I know that te asymptotic variance for the mle is in general1/nI(p), where I is

ID: 2951528 • Letter: I

Question

I know that te asymptotic variance for the mle is in general1/nI(p), where I is the information,I(p)=-E[d^2/dp^2(log(f(x|p))]

In this particular case, you first need to find deloglikelihoodfunction log(f(x|p) and than differentiate this twicew.r.t. p.

I get this:

(the sum goes from i=0 to n and xi's are the r.v.'s, and let X=1/nxi be the average)

log(f(x|p)) = [log(p) - log(1-p) +log(1-p)*xi]=n*log(p) - n*log(1-p) + X*n*log(1-p)

Differantiation w.r.t. p gives

d/dp(log(f(x|p)) = n/p+n/log(1-p)-X*n/(1-p)

(if we equalize the above to zero and solve for p, we get answer(b), namely p=1/X )

differentiate another time gives:

d^2/dp^2(log(f(x|p)=-n/p^2 +n/(1-p)^2-X*n/(1-p)^2

So I(p)

= - E(-n/p^2 +n/(1-p)^2-X*n/(1-p)2

=n/p^2-n/(1-p)^2+n/(1-p)^2*E(X)

For the geometric distribution, E(X)=1/p, so inserting ensimplifying gives:

I(p)=n/(p^2(1-p))

Now Var(p) = 1/nI(p), so Var(p) = p^2(1-p)/n^2

The difference between my answer and the answers provided hereand the book of Rice is that I have a factor 1/n to much!

Where did I go wrong?

It is hard to determine what your original distribution functionwas, can you post that so I can start from the beginning and try tohelp you?