Standard normal moments

Question

My textbook says, without proof, that $E(X^4)=3$, where $X\sim N(0,1)$. Is it so simple to obtain?

Answer 1

If you start with the moment generating function $M_{X}(t)$, one can quickly obtain arbitrary moments. The moment generating function for a standard normal distribution is given by $$ \begin{aligned} M_{X}(t) &= \operatorname E[\exp(tX)]\\ &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} dX \exp(tX)\exp\left(-\frac{1}{2}X^2\right)\\ &= \exp\left(\frac{1}{2}t^2\right), \end{aligned} $$ where I quoted the answer from here, but this can also be easily calculated by completing the square.

Because $$ \exp(tX) = 1 + t\,X + \frac{t^2\,X^2}{2!} + \frac{t^3\,X^3}{3!} + \cdots +\frac{t^n\,X^n}{n!} + \cdots, $$ subsequently, $$ M_X(t) = 1 + t \operatorname E (X) + \frac{t^2 \operatorname E (X^2)}{2!} + \frac{t^3\operatorname E (X^3)}{3!}+\cdots + \frac{t^n\operatorname E (X^n)}{n!}+\cdots. $$ Therefore by differentiating the moment generating function, you can obtain the desired moment. In particular, the fourth moment $\operatorname E[X^4]$ is given by $$ \begin{aligned} \operatorname E[X^4] &= \frac{d^4}{dt^4}M_{X}(t) \bigg\vert_{t=0}\\ &= \frac{d^4}{dt^4}\exp\left(\frac{1}{2}t^2\right) \bigg\vert_{t=0}\\ &= \exp\left(\frac{1}{2}t^2\right) (t^4 + 6 t^2 + 3) \bigg\vert_{t=0}\\ &= 3. \end{aligned} $$