Distribution of Binomial MLE and intervals

Question

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.

Answer 1

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 $\theta$

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

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

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

• 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