Given you have an independent random sample

siliciooy0j

siliciooy0j

Answered question

2022-03-24

Given you have an independent random sample X1,X2,,Xn of a Bernoulli random variable with parameter p, estimate the variance of the maximum likelihood estimator of p using the Cramer-Rao lower bound for the variance
So, with large enough sample size, I know the population mean of the estimator P^ will be p, and the variance will be:
Var[P^]=1nE[((/p) lnfx(X))2]
Now I'm having some trouble calculating the variance of P^, this is what I have so far:
since the probability function of X is binomial, we have:
fx(X¯)=ni=1nXi*pi=1nXi*(1-p)n-i=1nXi
so: lnfX(X)=ln(ni=1nXi)+i=1nXiln(p)+(n-i=1nXi)ln(1-p)
and: lnfX(X)p=i=1nXip-(n-i=1nXi)(1-p)=nX¯p-(n-nX¯)(1-p)
 

and: (lnfX(X)p)2=(nX¯p-(n-nX¯)(1-p))2=n2p2-2n2pX¯+n2X¯2p2(1-p)2
since E[X2]=μ2+σ2n, and for a Bernoulli random variable E[X]=μ=p=E[X] and Var[X]=σ2=p(1p):
E(lnfX(X)p)2=n2p2-2n2pE[X¯]+n2E[X¯2]p2(1-p)2=n2p2-2n2p2+n2(p2+p(1-p)n)p2(1-p)2=np(1-p)p2(1-p)2=np(1-p)
Therefore, Var[P^]=1nE[((/p) lnfx(X))2]=1nnp(1-p)=p(1-p)n2
However, I believe the true value I should have come up with is p(1p)n.

Answer & Explanation

Nyla Trujillo

Nyla Trujillo

Beginner2022-03-25Added 9 answers

You seem to have confused the formulas for the lower bound in the case where the sample is IID, and the more general case of not necessarily identically distributed variables.
The correct formula should be (in the unbiased case)
Var[p^]E(plogfX(x|p))2-1,
when the joint density is over the sample x=(x1,,xn). Note there is no factor of n in front. But when the sample is IID, this can be written
Var[p^]nE(plogfX(x|p))2-1,
where the density (notice no boldface!) is the univariate density for a single observation from the sample.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in College Statistics

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?