Fourth-moment of a mean of normal random variables.

Question

If $X_1, \dots, X_n$ are i.i.d r.v. of $X \sim N(\mu, 1)$, and $\hat{g}(X)=\bar{X}^2 - \frac{1}{n}$ is an unbiased estimator for $g(\mu) = \mu^2$. (I already checked that).

More, I've calculated the minimal variance, it's $\frac{4\mu^2}{n}$.

Now, I have to calculate $Var(\hat{g}(X))$ to check if it's equal to the minimal variance above.

So far I did: $Var(\hat{g}(x)) = Var(\bar{x}^2 - \frac{1}{n}) = E(\bar{x}^4) - E(\bar{x}^2)^2$.

Don't know how to calculate the fourth-moment.

Answer 1

Write $\overline{x}=\mu+\frac{Z}{\sqrt{n}}$ so $Z\sim N(0,\,1)$ and $$E(\overline{x}^2)=E\left(\mu^2+\frac{Z^2}{n}\right)=\mu^2+\frac{1}{n}$$ (note we never need odd powers of $Z$ when we calculate means), while $$E(\overline{x}^4)=E\left(\mu^4+\frac{6\mu^2Z^2}{n}+\frac{Z^4}{n^2}\right)=\mu^4+\frac{6\mu^2}{n}+\frac{3}{n^2}.$$(I'll discuss how to prove $EZ^4=3$ in a moment.) Hence $$E(\overline{x}^4)-E(\overline{x}^2)^2=\frac{4\mu^2}{n}+\frac{2}{n^2}.$$Note that $$EZ^4=\left.\int_{\Bbb R}\frac{z^4}{\sqrt{2\pi}}\exp(-\alpha z^2)dz\right|_{\alpha=1/2}=\partial_\alpha^2\left.\int_{\Bbb R}\frac{1}{\sqrt{2\pi}}\exp(-\alpha z^2)dz\right|_{\alpha=1/2}\\=\partial_\alpha^2\left.\frac{\alpha^{-1/2}}{\sqrt{2}}\right|_{\alpha=1/2}=\left.\frac{3\alpha^{-5/2}}{4\sqrt{2}}\right|_{\alpha=1/2}=3.$$

Answer 2

For $i=1,\dots,n$ let $U_{i}=X_{i}-\mu$ so that $\mathbb{E}\overline{U}=\mathbb{E}\frac{1}{n}\sum_{i=1}^{n}U_{i}=0$ and $\mathsf{Var}\left(\overline{U}\right)=\mathbb{E}\overline{U}^{2}=n^{-1}$.

Now set $V=n^{\frac{1}{2}}\overline{U}$ so that $V$ has standard normal distribution.

Then $\overline{X}=\overline{U}+\mu=n^{-\frac{1}{2}}V+\mu$ and consequently $\overline{X}^{2}=n^{-1}V^{2}+2n^{-\frac{1}{2}}\mu V+\mu^{2}$.

Then:

$$\begin{aligned}\mathsf{Var}\left(\overline{X}^{2}\right) & =\mathsf{Var}\left(n^{-1}V^{2}+2n^{-\frac{1}{2}}\mu V+\mu^{2}\right)\\ & =\mathsf{Var}\left(n^{-1}V^{2}+2n^{-\frac{1}{2}}\mu V\right)\\ & =n^{-2}\mathsf{Var}\left(V^{2}\right)+4n^{-\frac{3}{2}}\mu \mathsf{CoVar}\left(V^{2},V\right)+4n^{-1}\mu^{2}\mathsf{Var}\left(V\right)\\ & =n^{-2}\mathsf{Var}\left(V^{2}\right)+4n^{-\frac{3}{2}}\mu \mathsf{CoVar}\left(V^{2},V\right)+4n^{-1}\mu^{2} \end{aligned} $$

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Can you find $\mathsf {Var}(V^2)$ and $\mathsf{CoVar}(V^2,V)$ yourself?

For finding $\mathbb EV^4$ just have a look at Wikipedia.

(Also check me on mistakes)