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The problem is as follows:

Suppose we have have independent random variable $Y_1, Y_2, ... ,Y_n$, and they are uniformly distributed over the closed interval [0,1]. If $V$ and $W$ are the smallest and largest, respectively of the $Y_i$'s, what is the joint PDF of $V$ and $W$?

I am approaching this problem by first trying to find the pdf's of the marginal and the conditional, and then multiplying them to find the joint. So I am trying to find $f_W(w)$ and $f_{v|w}(v|w)$.

Finding $f_W(w)$ is pretty straightforward.

$W = \max(Y_1, Y_2, Y_3, ..., Y_n)$

$W \leq w$ iff $Y_k \leq w, \forall k$

$P(Y_k \leq w) = w$, because $Y_k$ is uniformly distributed on [0,1]

$\therefore P(W \leq w) = P(Y_1 \leq w)P(Y_2 \leq w)...P(Y_n \leq w) = w^n$

$\therefore F_w(w) = w^n, 0 \leq w \leq 1$

$\therefore f_w(w) = nw^{n-1}, 0 \leq w \leq 1$

I am unsure of how to continue from here to find $f_{v|w}(v|w)$

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    $\begingroup$ "I am approaching this problem by first trying to find the pdf's of the marginal and the conditional, and then multiplying them to find the joint" This is just complicating things for nothing. Instead, compute $$P(v<V<W<w)$$ for every $0<v<w<1$ and rejoice... $\endgroup$ – Did Sep 7 '18 at 15:01
  • $\begingroup$ Thanks for the comment. I am unsure of how I would turn that into a PDF? $\endgroup$ – Samyak Shah Sep 7 '18 at 15:13
  • $\begingroup$ "how I would turn that into a PDF?" Did you compute $P(v<V,W<w)$ and did you differentiate it twice? What did you get? $\endgroup$ – Did Sep 8 '18 at 6:29
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For the Max- and Min distribution we can derive the following two statements:

  • For the joint pdf every $\texttt{independent}$ random variable $Y_i$ must be smaller than $w$.

  • For the joint pdf every $\texttt{independent}$ random variable $Y_i$ must be larger than $v$.

That means $P(v<Y_i<w)=w-v \qquad \forall \ \ i =\{1,2,...,n \} $

Now use the independence to find the joint distribution when $n$ variables are involved.


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  • $\begingroup$ Thanks this helped me perfectly! $\endgroup$ – Samyak Shah Sep 11 '18 at 2:57
  • $\begingroup$ I´m glad to hear. You´re welcome. $\endgroup$ – callculus Sep 11 '18 at 14:03

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