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Suppose $U,V$ are i.i.d. random variables following Unif$(0,1)$, what is the conditional distribution of $U$ given $Z:=\max(U,V)$ ?

I tried writing $Z=\Bbb{I}\cdot V+(1-\Bbb{I})\cdot U$ where $\Bbb{I}=\begin{cases}1&U<V\\0&U>V\end{cases}$

But I am not getting anywhere.

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    $\begingroup$ If it helps: \begin{align*}\mathbb{P}(U \leq u, Z \leq z) &= \mathbb{P}(U \leq u, U \leq z, V \leq z) \\&= \mathbb{P}(U \leq \min(u,z), V \leq z) \\&= \mathbb{P}(U \leq \min(u,z)) \, \mathbb{P}(V \leq z)\end{align*} $\endgroup$
    – Tom
    Aug 27, 2016 at 20:25
  • $\begingroup$ @Tom SO, $$=\min (u,z)\cdot z$$? $\endgroup$
    – Qwerty
    Aug 27, 2016 at 21:26
  • $\begingroup$ For $u, z \in (0,1)$ this seems correct to me, unless I'm missing something. This isn't the conditional distribution, of course, just the joint CDF. $\endgroup$
    – Tom
    Aug 27, 2016 at 21:27
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    $\begingroup$ The conditional distribution of $U$ conditionally on $Z=z$ is the measure $\frac12\mathrm{Unif}(0,z)+\frac12\delta_z$, that is, $P(U=z\mid Z=z)=\frac12$ and, for every $u$ in $(0,z)$, $P(U\leqslant u\mid Z=z)=\frac{u}{2z}$. $\endgroup$
    – Did
    Aug 28, 2016 at 18:50
  • $\begingroup$ Cross-posted at stats.stackexchange.com/questions/232085/…. $\endgroup$ Jan 4, 2020 at 6:50

1 Answer 1

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Conditional CDF of $U$ given $Z$ is:

\begin{eqnarray*} F_{U|Z}(u|z) & = & \Pr(U\leq u|Z =z) \\ &= & \Pr(U\leq u|\max(U, V) =z) \\ &= & \Pr(U\leq u, U \leq V|\max(U, V) =z) + \Pr(U\leq u, U > V|\max(U, V) =z) \\ &=& \Pr(U\leq u|U\leq V, \max(U, V) =z)\Pr(U\leq V|\max(U, V) =z) \\ && + \Pr(U\leq u|U > V, \max(U, V) =z)\Pr(U > V|\max(U, V) =z) \\ &=& \Pr(U\leq u|U\leq V, V =z)\times \frac{1}{2} + \Pr(U\leq u|U > V, U =z)\times \frac{1}{2} \\ &=& \begin{cases} 0 & if \ u\leq 0 \\ \displaystyle\frac{u}{2z} & if \ 0 < u < z < 1 \\ 1 & if \ u\geq z \end{cases}\end{eqnarray*}

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