Likelihood ratio of Hardy–Weinberg proportions Consider a population with three kinds of individuals labeled $1, 2$, and $3$ occuring in the Hardy–Weinberg proportions $f(1,\theta)=\theta^2,f(2,\theta)=2\theta(1−\theta),f(3,\theta)=(1−\theta)^2$. For a sample $X_1, . . . , X_n$ from this population, let $N_1$, $N_2$, and $N_3$ denote the number of $X_j$ equal to $1$, $2$, and $3$, respectively. Let $0<\theta_0<\theta_1 <1$.
I want to show that the likelihood ratio $L(\mathbf{x},\theta_0,\theta_1)$ is an increasing function of $2N_1+N_2$. I was able to obtain the likelihood as:
$$L(p_1,p_2,p_3)=\frac{N!}{N_1!N_2!N_3!}p_1^{N_1}p_2^{N_2}p_3^{N_3}$$
where $N=N_1+N_2+N_3$ and the $p_j$'s are the corresponding $f(j,\theta)$.
Now I'm stuck on how to show this ratio, under $H_0$: HW proportions, would be equal to $$\theta^2=(\frac{2N_1+N_2}{N})^2$$
PS. The likelihood ratio is defined as:
$$L(\mathbf{x},\theta_0,\theta_1)=\frac{p(\mathbf{x},\theta_1)}{p(\mathbf{x},\theta_0)}$$
Any helps would be appreciated!
 A: The other approach from @Henry's uses a bit more information about Hardy-Weinberg equilibrium: the three types are those with 0, 1, or 2 copies of a particular allele out of 2, and HWE is the assumption that the two copies are independent.
Under HWE the distribution of the number of copies for an individual is $Binom(2,\theta)$, and so
$$2N_1+N_2\sim Binom(2N,\theta)$$
It's clear that the LR in this model is increasing in $2N_1+N_2$.
Write $Y=2N_1+N_2$. The likelihood ratio between two values $\theta_0$ and $\theta_1$ is just
$$\left(\frac{\theta_1}{\theta_0}\right)^Y\left(\frac{1-\theta_1}{1-\theta_0}\right)^{N-Y}$$
so if the first term is bigger than 1 and the second is smaller than 1, they both increase as $Y$ increases.
A: It is not totally clear what you are asking, but if you are looking for a maximum likelihood estimator for $\theta$ based on $$L(\theta \mid \mathbf{x}, HW) = \frac{(x_1+x_2+x_3)!}{x_1!\, x_2!\, x_3!} \left(\theta^2\right)^{x_1} \left(2\theta(1-\theta)\right)^{x_2} \left((1-\theta)^2\right)^{x_3} $$
then you want to differentiate this, or take the logarithm of this and then differentiate, to find the $\theta$ which maximises it.  The derivative when tidied up will be something like $\frac{2x_1+x_2 -2\theta(x_1+x_2+x_3)}{\theta(1-\theta)}$ possibly multiplied by the original likelihood if you did not take the logarithm, leading to a zero derivative and a maximum likelihood when $$\hat \theta = \frac{2x_1+x_2}{2(x_1+x_2+x_3)}$$
Your stated result about $\theta^2$ essentially squared this and is missing a factor of $2$ or $2^2$ in the denominator.  It should have been $$\hat \theta^2 = \left(\frac{2x_1+x_2}{2(x_1+x_2+x_3)}\right)^2$$
