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faltersainse [42]
3 years ago
14

Let x1, . . . , xn be independent exponential random variables having a common parameter λ. determine the distribution of min(x1

, . . . , xn)
Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
5 0

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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From the given recurrence, it follows that

a_{n+1} = 2a_n + 1 \\\\ a_{n+1} = 2(2a_{n-1} + 1) = 2^2a_{n-1} + 1 + 2 \\\\ a_{n+1} = 2^2(2a_{n-2}+1) + 1 + 2 = 2^3a_{n-2} + 1 + 2 + 2^2 \\\\ a_{n+1} = 2^3(2a_{n-3} + 1) + 1 + 2 + 2^2 = 2^4a_{n-3} + 1 + 2 + 2^2 + 2^3

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