Moment Generating Functions of Normal/Gaussian Random Variable: 3rd, 4th,..,kth Moment

Question

Derivation of Normal Random MGF:

I'm having trouble deriving the answers for $E\,[X^3]$ and $E\,[X^4]$ given the information from the image I've posted. In the image I understand how they setup the equation for normal random distribution $X$: given as $G(\theta)$.

What I am not understanding is how you can go from simply completing the square to finding the 4th moment. Am I simply misinterpreting the equation given in the image or do I need to complete more steps that are not stated in the image?

Answer 1

You've got that your MGF is \begin{equation} G(\theta) = \exp(\mu \theta + \frac{\sigma^2\theta^2}{2}) \end{equation} This means that you can get $E(X^n)$ for any $n \geq 1$, using the following \begin{equation} E(X^n) = \frac{d^n}{d \theta^n} G(\theta) \Big\vert_{\theta=0} =G^{(n)}(0) \tag{1} \end{equation} So, let's get the four derivatives \begin{align} G'(\theta) &= (\sigma^2 \theta + \mu)G(\theta)\\ G''(\theta) &= (\sigma^4 \theta^2 + 2\mu\sigma^2\theta +\sigma^2 + \mu^2)G(\theta)\\ G'''(\theta) &= (\sigma^2 \theta + \mu)(\sigma^4 \theta^2 + 2\mu\sigma^2\theta +3\sigma^2 + \mu^2)G(\theta)\\ G''''(\theta) &= (\sigma^8 \theta^4 + 4 \mu \sigma^6 \theta^3 + 6\sigma^4(\sigma^2 + \mu^2)\theta^2 + 4\mu\sigma^2(3\sigma^2 + \mu^2)\theta + 3\sigma^4 + 6\mu^2\sigma^2 + \mu^4)G(\theta) \end{align} Using equation $(1)$, we get that \begin{align} E(X) &= G'(0)\\ E(X^2) &= G''(0)\\ E(X^3) &= G'''(0)\\ E(X^4) &= G''''(0) \end{align} Replacing \begin{align} E(X) &= \mu \\ E(X^2) &= \sigma^2 + \mu^2\\ E(X^3) &= \mu(3\sigma^2 + \mu^2)\\ E(X^4) &= ( 3\sigma^4 + 6\mu^2\sigma^2 + \mu^4) \end{align}

Answer 2

The completing the square in the exponent is to show $$ G(\theta)=\mathbb{E}e^{\theta X}=\exp\left(\mu\theta+\frac12\sigma^2\theta^2\right) $$ because the integrand $e^{\theta x}f_X(x)$ is $\exp\left(\mu\theta+\frac12\sigma^2\theta^2\right)$ times the probability density function of a $N(\mu+\sigma^2\theta,\sigma^2)$ distribution.

Now you have $G(\theta)$, which you know is analytic about $\theta=0$, you can differentiate it as many times as you please and find the value of $\mathbb{E}[X^n]=G^{(n)}(0)$