Let Y1,Y2,...,Yn be a sample from a Gamma distribution with probability density function.f(y)
= .where beta > 0 What is the relevant likelihood? (Be sure you get this correct since your
answers to the other parts of the problem depend on this likelihood) Show that is sufficient for
beta . Find the maximum likelihood estimator for beta Find the minimum variance unbiased
estimator of beta .
Solution
(a)
Likelihood is given by
product over all i=1 to i=n, (beta)^(-2)y_ie^(-y_i/beta)
=(beta)^(-2n) prod y_i e^(-sum(y_i)/beta)
(b)
Hence sum(y_i) is sufficient statistics for beta since term related with beta is sum(y_i), from
factorization theorem.
(c)
Maximum likelihood estimator for beta is obtained by
log likelihood is -2n log(beta)+sum(log(y_i))-sum(y_i)/beta
differentiating we get,
-2n/beta+sum(y_i)/beta^2 and equating to zero
2n=sum(y_i)/beta
=> beta=sum(y_i)/2n
(d)
since sum(y_i) is sufficient statisics, any minimum variance unbaised estimator will be a
function of sum(y_i),
E(sum(y_i))=2(beta)n (from mean of gamma distribution)
=> E(sum(y_i)/2n)=beta
Hence UMVUE of beta is sum(y_i)/2n..
Let Y1,Y2,...,Yn be a sample from a Gamma distribution with probabili.pdf
1. Let Y1,Y2,...,Yn be a sample from a Gamma distribution with probability density function.f(y)
= .where beta > 0 What is the relevant likelihood? (Be sure you get this correct since your
answers to the other parts of the problem depend on this likelihood) Show that is sufficient for
beta . Find the maximum likelihood estimator for beta Find the minimum variance unbiased
estimator of beta .
Solution
(a)
Likelihood is given by
product over all i=1 to i=n, (beta)^(-2)y_ie^(-y_i/beta)
=(beta)^(-2n) prod y_i e^(-sum(y_i)/beta)
(b)
Hence sum(y_i) is sufficient statistics for beta since term related with beta is sum(y_i), from
factorization theorem.
(c)
Maximum likelihood estimator for beta is obtained by
log likelihood is -2n log(beta)+sum(log(y_i))-sum(y_i)/beta
differentiating we get,
-2n/beta+sum(y_i)/beta^2 and equating to zero
2n=sum(y_i)/beta
=> beta=sum(y_i)/2n
(d)
since sum(y_i) is sufficient statisics, any minimum variance unbaised estimator will be a
function of sum(y_i),
E(sum(y_i))=2(beta)n (from mean of gamma distribution)
=> E(sum(y_i)/2n)=beta
Hence UMVUE of beta is sum(y_i)/2n.
