Question

6.81 Let Yı, Y. ..., Y, be independent, exponentially distributed random variables with mean B. a Show that Y) = min(Y , Y2, ..., Y,) has an exponential distribution, with mean B/n. b If n = 5 and B = 2, find P(Ym <3.6).

Answer 1

Given : Y₁, Y₂, ..........Y_n be independent and exponentially distributed with mean $\beta$

Probability density function of Yi is

------------------(I)

0<fo<*g/k_3 = (h)!

and its distribution function is

Fy) = 1-e-4/8

= First order statistic = smallest order statistic

Y(1) = min(y1, .......... Yn)

By using results of order statistic

The probability distribution of first order statistic is

fyu (y) = n* [1 – F(y)]n-1 * f(y)

Hence distribution of Y₍₁₎ is

fyu) (y) = n* [1 – (1 – (*)n-1 (*e=1/

, (+-) - = ()***

----------------------(II)

fway (9) = *

From (I) and (II)

The distribution of Y₍₁₎ is exponential with mean $\beta /n$

Hence

Y(1) ~ Erp(Mean = B/n)

b) Given: n= 5 and beta = 2

Hence

Υ1) Ετp Mean = 0.4)

The p.d.f of Y₍₁₎ is

0 < (1)*mp_3* + 0 = () (RI

P(Y(1) < 3.6) = *e-4/0.4 * du o 0.4

12-=(985 01d

P(Y(1) < 3.6) = 1-

.

P(Y1) <3.6) = 0.9998