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Let $X$ and $Y$ be independent and both have distribution $F$. Suppose that $F$ has density $f$ wrt Lebesgue measure. Let:

$Z\doteq\text{max}\{X,Y\}$

The distribution function of $Z$ is $F^{2}(z)$, and moreover $Z$ has density $f_{Z}(z)=2F(z)f(z)$, $z\in\mathbb{R}$.

What is the conditional distribution function of $X$ given $Z=z$, $F(x|z)=?$

md2perpe
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gva
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  • Changing the tag 'distribution-theory' to 'probability-distributions'. Please read the description of the tags. https://math.stackexchange.com/tags/distribution-theory/info https://math.stackexchange.com/tags/probability-distributions/info – md2perpe May 01 '20 at 15:15

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Well, first of all $X$ has $50 \%$ chance being equal to $Z$ and the remaining $50 \%$ chance of being a random variable from its distribution conditioned on being less than $Z$. So it's not continuous nor discrete - it will have PDF up to $Z$ which is a scaled version of its usual PDF in such a way that the area is $\frac{1}{2}$ and the remaining half is just devoted to being equal to $Z$.

Bartek
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  • Yes, I agree with the intuition. It will look like $F(x|z)=\dfrac{F(x)}{2F(z)}$ for $x<z$ and 1 for $x\ge z$. However, I would like to derive this in a mathematically rigorous way. – gva May 01 '20 at 11:47