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Let $X_1, \ldots, X_n$ a random sample of a Uniform(0,1), I want to show which the joint distribution of $(X_1,X_{(n)})$ is. I do the following:

$$ P(X_1\leq x, X_{(n)}\leq y)=P(X_1\leq x, X_1\leq y, \ldots , X_n\leq y)=$$ $$ P(X_1\leq \min(x,y), X_2\leq y, \ldots, X_n\leq y)=P((X_1\leq \min(x,y))P(X_2\leq y)\ldots P( X_n\leq y)=$$ $$ = \min(x,y)y^{n-1}$$

When I compute the density function (by derivation), it is:

$f(x,y)=(n-1)y^{n-2}$ in $0\leq x\leq y\leq 1$

and it doesn't integrate $1$ but $\frac{n-1}{n}$. I don't know where the problem is.

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  • $\begingroup$ I think this question has been asked and answered previously but cannot find it. $\endgroup$
    – Dilip Sarwate
    Nov 5, 2014 at 18:15

1 Answer 1

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Note that here $P(X_1=X_{(n)})=\frac{1}{n}$ and addition with $\frac{n-1}{n}$ gives $1$.

In fact this distribution has no PDF. This because the set $\{(x,y)\in (0,1)^2\mid x=y\}$ has positive probability $\frac{1}{n}$ and Lebesguemeasure $0$.

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  • $\begingroup$ Thanks. I hadn't notice. So then, how can I prove that $X_1 | X_{(n)} \sim Uniform(0,X_{(n)})$ ? $\endgroup$
    – rcg90
    Nov 5, 2014 at 18:15
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    $\begingroup$ Firstly: that is another question. Secondly: that is not true because $P\left(X_{1}=x\mid X_{\left(n\right)}=x\right)=\frac{1}{n}$. $\endgroup$
    – drhab
    Nov 5, 2014 at 18:25
  • $\begingroup$ Ok thank you! I'll ask in a new question. $\endgroup$
    – rcg90
    Nov 5, 2014 at 18:28

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