if (X_{1},...,X_{n}) are independent random variables, each with the density function f, show that the joint density of (X_{(1)},...,X_{(n)} is n!f (x_{1})..f(x_{n}), x_{1}<x_{2}<...<x_{n}

nagasenaz

nagasenaz

Answered question

2021-02-13

if (X1,...,Xn) are independent random variables, each with the density function f, show that the joint density of (X(1),...,X(n) is n!f(x1)f(xn)

Answer & Explanation

Yusuf Keller

Yusuf Keller

Skilled2021-02-14Added 90 answers

Calculation:
Assume that the set of random variables X1,X2,X3,...,Xn are same as the set of random
variables X(1),X(2),....,X(n) because each Xi is equal to the one of the Xj′s.
The random variables are in increasing order. That is, X(1)≤X(2)≤...≤X(n) Therefore,
density is non zero when X1,X2,X3,...,Xn.
The joint density of X1,X2,X3,...,Xn is,
fX1,...,xn(x1,x2,....,xn)=f(x1)f(x2)...f(xn) (since X_i are all independent)
From the above result, it can be observed that the joint density of X(1),...X(n) is
π cysX1,22,An)=a!F Car)f G2)efG∈)3x1 Sx 2SS∈
fX1,...,xn(x1,x2,....,xn)=n!(f(x1)f(x2)),x1≤x2..≤xn.

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