Find the joint distribution of the continuous order statistics?

Question

Take $n$ independent variables, ${X_1, X_2,\dots, X_n}$, which are uniformly distributed over the interval $(0,1)$.

Then, introduce the variable $M=\min(X_1, X_2,\dots, X_n)$ and the variable $N=\max(X_1, X_2,\dots, X_n)$.

Now, we want to find the joint distribution of a pair $(M,N)$ and the CDF and density for each respective variable.

Alright, so here's what I think. These follow the definitions given in order statistics, where $X_{(1)}=\min(X_1, X_2,\dots, X_n)=M$ and $X_{(n)}=\max(X_1, X_2,\dots, X_n)=N$. So the joint density of the variables ${X_1, X_2,\dots, X_n}$ is $H(x_1, x_2, \dots, x_n) = P(X_1<x_1, X_2<x_2, \dots, X_n<x_n)$, right?

I'm not totally sure about this. It's difficult for me to conceptualize this distribution.

Answer 1

A general aproach is:

$\begin{eqnarray} F_M(m)&=&P(M\le m)\\ &=&P(X_1\le m,...,X_n\le m)\\ &=&P(X_1\le m)...P(X_n\le m)\\ &=&P(X_1\le m)^n\\ &=&F_X(m)^n \end{eqnarray}$

Thus $f_M(m)=nF_X(m)^{n-1}f_X(m)$. In the same spirit, $F_N(m)=1-(1-F_X(m))^n$. and $f_N(m)=n(1-F_X(m))^{n-1}f_X(m)$.

On the other hand, suppose $s\le t$.

$\begin{eqnarray}F_{(N,M)}(s,t)&=&P(N\le s,M\le t)\\ &=&P(N\le s|M\le t)P(M\le t)\\ &=&(1-P(N> s|M\le t))P(M\le t)\\ &=&P(M\le t)-P(N> s|M\le t)P(M\le t)\\ &=&P(M\le t)-P(N>s,M<t)\\ &=&P(M\le t)-P(s<X_1<t,...,s<X_n<t)\\ &=&P(M\le t)-P(s<X_1<t)...,P(s<X_n<t)\\ &=&P(M\le t)-P(s<X_1<t)^n\\ &=&P(M\le t)-[P(X_1<t)-P(X_1<s)]^n\\ &=&F_M(t)-[F_X(t)-F_X(s)]^n\\ &=&F_X(t)^n-[F_X(t)-F_X(s)]^n\\ \end{eqnarray}$

Finally $f_{N,M}=n(n-1)f_X(t)f_X(s)[F_X(t)-F_X(s)]^{n-2}$, always that it exists.

Since $X\sim U(0,1)$, then $F_X(x)=x1_{[0,1]}(x)+1_{(1,\infty)}(x)$ and $f_X(x)=1_{[0,1]}(x)$.

Therefore:

1)$F_M(t)=t^n1_{[0,1]}(t)+1_{(1,\infty)}(t)$

2)$f_M(t)=n[t^{n-1}1_{[0,1]}(t)+1_{(1,\infty)}(t)]1_{[0,1]}(t)$

3)$F_N(t)=1-(1-t1_{[0,1]}(t)-1_{(1,\infty)}(t))^{n}$

4)$f_N(t)=n(1-t1_{[0,1]}(t)-1_{(1,\infty)}(t))^{n-1}1_{[0,1]}(t)$

Can you finish?

Answer 2

You want $f_{X_{(n)},X_{(1)}}(M,N)$

To hit the favoured space:

• one of the $n$ variables needs to be the maximum value $M$, with density $f_X(M)$.

• given that, then one of the remaining $n-1$ variables then needs to be the minimum value $N$ with density $f_X(N)$.

• given that, then each of the remaining $n-2$ variables to lie between $(N;M)$, and each of these occur with probability mass $(F_X(M)-F_X(N))$

Put it together:

$$\begin{align}f_{X_{(n)},X_{(1)}}(N,M) \; & =\; n\;(n-1)\; f_{X}(N)\; f_{X}(M)\; (F_X(M)-F_X(N))^{n-2} \\ & = \; n\,(n-1)\,(M-N)^{n-2}\end{align}$$

Answer 3

Not quite. Notice that $$P(M\in dm,N\in dn) = P(X_{(1)} \in dm,X_{(n)}\in dn)$$ For a moment, let's pretend that $X_1 \in dm$ and $X_n\in dn$. This forces $X_2,\dotsc,X_{n-1}\in (m-n)$. This equates to \begin{align*} P(X\in dm) &= f_{X_1}(m)\,dm = 1\,dm\\ P(X\in dn) &= f_{X_n}(n)\,dn = 1\,dn\\ P(m< X_k<n) &=\frac{n-m}{1} = n-m \end{align*} But $X_1$ didn't have to be the smallest. You have $\binom{n}{1}$ choices. Then the number of ways to choose a largest is $\binom{n-1}{1}$. Every thing else goes in the middle. Thus $$P(M\in dm,N\in dn) = \binom{n}{1}\binom{n-1}{1}(1\,dm)(n-m)^{n-2}(1\,dn).$$ This gives the density $$f_{M,N}(m,n) = n(n-1)(n-m)^{n-2}.$$

You could try to integrate this, or you can use a similar counting arguement to find the CDF, if you recall that $$P(X\leq x, Y\leq y) = P(Y\leq y)-P(X>x,Y\leq y).$$