Let X1Xn denote a random sample from a Poisson Distribution

Let X1...Xn denote a random sample from a Poisson Distribution with parameter gamma. Find the Maximum likelihood estimator for gamma. Show that the variance of the maximum likelihood estimator achieves the Cramer-Rao lower bound and hence the MLE is also the MVUE.

Solution

Given that let X1...Xn denote a random sample from a Poisson Distribution with parameter gamma.

Xi ~ P()

The p.m.f.. of Poisson distribution is,

P(X = x) = (e^- * ()^x ) / x! ) x   0.

To find a maximum likelihood estimator of ,

Xi\'s are i.i.d. random variables.

Consider likelihood function of X,

L = f(Xi,) (i is from 1 to n)

=   (e^- * ()^x ) / x! ) (i is from 1 to n)

=  (e^- * ()^x1 ) / x1! ) + (e^- * ()^x2 ) / x2!) +   (e^- * ()^x3 ) / x3!) +.............+ (e^- * ()^xn ) / xn!)

= e ^-n * xi / xi !    (i is from 1 to n)

loglikelihood function would be,

ln(L) = ln [  e ^-n * xi / xi ! ]    (i is from 1 to n)

= -n + xi ln() -   ln(yi !)   (i is from 1 to n)

differantiate ln(L) wi4th respect to and equate it to 0.

ln(L) / = -n + xi / - 0

-n +  xi / = 0

^ = xi / n

Where   ^ is the malximum likelihood estimator of   .

Show that the variance of the maximum likelihood estimator achieves the Cramer-Rao lower bound and hence the MLE is also the MVUE.

We have seen that Xbar is an maximum likelihood estimator of .

Consider ln [f(x,)] = ln [  (e^- * ()^x ) / x! ]

= - + x ln() - ln(x!)

taking partial derivative with respect to ,

ln(x,) / = x / - 1 = (x-) /

squaring and taking expectation of both sides,

E { [ ln(x,) / ]² } = E(x - )² / ²

= ² /  ²

= /  ²

= 1 /

We see that the Rao-Cramer lower bound is /n, which is the variance of Xbar . Hence Xbar is an efficient estimator of .