# MFG Normal Distribution

### Proof of moment Generating Function of the Normal Distribution

$X \sim N(\mu, \sigma^2)$

$M_X(t)=E(e^{Xt})=\frac{1}{\sqrt{2 \pi}\sigma} \displaystyle\int_{-\infty}^{\infty}e^{xt}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx$

Just think multiply probability by function value and add up

Let $A=\frac{1}{\sqrt{2 \pi}\sigma} \displaystyle\int_{-\infty}^{\infty}e^{xt}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx$

We now want to re-arrange integral so the part inside the integral sums to 1 and we factor everything else outside

Consider

$xt-\frac{(x-\mu)^2}{2\sigma^2}$ i.e. the exponent

$=xt-\frac{x^2-2\mu x +\mu^2}{2\sigma^2}$

$=-\frac{1}{2\sigma^2} \left[-2 \sigma^2 x t + x^2 - 2\mu x + \mu^2 \right]$

$=-\frac{1}{2\sigma^2} \left[x^2 + x(-2\sigma^2 t - 2\mu) + \mu^2 \right]$

$=-\frac{1}{2\sigma^2} \left[(x-\sigma^2 t - \mu)^2+ \mu^2 -(\sigma^2 t + \mu)^2 \right]$

$A=\frac{1}{\sigma \sqrt{2 \pi}} e^{\frac{\mu^2-(\sigma^2 t + \mu)^2}{-2 \sigma^2}} \displaystyle\int_{-\infty}^{\infty}e^{\frac{-(x-\sigma^2 t - \mu)^2}{2\sigma^2}}dx$

so the term inside the integral is now just a standard normal distribution translated along the x-axis

$A=e^{\frac{\mu^2-(\sigma^2 t + \mu)^2}{-2 \sigma^2}}$

$A=e^{\frac{\sigma^2 t^2}{2} + \mu t}$