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 2021-02-13 Answered

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)

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Expert Answer

Yusuf Keller
Answered 2021-02-14 Author has 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 Xjs.
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)),x1x2..xn.

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