the question asks to find the crlb for an unbiased estimator of \theta in the function p(x; \theta)= \theta(1-\theta)^(x-1)

it is a geometric function with parameter \theta it represents the number of attacks in basketball until the first basket.
the first question asked, given sample X_{1},,,X_{n} representing no. of attacks determine \hat{\theta}, the maximum likilihood estimator of \theta.

i think i got the first part right. I did \sum\log\theta(1-\theta)^(x-1)
\sum\log\theta+\sum(x_{i}-1)\log(1-\theta) log'(x;\theta)=0
\frac{n}{\theta}+\frac{\sum(x_{i}-1)}{1-\theta}=0

i split it up at this point,,and got,,

\frac{n}{\theta}+\frac{\sum(x_{i})}{1-\theta}-\frac{n}{1-\theta}=0
n(1-\theta)+\sum(x_{i})\theta-n\theta=0
\theta=\frac{-n}{-2n+\sum(x_{i})} could someone double check this is correct

****for the crlb the formula for it is \frac{1}{nl(\theta)} where l(\theta)=-E[\frac{d^2log(x;\theta)}{d^2\theta}]

this is my working but i get stuck at the end where I presume im making a fatal error****...

from the first derivative before i already had \frac{n}{\theta}+\frac{\sum(x-1)}{1-\theta}

taking the second derivative gives and negative mean gives:-

-E[\frac{-n}{\theta^2}-\frac{\sum(x-1)}{(1-\theta)^2}]................................................yi elds -E[\frac{-n+2\thetan-n\theta^2-\sumx\theta^2+\sumx\theta^2}{\theta^2+1-2\theta+\theta^2}] which i cannot simplify,,,,is it piossible to put the E on the denominator at this point...any help would be much appreciated! for now thankyou...it is worth noting i need a positive answer for [tex]l(\theta)[/amth] because in the next part of the question i have to give a 95 % c.i for n=5 x=1,2,2,4,1 because i use the square root \hat{\theta)+\- etcetc