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I know that there are several but I want to ask if this approach is correct:

Given $X_1,X_2... X_n$ iid continuous RVs

The objective is to find the joint pdf $f(Z,W)$ where $Z=Min(X_i's)$ and $W=Max(X_i's)$

$Pr(Z<z,W<w)=Pr(X_i>z\forall i,X_i<w \forall i)=Pr(z<X_i<w \forall i)=[Pr(z<X<w)]^n$

and the pdf is just the derivative of $[Pr(z<X<w)]^n$

With the input from the comment below, I revised my approach to as follows:

$Pr(Z<z,W<w)=P(W<w)-Pr(Z>z,W<w)$

Where $Pr(Z>z,W<w)=Pr(X_i>z\forall i,X_i<w \forall i)=Pr(z<X_i<w \forall i)=[Pr(z<X<w)]^n$

So,

$Pr(Z<z,W<w)=P(W<w)-[Pr(z<X<w)]^n$

I worked on it using the corrected relationship in my post and then checked my answer using the wikipedia link provided and ended up with 2 slightly different answers:

My work:

$f(z,w)=\frac{n(n-1)}{\theta^n}(w-z)^{n-2}$

Plugging in values in the wikipedia expression

$f(z,w)=\frac{n(n-1)}{\theta^n}(w-z+2-\theta)^{n-2}$

Are any of these two correct?

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  • $\begingroup$ Wikipedia gives the general case of the joint distribution of two order statistics for iid sampling from a continuous distribution in its article on order statistics, in the section titled The joint distribution of the order statistics of an absolutely continuous distribution. $\endgroup$
    – Glen_b
    Commented Jul 7, 2017 at 3:40
  • $\begingroup$ @Glen_b Thanks. I'm trying to avoid needing to memorize too many things though. Is there still anything wrong with my new representation for $Pr(Z<z, W<w)$? $\endgroup$
    – user164144
    Commented Jul 7, 2017 at 4:16
  • $\begingroup$ I'm not sure that there's anything to memorize in my suggestion. You can simply obtain the solution to your specific problem from the solution to a more general one by substitution. $\endgroup$
    – Glen_b
    Commented Jul 7, 2017 at 4:20
  • $\begingroup$ I meant the form of the joint order stat pdf in the wiki page that you have which even when limited to f(min,max) is quite long (unless I am not understanding your suggestion correctly). $\endgroup$
    – user164144
    Commented Jul 7, 2017 at 4:24
  • $\begingroup$ Again, I wasn't suggesting you memorize it; I was suggesting you use substitution into it. $\endgroup$
    – Glen_b
    Commented Jul 7, 2017 at 9:00

1 Answer 1

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If all $X_i$ are greater than $z$, the minimum will be greater than $z$ not less than. It is correct that the maximum is less than $w$ if all $X_i$ is less than $w$. So the last two equalities in your probability statements don't hold.

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  • $\begingroup$ Will the entire equality hold if I change the first statement to $Pr(Z>z,W<w)$? $\endgroup$
    – user164144
    Commented Jul 7, 2017 at 3:27

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