Moment Generating Function of Normal
MGF,
\[M_X(t) = \mathbb E[e^{tX}], t\in \mathbb R\]
For \(X\sim N(\mu, \sigma^2)\),
\[\begin{aligned} M_X(t) &= \int_{-\infty}^\infty{e^{tx} \frac{1}{\sqrt{2\pi} \sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}}dx \\ &= \int_{-\infty}^\infty{e^{t(a\sigma + \mu)} \frac{1}{\sqrt{2\pi} }e^{-\frac{a^2}{2}}}da \qquad \text{substitute x to a} \\ &= \int_{-\infty}^\infty{ \frac{1}{\sqrt{2\pi} } \exp{ \left( \frac{-(a-\sigma t)^2 +\sigma^2t^2+2\mu t}{2} \right)} }da \\ &= \exp { \left(\mu t + \frac{\sigma^2t^2}{2} \right)} \int_{-\infty}^\infty{ \frac{1}{\sqrt{2\pi} } \exp{ \left( \frac{-(a-\sigma t)^2}{2} \right)} }da \qquad \text{by the fact that gaussian cdf = 1} \\ &= \exp { \left(\mu t + \frac{\sigma^2t^2}{2} \right)} \end{aligned}\]
when \(\mu = 0\)
\[\mathbb E [e^X] = M_X(1) = e^{\mu + \sigma^2 /2} = e^{\sigma^2 /2}\]
Exponential Martingale
Theorem: Let \(W(t), t\geq 0\) be a Brownian motion with filtration \(\mathcal{F}_t, t\geq0\), and let \(\sigma\) be a constant. The process \(Z(t), t\geq 0\) of such form is a martingale. \[Z(t) = \exp{\left\{ \sigma W(t) - \frac12 \sigma^2 t \right\}}\]
Proof:
\[\begin{aligned} &\mathbb E [Z(t)|\mathcal F (s)] \\ &= \mathbb E\left[\exp{\left\{ \sigma W(t) - \frac12 \sigma^2 t \right\}} \bigg| \mathcal F(s) \right] \\ &= \mathbb E\left[\exp{\left\{ \sigma( W(t) - W(s)) + \sigma W(s)- \frac12 \sigma^2 t \right\}} \bigg| \mathcal F(s) \right] \\ &= \mathbb E\left[\exp{\left\{ \sigma( W(t) - W(s))\right\}} \cdot \exp{\left\{ \sigma W(s) - \frac12 \sigma^2 t \right\}} \bigg| \mathcal F(s) \right] \\ &= \exp{\left\{ \sigma W(s) - \frac12 \sigma^2 t \right\}} \cdot \mathbb E\left[\exp{\left\{ \sigma( W(t) - W(s))\right\}} \bigg| \mathcal F(s) \right] \\ &= \exp{\left\{ \sigma W(s) - \frac12 \sigma^2 t \right\}} \cdot \mathbb E\left[\exp{\left\{ \sigma( W(t) - W(s))\right\}} \right] \\ &= \exp{\left\{ \sigma W(s) - \frac12 \sigma^2 t \right\}} \cdot \exp{\left\{ \frac{\sigma^2(t-s)}{2} \right\}} \\ &= \exp{\left\{ \sigma W(s) - \frac12 \sigma^2 s \right\}} \\ &= Z(S) \end{aligned}\]