# 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!

• How did you go from having $\theta$ to having $\theta_0$ and $\theta_1$, to having $p_1$, $p_2$, and $p_3$? Commented Apr 21, 2021 at 22:40
• @AryaMcCarthy I am guessing that $\theta_0$ and $\theta_1$ are possible values of $\theta$. Meanwhile the use of $L(\mathbf{x},\theta_0,\theta_1)$ and $L(p_1,p_2,p_3)$ also looks confusing, with the former possibly being the likelihood ratio while the latter is just the likelihood. Commented Apr 21, 2021 at 22:49
• @Henry Yes, sorry, that's the likelihood, I edited it. Commented Apr 21, 2021 at 22:56
• @AryaMcCarthy Yes Henry is correct. The likelihood ratio is defined as $L(\mathbf{x},\theta_0,\theta_1)=\frac{p(\mathbf{x},\theta_1)}{p(\mathbf{x},\theta_0)}$. And for $p_j$'s that's just a simplification in notation I made so I don't need to write the $\theta$ terms each time. Commented Apr 21, 2021 at 22:57

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 $$\left(\frac{\theta_1}{\theta_0}\right)^Y\left(\frac{1-\theta_1}{1-\theta_0}\right)^{2N-Y}$$ so if the first term is bigger than $$1$$ and the second is smaller than $$1$$, they both increase as $$Y$$ increases.

• Okay, I get some ideas here but still don't understand how to conclude the LR ratio is increasing. Commented Apr 22, 2021 at 21:39

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 $$\hat \theta^2 = \left(\frac{2x_1+x_2}{2(x_1+x_2+x_3)}\right)^2$$