# Distribution of Binomial MLE and intervals

If $$\theta$$ is the frequency of an allele causing a Mendelian recessive disease, then the probability that an individual is affected is $$\theta^2$$. A random sample of size $$n$$ individuals is taken form a very large population, and $$x$$ individuals are observed to be affected with the disease.

a. What is the maximum likelihood estimator of $$\theta$$, and what is its approximate distribution when the sample size is large?

b. In small samples, is the estimator for $$\theta$$ an UMVUE (uniform minimum variance unbiased estimator)?

c. Use two approaches to construct a $$95\%$$ confidence interval for $$\theta$$.

d. Use two approaches to construct a $$95\%$$ confidence interval for $$\theta^2$$.

For part a I got the mle for $$\theta^2$$ as $$\bar{X}$$, so by the invariance property $$\hat{\theta}$$ = $$\sqrt{\bar{X}}$$. I could think of the asymptotic normality of MLE. So $$\hat{\theta^2} \sim N(\theta, \frac{1}{I(\theta^2)})$$. $$I(\theta^2)$$ is $$\frac{n}{\theta^2(1-\theta^2)}$$.

Now CRLB for $$\sqrt{\theta^2}$$ is $$\frac{\frac{1}{2\theta}^2}{\frac{n}{\theta^2(1-\theta^2)}} = \frac{1-\theta^2}{4n}$$ But I am not really confident with this result.

For part b, I could think of $$\hat{\theta}$$ losing normality when $$n$$ is not large. So it will not achieve the CRLB. Any more reasons you could think of are welcomed!

For part c and d I suppose the difference requires me to apply the invariance property. But one method could be using the standardised MLE: $$\left[\hat{\theta} - z_{0.975}\frac{1}{\sqrt{I(\theta)}}, \hat{\theta} + z_{0.975}\frac{1}{\sqrt{I(\theta)}}\right]$$ I am unsure about other method.

• Looks like you have $x\sim \mathsf{Bin}(n,\theta^2)$, so that MLE of $\theta^2$ is $x/n$. Part of the same question: math.stackexchange.com/q/2313173/321264. Then for (b), you only have to check whether the estimator is unbiased or not. Also see en.wikipedia.org/wiki/…. Jun 14, 2020 at 16:14

## 1 Answer

There is a little error in the beginning. The text says that $$\theta$$ is the frequency.... so the model is Bernulli. In fact reading the text

If $$\theta$$ is the frequency of an allele causing a Mendelian recessive disease, then the probability that an individual is affected is $$\theta^2$$

$$\theta$$ is the % of the population with specific allele (benulli model). Then, to show the disease, the two alleles must be present in the gene...of course with probability $$\theta^2$$, a function $$g(\theta)$$ of the parameter.

Formally, the basic Statistical Model is the following

$$f(x;\theta)=\theta^x(1-\theta)^{1-x}\mathbb{1}_{\{0;1\}}(x)$$

then the probability to be affected is $$g(\theta)=\theta^2$$, a function of $$\theta$$

Then the MLE estimator for $$\theta$$ is $$\hat{\theta}=\bar{X}_n$$

Now you surely know that it achieves the CRLB...

As Confidence interval of $$\theta$$ is concerned, for big samples you can use the following pivotal quantity

$$\frac{\hat{\theta}-\theta}{\sqrt{\frac{(\theta(1-\theta)}{n}}}$$

and

• solve in $$\theta$$

$$-z<\frac{\hat{\theta}-\theta}{\sqrt{\frac{(\theta(1-\theta)}{n}}}

• Substitute $$\theta(1-\theta)$$ with its estimation $$\hat{\theta}(1-\hat{\theta)}$$ and then solve in $$\theta$$

$$-z<\frac{\hat{\theta}-\theta}{\sqrt{\frac{\hat{\theta}(1-\hat{\theta)}}{n}}}

• for small samples use the binomial.

For the CI of $$g(\theta)=\theta^2$$ it is enough to observe that $$g$$ is monotone..

With this hints I think you can proceed successfully by yourself

• Hi thank you for the hints. Do you mean the bernoulli rvs have parameter $\theta$ or $\theta^2$? Is it okay if I just compute the MLE for $\theta$ in this case? Jun 14, 2020 at 10:31
• @siegfried : the parameter of the model is $\theta$. Jun 14, 2020 at 10:32
• @siegfried : $\theta^2$ is a particular function of the parameter stated by text, estimating with this function the probability for the disease.... Jun 14, 2020 at 10:34
• I do remember the Bernoulli parameter would be the 'success' probability, so as I understand Xi = 1 if the ith person is infected, with probability theta^2 of an individual getting infected. The sum of Xi's is thus a Binomial rv with parameters (n, $\theta^2$), as in the question x among n individuals infected. I am not sure if the frequency $\theta$ is the parameter... Jun 14, 2020 at 10:41
• @siegfried : "the frequency of an allele causing a mendelian recessive disease..." is a bernulli $B(\theta)$ model....this is my opinion Jun 14, 2020 at 10:46