Originally Posted by

**Moo** Yes, n=12.

Oh, ****** !!!

$\displaystyle {12 \choose x}$ is a constant with respect to $\displaystyle \theta$, so there is no problem for differentiation.

Sorry for any confusion I could've made...

Your Fisher info is weird... How did you get it ?

After taking the logarithm, we have :

$\displaystyle \log f(x,\theta)=\sum_{k=2}^n \log(k)-\sum_{k=2}^x \{\log(k)+\log(n-k)\}+x\log \theta+(12-x) \log(1-\theta)$

After differentiating once, we get

$\displaystyle \frac{\partial}{\partial \theta} f(x,\theta)=\frac x\theta-\frac{12-x}{1-\theta}$

After differentiating twice, we get

$\displaystyle \frac{\partial^2}{\partial \theta^2} f(x,\theta)=-\frac{x}{\theta^2}-\frac{12-x}{(1-\theta)^2}$

Under some conditions (which I think hold here), $\displaystyle \mathcal{I}(\theta)=-\mathbb{E}\left[\frac{\partial^2}{\partial \theta^2} f(X,\theta)\right]$

Which gives, if I'm not mistaking, $\displaystyle 12 \left(\frac 1\theta+\frac{1}{1-\theta}\right)$