Let $ X_1,X_2,…,X_n$ be i.i.d. $N(\theta_1, \theta_2)$, please prove that $E[(X_1-\theta_1)^4] = 3\theta_2^2$ If $X_{1}$, $X_{2}$, ..., $X_{n}$ is sampled from $N(\theta_1, \theta_2)$, how can I prove that $E [(X_{1} - \theta_1)^{4}] = 3 \theta_2^{2}$?
I started off this question finding the completely sufficient statistics. This can be done because the normal distribution of of exponential class. The results are $Y_1 = \sum\limits_iX_{i}$ and $Y_2 =\sum\limits_iX_{i}^{2}$
Therefore I know $E(\bar X) =\theta_1$, but how can I come up with the $\theta_2$?
Also, because of how the question look, I think I would need to use the formula: $\operatorname{Var}(X) = E[X^{2}] - E(X)^{2}$
Can someone give me some ideas on how to proceed? Help is much appreciated!
 A: Fact 1: If the distribution of $X$ is $\mathfrak N(\mu,\sigma^2)$, then $X=\mu+\sigma Y$ where the distribution of $Y$ is $\mathfrak N(0,1)$.
Fact 2: If the distribution of $X$ is $\mathfrak N(0,1)$, then $\mathbb E(X^4)=3$.
Conclusion: If the distribution of $X$ is $\mathfrak N(\mu,\sigma^2)$, then $\mathbb E((X-\mu)^4)=3\sigma^4$.
Can you prove these and reach the conclusion from them?
Hint for fact 2: For every integer $n$, $x^{2n}\mathrm e^{-x^2/2}=-u(x)v'(x)$ with $u(x)=x^{2n-1}$ and $v(x)=\mathrm e^{-x^2/2}$. Then $u'(x)=(2n-1)x^{2n-2}$ and $u(x)v(x)\to0$ when $x\to\pm\infty$, hence an integration by parts yields that, for every $X$ with distribution $\mathfrak N(0,1)$,
$$
\sqrt{2\pi}\mathbb E(X^{2n})=-\int u(x)v'(x)\mathrm dx=\int u'(x)v(x)\mathrm dx=\sqrt{2\pi}(2n-1)\mathbb E(X^{2n-2}).
$$
Starting from $\mathbb E(X^2)=1$ (or, for the adventurous ones, from $\mathbb E(X^0)=1$), one gets $\mathbb E(X^4)=3$ and, more generally, $\mathbb E(X^{2n})=(2n-1)(2n-3)\cdots3\cdot1$ for every nonnegative integer $n$.
A: That $X_1\sim N(\theta_1,\theta_2)$ implies that $(X_1-\theta_1)\sim N(0,\theta_2)$
The characteristic function of $N(0,\theta_2)$ is $\varphi(t)=\exp\left(-\frac12\theta_2t^2\right)$. Therefore the fourth moment is determined as
$$
E\left[(X_1-\theta_1)^4\right]=
i^{-4}\varphi^{(4)}(0)
= \left[(\theta_2^4t^4 - 6\theta_2^3t^2  + 3\theta_2^2)\varphi(t)\right]_{t=0}
= 3\theta_2^2.
$$
A: $$E[g(X)] = \int_{-\infty}^\infty g(x)\frac{1}{\sqrt{2 \pi \theta_2^2}} \exp \left( \frac{(x-\theta_1)^2}{2 \theta_2^2}\right) \mathrm{d}x$$
$$E[(X - \theta_1)^n] = \int_{-\infty}^\infty (x-\theta_1)^n\frac{1}{\sqrt{2 \pi \theta_2^2}} \exp \left( \frac{(x-\theta_1)^2}{2 \theta_2^2}\right) \mathrm{d}x$$
Now let $n=4$, and integrate.
