# Root estimator to get variance?

I have an exam tomorrow and I'm seeking advice on if I can square root the variance of the mle $\theta^2$ to get te variance of $\theta$. Does the invariance property apply here?

1. Question:

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

What is the approximate distribution of the MLE when the sample size is large?

I've gotten the mle to be $\sqrt{ \frac{X}{n}}$. Hence,

The mean is:

$E[\hat{\theta}] = \sqrt{E[X]/n} = \theta$

Which is easy cause the E just goes into the square root. However, I dont think I can do this for the variance right? So my question is, can I do:

$Var[\hat{\theta^2}] = Var[\frac{X}{n}] = \frac{1}{n^2}Var[X] = \frac{p(1-p)}{n}$ where p = $\theta^2$

Then,

$Var[\hat{\theta}] = \sqrt{Var[\hat{\theta^2}]} = \sqrt{\frac{p(1-p)}{n}}$

Or do I have to find the distribution for Root of X?

Thank you!

• Have you learned about the delta method yet? en.wikipedia.org/wiki/Delta_method – user408433 Jun 7 '17 at 14:08
• For any non-degenerate positive r.v. $E[\sqrt{X}]\neq \sqrt{E[X]}$. – NCh Jun 7 '17 at 14:57
• @Bey Could you explain how to use it in the context of this? :) – Wboy Jun 7 '17 at 23:48
• See my answer below. As pointed out by @NCh, non-linear transforms of a random variable do not transform the mean in the same way (this is Jensen's inequality). – user408433 Jun 8 '17 at 15:15

If you're familiar with the Delta Method, then we can apply it here:

Following your lead, let $p = \theta^2$, $n$ be the sample size and $x$ the number of individuals with the disease. then we know the mle for $p$ is $\hat{p}_n=\frac{x}{n}$. As long as $p \notin \{0,1\}$ then the mle will be asymptotically normal:

$$\sqrt{n}[\hat{p}_n-p] \xrightarrow{D}\mathcal{N}(0,p(1-p))$$

Let's define $g(p) = \sqrt{p} \implies g'(p) = \frac{1}{2\sqrt{p}}$. By the Delta Method we get the asymptotic distribution of $\sqrt{\hat{p}_n}$:

$$\sqrt{n}[\sqrt{\hat{p}}_n-\sqrt{p}] \xrightarrow{D}\mathcal{N}\left(0,p(1-p)\left[\frac{1}{2\sqrt{p}}\right]^2\right) = \mathcal{N}\left(0,\frac{1-p}{4}\right)$$

Now, since $\frac{x}{n}=\hat{p} = \widehat{\theta^2}$ is a maximum likelihood estimator, then we can indeed invoke invariance of mles to say that $\hat{\theta}_n = \sqrt{\frac{x}{n}}$ is the mle of $\theta$. The asymptotic distribution of $\hat{\theta}_n$ is given by the formula above by substituting $\theta = \sqrt{p}$:

$$\sqrt{n}[\hat{\theta}_n-\theta] \xrightarrow{D} \mathcal{N}\left(0,\frac{1-\theta^2}{4}\right) \implies \hat{\theta} \;\dot{\sim}\; \mathcal{N}\left(\theta,\frac{1-\theta^2}{4n}\right)$$