{1}_{\{U>x\}}\\ &=(1-e^{-x})\mathbb{1}_{\{U=x\}} + e^{-x}\mathbb{1}_{\{U>x\}} \end{align*}
that seems to be correct since it integrates 1. Then, given that $f_X(x)=e^{-x}\mathbb{1}_{\{x>0\}}$ and $f_U(u)=2e^{-u}(1-e^{-u})\mathbb{1}_{\{u>0\}}$, I computed $P(X=x\mid U=u)$ as follows
\begin{align*} P(X=x\mid U=u) &= \frac{P(U=u\mid X=x)f_X(x)}{f_U(u)}\\ &=\frac{e^{-x}(1-e^{-x})\mathbb{1}_{\{u=x\}} + e^{-2x}\mathbb{1}_{\{u>x\}}}{2e^{-u}(1-e^{-u})\mathbb{1}_{\{u>0\}}}\\ &=\frac{e^{-2x}}{2e^{-u}(1-e^{-u})}\mathbb{1}_{\{x<u\}} + \frac{1}{2} \mathbb{1}_{\{x=u\}}. \end{align*}
As you can see that the $1/2$ mass point in $X=u$ appears, however the first adding does not integrate $1/2$. This procedure worked for me when $X,Y\sim U(0,1)$ i.d.d r.v.'s. Where is the error?
