Consistency of a estimator, limit of the variance

Question

Let be $X \sim \mathcal{N}(\theta,1)$. It is easy to see that the maximum likelihood estimator (MLE) for $\theta^2$ is $T =\overline{X}^2$ because of the invariance of the MLE. Then I want to know if $T$ is consistent. Since $X \sim \mathcal{N}(\theta,1)$, then $\overline{X} \sim \mathcal{N}(\theta,1/\sqrt{n})$. Therefore $$ E(\overline{X}^2) = V(\overline{X}) + E(\overline{X})^2 = \frac{\sigma}{n} + \theta^2. $$ This say us that $\lim_{n \to \infty}E(\overline{X}^2) = \theta^2$. Now, I want to see that $$ \lim_{n \to \infty}V(\overline{X}^2) = 0. $$ I can't prove this and I can't apply the previous trick. Thank you!

Answer 1

We have to find the forth central moment of a Gaussian then. We have $\overline{X} \sim N(\theta,\frac{1}{n})$ and we have the fact that if $r \sim \mathcal{N}(\mu,\sigma^2)$ (Gaussian with mean $\mu$ and variance $\sigma^2$, standard deviation $\sigma$) then $\mathbb{E}\{(r-\mu)^4\}=3\sigma^4$

(got that from Wiki https://en.wikipedia.org/wiki/Normal_distribution#Moments).

Therefore

$$\lim_{n \to \infty}V(\overline{X}^2)=\lim_{n \to \infty} 3 \left(\frac{1}{\sqrt{n}} \right)^4=\lim_{n \to \infty} \frac{3}{n^2} = 0$$