Genotype AA, Aa, and aa occur with probabilities [$\theta^2, 2\theta(1-\theta),(1-\theta)^2$].
A multinomial sample of size n has frequencies ($n_1, n_2, n_3$).
I try to derive a Fisher information.
$l(\theta) = (\theta^2)^{n_1} * (2\theta(1-\theta))^{n_2} * ((1-\theta)^2)^{n_3}$
$log l(\theta) = 2 n_1 log\theta + n_2(log2+log\theta+log(1-\theta))+2n_3log(1-\theta)$
${\frac {\partial} {\partial \theta}}$ $log l(\theta)=2n_1{\frac{1}{\theta}}+n_2({\frac{1}{\theta}}-{\frac{1}{1-\theta}})-2n_3{\frac{1}{1-\theta}} = \frac{2n_1+n_2}{\theta}-{\frac{n_2+2n_3}{1-\theta}}$
${\frac {\partial} {\partial \theta}}$$log l(\theta)=0, \hat\theta_{MLE}={\frac{2n_1+n_2}{2(n_1+n_2+n_3)}}$
${\frac {\partial^2} {\partial \theta^2}}$ $log l(\theta)={\frac {\partial^2} {\partial \theta^2}}$ $({\frac{2n_1+n_2}{\theta}}-{\frac{n_2+2n_3}{1-\theta}})=-{\frac{2n_1+n_2}{\theta^2}}-{\frac{n_2+2n_3}{(1-\theta)^2}}$
$-E(-{\frac{2n_1+n_2}{\theta^2}}-{\frac{n_2+2n_3}{(1-\theta)^2}})=E({\frac{2n_1+n_2}{\theta^2}}-{\frac{n_2+2n_3}{(1-\theta)^2}})$
And then how can drive the Fisher Information?
For reference,
$\hat\mu_1=n_1 * \hat\theta^2 = {\frac{n_1(2n_1-n_2)^2}{4(n_1+n_2+n_3)^2}}$
$\hat \mu_2=n_2 * 2\hat\theta(1-\hat\theta)={\frac{n_2(2n_1-n_2)(3n_2+2n_3)}{2(n_1+n_2+n_3)^2}}$
$\hat\mu_3=n_3 * (1-\hat\theta)^2={\frac{n_3(3n_2+2n_3)^2}{4(n_1+n_2+n_3)^2}}$