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

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

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

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

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

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

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$\displaystyle \frac{n}{\theta}+\frac{\sum(x-1)}{1-\theta}$

taking the second derivative gives and negative mean gives:-

$\displaystyle -E[\frac{-n}{\theta^2}-\frac{\sum(x-1)}{(1-\theta)^2}]$................................................yi elds$\displaystyle -E[\frac{-n+2\thetan-n\theta^2-\sumx\theta^2+\sumx\theta^2}{\theta^2+1-2\theta+\theta^2}]$ 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 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 $\displaystyle \hat{\theta)+\- $etcetc