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!

  • $\begingroup$ How did you go from having $\theta$ to having $\theta_0$ and $\theta_1$, to having $p_1$, $p_2$, and $p_3$? $\endgroup$ Apr 21, 2021 at 22:40
  • $\begingroup$ @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. $\endgroup$
    – Henry
    Apr 21, 2021 at 22:49
  • $\begingroup$ @Henry Yes, sorry, that's the likelihood, I edited it. $\endgroup$
    – statwoman
    Apr 21, 2021 at 22:56
  • $\begingroup$ @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. $\endgroup$
    – statwoman
    Apr 21, 2021 at 22:57

2 Answers 2


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.

  • $\begingroup$ Okay, I get some ideas here but still don't understand how to conclude the LR ratio is increasing. $\endgroup$
    – statwoman
    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 $$\hat \theta^2 = \left(\frac{2x_1+x_2}{2(x_1+x_2+x_3)}\right)^2$$


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