Let x1, . . . , xn be independent exponential random variables having a common parameter λ. determine the distribution of min(x1 , . . . , xn)
1 answer:
Given fx (x) = λe^λx
Fx (x) = 1 – e^-λx x…0
To find distribution of Min (X1,….Xn)
By applying the equation
fx1 (x) = [n! / (n – j)! (j – 1)!][F(x)]^j-1[1-F(x)]^n-j f(x)
For minimum j = 1
[Min (X1,…Xn)] = [n!/(n-1)!0!][F(x)]^0[1-(1-e^-λx)]^n-1λe^-λx
= ne^[(n-1) λx] λe^(λx)
= nλe^(-λx[1+n-1])
= nλe^(-nλx)
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