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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?

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    $\begingroup$ The conditional law does not have a density, because $\Pr(X=u|U=u) = 1/2$. Did you take that into account? If you did, how did you proceed to find the rest of the solution? If we don't know what you've tried, we can't help you. $\endgroup$
    – whuber
    Jan 9 '17 at 20:42
  • $\begingroup$ Also put the self study tag on the question. $\endgroup$ Jan 9 '17 at 20:51
  • $\begingroup$ @whuber : Can you write $\Pr(X=u\mid U=u)$ instead of $\Pr(X=u|U=u)$? $\endgroup$ Jan 10 '17 at 2:16
  • $\begingroup$ @Christian Please show what you did. $\endgroup$
    – Glen_b
    Jan 10 '17 at 11:40
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    $\begingroup$ @Michael Thank you for the tip! \mid is more convenient than the \,|\, phrase I had been using to kern the character. $\endgroup$
    – whuber
    Jan 10 '17 at 15:07

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