expectation of a nonlinear function of a Gaussian random variable

Question

I am reading a book on Monte Carlo Simulation and I want to know where the formula below come from.

$$E\left[ X^4\right] {\text{ }} = 3\sigma _x^4 + 6\sigma _x^2\mu _x^2 + \mu _x^4$$

Suppose $X$ is a Gaussian random variable: $X \sim N\left({\mu_x},\sigma _x^2\right)$. Thanks.

Answer 1

This follows from the computation of $$ \mathbb{E}[X^4] = \frac{1}{\sqrt{2\pi}\sigma_x}\int_{-\infty}^\infty x^4 e^{-\frac{(x-\mu_x)^2}{2\sigma_x^2}}dx $$ for instance (the definition of the 4th raw moment of a Gaussian r.v.). See e.g. this page on Wolfram Mathworld, Eq. (31) and (36).

Answer 2

The easiest way is to use moment generating function. MGF of Gaussian random variable is \begin{align} M_X(t)=e^{t\mu_x+\frac{1}{2}t^2\sigma_x^2}=\mathsf{E}(e^{tX})~\text{By def.}\\ \end{align} Now using Taylor series of exponential function, expand both RHS and LHS of above equation, and equate the 4th order terms on both sides you will get 4th moment of Gaussian random variable. i.e., \begin{align} 1+t\mu_x+\frac{1}{2}t^2\sigma_x^2+\left(t\mu_x+\frac{1}{2}t^2\sigma_x^2\right)^2\frac{1}{2!}+\ldots =1+t\mathsf{E}(X)+\frac{t^2}{2!}\mathsf{E}(X^2)+\frac{t^3}{3!}\mathsf{E}(X^3)+\frac{t^4}{4!}\mathsf{E}(X^4)+\ldots \end{align} Comparing 4th power of $t$ on both sides gives the answer.