nagasenaz

2021-02-13

if $\left({X}_{1},...,{X}_{n}\right)$ are independent random variables, each with the density function f, show that the joint density of $\left({X}_{\left(1\right)},...,{X}_{\left(n\right)}$ is $n!f\left({x}_{1}\right)f\left({x}_{n}\right)$

Yusuf Keller

Calculation:
Assume that the set of random variables ${X}_{1},{X}_{2},{X}_{3},...,Xn$ are same as the set of random
variables ${X}_{\left(1\right)},{X}_{\left(2\right)},....,{X}_{\left(n\right)}$ because each Xi is equal to the one of the ${X}_{{j}^{\prime }s}$.
The random variables are in increasing order. That is, ${X}_{\left(1\right)}\le {X}_{\left(2\right)}\le ...\le {X}_{\left(n\right)}$ Therefore,
density is non zero when ${X}_{1},{X}_{2},{X}_{3},...,{X}_{n}$.
The joint density of ${X}_{1},{X}_{2},{X}_{3},...,{X}_{n}$ is,
$f{X}_{1},...,{x}_{n}\left({x}_{1},{x}_{2},....,{x}_{n}\right)=f\left({x}_{1}\right)f\left({x}_{2}\right)...f\left({x}_{n}\right)$ (since X_i are all independent)
From the above result, it can be observed that the joint density of ${X}_{\left(1\right)},...{X}_{\left(n\right)}$ is

$f{X}_{1},...,{x}_{n}\left({x}_{1},{x}_{2},....,{x}_{n}\right)=n!\left(f\left({x}_{1}\right)f\left({x}_{2}\right)\right),{x}_{1}\le {x}_{2}..\le {x}_{n}$.

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