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 $\displaystyle N(\mu, \sigma^2)$ is to find the moment for a standard normal $\displaystyle N(0,1)$ and then use the formula for the linear transformation of a moment. Given $\displaystyle (\Omega, F, P)$, we have a probability space.
Lemma: Let $\displaystyle X: \Omega \longrightarrow R$ be an absolutely continuous random variable whose moment-generating function is $\displaystyle M_x (s)$. Then if $\displaystyle Y = aX +b$, then $\displaystyle M_y (s) = e^{sb} M_x (sa)$
Proof: Let a second random variable $\displaystyle Y = aX + b$. Then the moment generating function for $\displaystyle Y$ is $\displaystyle M_y (s) = E(e^{sY})$
$\displaystyle = E(e^{s(aX+b)})$
$\displaystyle = E(e^{saX+sb})$
$\displaystyle = E(e^{saX} e^{sb})$
$\displaystyle = e^{sb} E(e^{saX})$
$\displaystyle = e^{sb} M_x (sa)$
Therefore, if $\displaystyle Y = aX +b$, then $\displaystyle M_y (s) = e^{sb} M_x (sa)$ - this completes the lemma.
Consider $\displaystyle X$, which is standard normally distributed with mean 0 and variance 1, so that $\displaystyle f_x (x) = \frac{1}{\sqrt{2 \pi}} e^{-x^2/2}$
The moment generating function for $\displaystyle X$ is calculated by $\displaystyle M_x (s) = E(e^{sx})$
$\displaystyle = \int_{-\infty}^{\infty} e^{sx} \frac{1}{\sqrt{2 \pi}} e^{-x^2/2}$
$\displaystyle = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{sx -x^2/2}$
$\displaystyle = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{- \frac{1}{2} ((x-s)^2 - s^2)}$
$\displaystyle = e^{s^2/2} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{- \frac{1}{2} (x-s)^2}$
$\displaystyle = e^{s^2/2}$
Finally, consider $\displaystyle Y$, which is normally distributed with mean $\displaystyle \mu$ and variance $\displaystyle \sigma$, so that $\displaystyle Y = aX +b = \sigma X + \mu$
Recalling that $\displaystyle M_y (s) = e^{sb} M_x (sa)$ from our lemma, we have $\displaystyle M_y (s) = e^{s \mu} M_x (s \sigma) = e^{s \mu} e^{s^2 \sigma^2/2} = e^{s \mu + s^2 \sigma^2/2} \qquad QED$