1. Stat
Final Lab

2. Q1a: Write down the log likelihood function.
Find the density f of xi first by taking
the 1st derivative of the cdf F.
Take the natural log of f

3. Density f of xi:
Natural log f of xi:
Log likelihood of all observed data:

4. rweibull(n, shape=a, scale=b)
Q1b: For n=200, write R code to generate data from the Weibull distribution with α=2.3. Then comput the MLE of α by using ‘optim’ in R.
In R, use
rweibull(n, shape=a, scale=b)
to generate data from the Weibull distribution.
Check the definition of the Weibull distribution in R by
?rweibull
because the density of the Weibull defined in R may be different from the Weibull density in our assignment.

5. f(x) = (a/b) (x/b)^(a-1) exp(- (x/b)^a)
In R, the Weibull distribution with
shape parameter a and scale parameter b has density given by
f(x) = (a/b) (x/b)^(a-1) exp(- (x/b)^a)
for x >= 0.
The cdf is
F(x) = 1 - exp(- (x/b)^a)
on x >= 0,
E(X) = b Gamma(1 + 1/a)
Var(X) = b^2 * (Gamma(1 + 2/a) - (Gamma(1 + 1/a))^2).
Compare with the definition of the cdf of the Weibull distribution in our assignment.
where x > 0.
α=a=shape, 1/5=b=scale

6. α=a=shape, 1/5=b=scale Use ‘optim’ to find the MLE of α. n = 200
alpha=2.3
beta=1/5
x = rweibull(n, shape=alpha, scale=beta)
Log likelihood of all observed data:
Use ‘optim’ to find the MLE of α.
f=function(y) {
h=n*log(y)+(y-1)*sum(log(5*x))-5^y*sum(x^y) }
Then use optim(par, f), where par is an initial value for the parameters to be optimized over.

7. a1 = optim(2, f, method = "BFGS")$par
f=function(y) {
h=n*log(y)+(y-1)*sum(log(5*x))-5^y*sum(x^y) } -h
optim(par, fn, …)
par: Initial values for the parameters to be optimized over.
fn: A function to be minimized.
a1 = optim(2, f, method = "BFGS")$par
Get the MLE from “optim”:

8. Problems!! The approximate 95% C.I. for α is where
Q1c and d: Find an approximate 95% C.I. for α.
where
Problems!!
The approximate 95% C.I. for α is

9. Natural log f of X:

10. Shit!! ? Natural log f of X: The 1st derivative of the log f:
The Fisher information at the MLE:
The Observed Fisher information at the MLE:
Shit!!
has no closed form. How to find ?
Even we know the closed form, then how to find ?
The 2nd derivative of the log f:

11. where z0.975 is the 97.5th quantile of the standard normal density and

12. # MLE of alpha
a1 = optim(2, f, method = "BFGS")$par
# Observed Fisher Information
I1= (a1)^(-2)+ 5^(a1)*mean(x^(a1)*(log(5*x))^2)
# 1.96 is (approximately) the 97.5th quantile of standard normal
sd1 =1.96*sqrt(1/(n*I1))
# The approximate 95% C.I. for alpha
CIa =c(a1-sd1, a1+sd1 )
CIa

13. lambda1=optim(6, f2, method="BFGS")$par # Ques 2d:
# Ques 2a:
n =200
x =rweibull(n,2.3,1/5)
f2 =function(lambda) {
h=n*log(2.3)+n*2.3*log(lambda)+(1.3)*sum(log(x))-lambda^(2.3)*sum(x^(2.3)) -h }
# MLE of lambda
lambda1=optim(6, f2, method="BFGS")$par
# Ques 2d:
# Observed Fisher Information
I2 = 2.3/(lambda1)^2+2.3*(2.3-1)*(lambda1)^(2.3-2)*mean(x^2.3)
sd2=1.96*((n*I2)^(-1/2))
# The approximate 95% C.I. for alpha
CIlambda = c(lambda1-sd2, lambda1+sd2)
CIlambda