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Chapter 4: Problem 10

Let \(Y_{1}

Short Answer

Expert verified
The distribution function of \(Y_{1}\) from a Weibull distribution is: \[F_{Y_{1}}(y)= 1-[1-(1-e^{-位y^k})]^n\] and its pdf is \[f_{Y_1}(y) = n位k y^{k-1}e^{-位y^k} [1-(1-e^{-位y^k})]^{n-1}\].

Step by step solution

01

Understand the properties of Weibull distribution

A random variable \(X\) has a Weibull distribution, denoted as \(X鈭糤eibull(位, k)\), if its pdf is given by: \[f_X(x)= 位k x^{k-1}e^{-位x^k}\] for \(x > 0\), and 0 otherwise, where \(位 > 0\) is the scale parameter and \(k > 0\) is the shape parameter. And its CDF is given by: \[F_X(x)=1-e^{-位x^k}\] for \(x > 0\), and 0 otherwise.
02

Finding the distribution function(CDF) of \(Y_1\)

In a random sample of size \(n\) from a population, \(Y_1\) represents the smallest order statistic. The probability that \(Y_1\) is greater than a certain value \(y\) is the probability that all \(n\) values are greater than \(y\). For a Weibull random variable, this is \[(1 - F_X(y))^n\]. Hence, the CDF of \(Y_1\), which is the probability that \(Y_1\) is less than or equal to \(y\), is obtained by subtracting the above value from 1. Hence, \[F_{Y_{1}}(y) = 1 - (1 - F_X(y))^n = 1- [1-(1-e^{-位y^k})]^n\].
03

Finding the pdf of \(Y_1\)

The pdf is the derivative of the CDF. Differentiating \(F_{Y_{1}}(y)\) with respect to \(y\) using the chain rule, we get: \[f_{Y_1}(y) = n位k y^{k-1}e^{-位y^k} [1-(1-e^{-位y^k})]^{n-1}\]. This is the pdf of the smallest order statistic of a random sample from a Weibull distribution.

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