1. ## normal distribution (mgf)

Hi

Can someone please explain the process in creating the moment generating function for the normal distribution? i can do it for the other special distributions but not the normal.

thanks.

2. Originally Posted by gvidfhi
Hi

Can someone please explain the process in creating the moment generating function for the normal distribution? i can do it for the other special distributions but not the normal.

thanks.
The easiest method to derive the moment-generating function of a general normal distribution $N(\mu, \sigma^2)$ is to find the moment for a standard normal $N(0,1)$ and then use the formula for the linear transformation of a moment. Given $(\Omega, F, P)$, we have a probability space.

Lemma: Let $X: \Omega \longrightarrow R$ be an absolutely continuous random variable whose moment-generating function is $M_x (s)$. Then if $Y = aX +b$, then $M_y (s) = e^{sb} M_x (sa)$

Proof: Let a second random variable $Y = aX + b$. Then the moment generating function for $Y$ is $M_y (s) = E(e^{sY})$
$= E(e^{s(aX+b)})$
$= E(e^{saX+sb})$
$= E(e^{saX} e^{sb})$
$= e^{sb} E(e^{saX})$
$= e^{sb} M_x (sa)$

Therefore, if $Y = aX +b$, then $M_y (s) = e^{sb} M_x (sa)$ - this completes the lemma.

Consider $X$, which is standard normally distributed with mean 0 and variance 1, so that $f_x (x) = \frac{1}{\sqrt{2 \pi}} e^{-x^2/2}$

The moment generating function for $X$ is calculated by $M_x (s) = E(e^{sx})$
$= \int_{-\infty}^{\infty} e^{sx} \frac{1}{\sqrt{2 \pi}} e^{-x^2/2}$
$= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{sx -x^2/2}$
$= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{- \frac{1}{2} ((x-s)^2 - s^2)}$
$= e^{s^2/2} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{- \frac{1}{2} (x-s)^2}$
$= e^{s^2/2}$

Finally, consider $Y$, which is normally distributed with mean $\mu$ and variance $\sigma$, so that $Y = aX +b = \sigma X + \mu$

Recalling that $M_y (s) = e^{sb} M_x (sa)$ from our lemma, we have $M_y (s) = e^{s \mu} M_x (s \sigma) = e^{s \mu} e^{s^2 \sigma^2/2} = e^{s \mu + s^2 \sigma^2/2} \qquad QED$