Question

The Weibull distribution was introduced in Sect. 3.5.

(a) Find the inverse cdf for the Weibull distribution.

(b) Write a program to simulate n values from a Weibull distribution. Your program should have three inputs: the desired number of simulated values n and the two parameters α and β. It should have a single output: an n x 1 vector of simulated values.

(c) Use your program from part (b) to simulate 10,000 values from a Weibull(4, 6) distribution and estimate the mean of this distribution. The correct value of the mean is 6Γ(5/4) = 5.438; how close is your sample mean?

Answer 1

Solution:

If X follows weibull distribution then cumulative distribution function of x is ,

$F(x;k,\lambda) = 1- e^{-(x/\lambda)^k}\,$

$\therefore 1-F=e^{(x/\lambda )^{k}}$

therefore $\therefore {(x/\lambda )^{k}}=ln(1-F) \Rightarrow x=\lambda(ln(1-F)) ^{1/k}$

$\Rightarrow x=\lambda(ln(1-F)) ^{1/k}$is inverse cdf of random variable x

here F~U(0,1)

Therefore considering random sample from U(0,1) we can generate random sample x

consider following code in matlab

n=10000
k=4
l=6
for i=1:n
F(i)=unifrnd(0,1);
x(i)=l*(-log(1-F(i)))^(1/k);
end
m=mean(x)

m=5.4708

this implies populaio mean is equal to sample mean