(6.11.a) Let $X_1,X_2,X_3,X_4,X_5$ be independent continuous random variables having a common distribution function $F$ and density function $f$, and set
$$ I=\pr{X_1<X_2<X_3<X_4<X_5} $$
Show that $I$ does not depend on $F$.
Hint: Write $I$ as a five-dimensional integral and make the change of variables $u_i=\cdf{x_i},i=1,…,5$.
Solution Since the $\set{X_i}$ are IID continuous, their joint density can be factored:
$$ \pdfa{x_1,x_2,x_3,x_4,x_5}{X_1,X_2,X_3,X_4,X_5}=\pdf{x_1}\wts\pdf{x_2}\wts\pdf{x_3}\wts\pdf{x_4}\wts\pdf{x_5} $$
Hence
$$ I=\pr{X_1<X_2<X_3<X_4<X_5} $$
$$ =\mathop{\iiiint\int}_{x_1<x_2<x_3<x_4<x_5}\pdf{x_1}\wts\pdf{x_2}\wts\pdf{x_3}\wts\pdf{x_4}\wts\pdf{x_5}dx_1dx_2dx_3dx_4dx_5 $$
$$ =\int_{-\infty}^{\infty}\int_{-\infty}^{x_5}\int_{-\infty}^{x_4}\int_{-\infty}^{x_3}\int_{-\infty}^{x_2}\pdf{x_1}\wts\pdf{x_2}\wts\pdf{x_3}\wts\pdf{x_4}\wts\pdf{x_5}dx_1dx_2dx_3dx_4dx_5 $$
$$ =\int_{-\infty}^{\infty}\pdf{x_5}\int_{-\infty}^{x_5}\pdf{x_4}\int_{-\infty}^{x_4}\pdf{x_3}\int_{-\infty}^{x_3}\pdf{x_2}\int_{-\infty}^{x_2}\pdf{x_1}dx_1dx_2dx_3dx_4dx_5 $$
$$ =\int_{-\infty}^{\infty}\pdf{x_5}\int_{-\infty}^{x_5}\pdf{x_4}\int_{-\infty}^{x_4}\pdf{x_3}\int_{-\infty}^{x_3}\pdf{x_2}\cdf{x_2}dx_2dx_3dx_4dx_5 $$
where the last equation holds because $\cdf{x_2}=\int_{-\infty}^{x_2}\pdf{x_1}dx_1$. Now let’s make a change of variable:
$$ u_2=\cdf{x_2}\dq du_2=F'(x_2)dx_2=\pdf{x_2}dx_2\dq u_2du_2=\pdf{x_2}\cdf{x_2}dx_2 $$
$$ u_2^0=\cdf{x_2^0}=\cdf{-\infty}=0\dq u_2^1=\cdf{x_2^1}=\cdf{x_3} $$
$$ =\int_{-\infty}^{\infty}\pdf{x_5}\int_{-\infty}^{x_5}\pdf{x_4}\int_{-\infty}^{x_4}\pdf{x_3}\int_{0}^{\cdf{x_3}}u_2du_2dx_3dx_4dx_5 $$
$$ =\int_{-\infty}^{\infty}\pdf{x_5}\int_{-\infty}^{x_5}\pdf{x_4}\int_{-\infty}^{x_4}\pdf{x_3}\frac12\prn{\cdf{x_3}}^2dx_3dx_4dx_5 $$
$$ u_3=\cdf{x_3}\dq du_3=F'(x_3)dx_3=\pdf{x_3}dx_3\dq u_3^2du_3=\pdf{x_3}\prn{\cdf{x_3}}^2dx_3 $$
$$ u_3^0=\cdf{x_3^0}=\cdf{-\infty}=0\dq u_3^1=\cdf{x_3^1}=\cdf{x_4} $$
$$ =\frac12\int_{-\infty}^{\infty}\pdf{x_5}\int_{-\infty}^{x_5}\pdf{x_4}\int_0^{\cdf{x_4}}u_3^2du_3dx_4dx_5 $$
$$ =\frac12\int_{-\infty}^{\infty}\pdf{x_5}\int_{-\infty}^{x_5}\pdf{x_4}\frac13\prn{\cdf{x_4}}^3dx_4dx_5 $$
$$ u_4=\cdf{x_4}\dq du_4=F'(x_4)dx_4=\pdf{x_4}dx_4\dq u_4^3du_4=\pdf{x_4}\prn{\cdf{x_4}}^3dx_4 $$
$$ u_4^0=\cdf{x_4^0}=\cdf{-\infty}=0\dq u_4^1=\cdf{x_4^1}=\cdf{x_5} $$
$$ =\frac12\frac13\int_{-\infty}^{\infty}\pdf{x_5}\int_{0}^{\cdf{x_5}}u_4^3du_4dx_5 $$
$$ =\frac12\frac13\int_{-\infty}^{\infty}\pdf{x_5}\frac14\prn{\cdf{x_5}}^4dx_5 $$
$$ u_5=\cdf{x_5}\dq du_5=F'(x_5)dx_5=\pdf{x_5}dx_5\dq u_5^4du_5=\pdf{x_5}\prn{\cdf{x_5}}^4dx_5 $$
$$ u_5^0=\cdf{x_5^0}=\cdf{-\infty}=0\dq u_5^1=\cdf{x_4^1}=\cdf{\infty}=1 $$
$$ =\frac12\frac13\frac14\int_0^1u_5^4du_5=\frac12\frac13\frac14\frac15=\frac1{5!} $$
Since we used only those general properties of CDF’s and PDF’s and since the computed integral doesn’t depend on $F$, then $I$ doesn’t depend on $F$.
(6.11.b) Evaluate $I$.
Solution Done in part (a).
(6.11.c) Give an intuitive explanation for your answer to (b).
Solution This is the result of symmetry. Let’s first consider just two variables. Since $X_1$ and $X_2$ are independent and have the same distribution, it is just as likely that $X_1<X_2$ as it is that $X_2<X_1$.
Similarly, since the $\set{X_i,i=1,…5}$ are independent variables with the same distribution, the likelihood of any one ordering (e.g. $X_1<X_2<X_3<X_4<X_5$) is the same as any other ordering. For example
$$ \pr{X_1<X_2<X_3<X_4<X_5}=\pr{X_1<X_2<X_3<X_5<X_4}=\pr{X_5<X_4<X_3<X_2<X_1} $$
There are exactly $5!$ possible orderings of the $\set{X_i}$, and they have equal probabilities that sum to $1$, so each has probability $\frac1{5!}$. That is, if $\set{O_i,i=1,…,5!}$ is the collection of all possible orderings and if $p$ is the probability of any one of the possible orderings, then
$$ 1=\sum_{i=1}^{5!}\pr{O_i}=\sum_{i=1}^{5!}p=5!p\iff p=\frac1{5!} $$