MLE of fourth moment of normal distribution Take $X\sim N(0,\theta)$, and let $\phi = E(X^4)$, the fourth moment. What is its MLE, $\hat{\phi}$, and what is the asymptotic distribution of $\sqrt{n}(\hat{\phi} - \phi) $ as $n\to \infty$? Any help with this question would be appreciated, as I really can't think of where to start!! Many thanks.
I have integrated the expression for $E(X^4)$ by parts three times to get $\phi = 6\theta^3$, which gives $\hat{\phi} = 6\hat{\theta}^3$ as the function is continuous (is this correct?). However, I am not sure what the MLE of $\theta$ should be, and what that makes the asymptotic distribution $\sqrt{n}(\hat{\phi} - \phi) $.
 A: I assume you mean $\theta=\operatorname E(X^2)$.
The fourth moment is
$$
\operatorname E(X^4) = 3\theta^2.
$$
If you can find the MLE $\hat\theta$ for $\theta$, then the MLE for $3\theta^2$ is just $3\hat\theta^2$.  Something useful to know about MLEs is that if $g$ is a function, and which function $g$ is does not depend on any parameters being estimated, then the MLE of $g(\alpha)$ is $g(\hat\alpha)$ where $\hat\alpha$ is the MLE of $\alpha$.
(The asymptotic distribution is something I'll look at later.  Probably the delta method will work.)
PS: I see you're saying you got $6\theta^3$.  Having a third power there doesn't make sense.  Let $\varphi$ be the standard normal density, so the standard distribution is
$\varphi(x)\,dx$.  Then the $N(0,\theta)$ distribution (assuming you mean by that that the variance, not the standard deviation, is $\theta$, is
$$
\varphi\left(\frac{x}{\sqrt{\theta}}\right)\,\frac{dx}{\sqrt{\theta}}.
$$
So
\begin{align}
\operatorname E(X^4) & = \int_{-\infty}^\infty x^4 \varphi\left(\frac{x}{\sqrt{\theta}}\right)\,\frac{dx}{\sqrt{\theta}} = \theta^2 \int_{-\infty}^\infty \frac{x^4}{\theta^2} \varphi\left(\frac{x}{\sqrt{\theta}}\right)\,\frac{dx}{\sqrt{\theta}} \\[10pt]
& = \theta^2 \int_{-\infty}^\infty u^4 \varphi(u)\,du = \theta^2 \operatorname E (Z^4)
\end{align}
where $Z\sim N(0,1)$.  So it has to be proportional to $\theta^2$.
PPS:  OK, let's look at the integral
\begin{align}
& \phantom{={}}\int_{-\infty}^\infty z^4\varphi(z)\,dz = 2\int_0^\infty z^4\varphi(z)\,dz \\[8pt]
& = \frac{2}{\sqrt{2\pi}}\int_0^\infty z^3 \exp(-z^2/2)\Big(z\,dz\Big) \\[8pt]
& = \frac{2}{\sqrt{2\pi}} \int_0^\infty \sqrt{2w\,{}}^3 \exp(-w)\,dw \\[8pt]
& = \frac{2}{\sqrt{2\pi}}\sqrt{2}^3\Gamma(5/2) = \frac{1}{\sqrt{2\pi}}\sqrt{2}^3\cdot\frac12\cdot\frac32\cdot\Gamma(1/2) \\[8pt]
& = \frac{2}{\sqrt{2\pi}}\cdot\sqrt{2}^3\cdot\frac12\cdot\frac32\cdot\sqrt{\pi} \\[10pt]
& = 3.
\end{align}
