Conditional expectation of geometric brownian motion

Question

Given a geometric Brownian motion $S ( t ) = e ^ { \mu t + \sigma B ( t ) }$, I'm trying to calculate $E [ S ( t ) | \mathcal { F } ( s ) ]$ where $\mathcal { F } ( s )$ is the history of the process. Here is my try:

This is conditioned on history of the process $\mathcal { F } ( s )$, so we need to rewrite $B(t)$ as $B ( s ) + ( B ( t ) - B ( s ) )$

\begin{align*} S ( t ) &= e ^ { \mu t + \sigma B ( t ) }\\ &= e ^ { \mu t + \sigma (B ( s ) + ( B ( t ) - B ( s ) )) }\\ &= e ^ { \mu t + \sigma B ( s ) + \sigma\left( B ( t ) - B ( s ) \right) } \end{align*}

\begin{align*} \mathbb{E}[S ( t )|\mathcal { F } ( s )] &= \mathbb{E}[e ^ { \mu t + \sigma B ( s ) + \sigma\left( B ( t ) - B ( s ) \right) }]\\ &= \mathbb{E}[e ^ { \mu t + \sigma B ( s )}e^{\sigma\left( B ( t ) - B ( s ) \right) }]\\ \end{align*}

Edit: Now here is my problem: I see that many online solutions proceed as following

$$\mathbb{E}[S ( t )|\mathcal { F } ( s )] = e ^ { \mu t + \sigma B ( s )}\mathbb{E}[e^{\sigma\left( B ( t ) - B ( s ) \right) }]=e ^ { \mu t + \sigma B ( s )}e^{\sigma^{2}(t-s)/2}$$

But I don't understand 2 things:

1. How the first term comes out of expectation.

2. What does it mean when we say Using moment generating function, we know that $$\mathbb{E}[e^{\sigma B_t}]=e^{\frac{1}{2}\sigma^2t},\qquad \sigma\in\mathbb{R}.$$

Answer 1

$\mathcal F(s)$ is the filtration of $B(t)$ for $t<s$, hence, $B(s)$ is $\mathcal F(s)$-measurable; this means that, for any measurable function $f(\cdot)$, $$\mathbf E[ f\big(B(s)\big)|\mathcal F(s)] = f\big(B(s)\big).$$

To compute $$\mathbf E[\mathrm e^{\mu t + \sigma B(s)}\mathrm e^{\sigma (B(t) - B(s))}| \mathcal F(s)]$$ we use two facts

1. $\mathrm e^{\mu t + \sigma B(s)}$ is $\mathcal F(s)$-measurable, so it goes out of the conditional expectation (it acts as a constant);
2. Brownian motion has independent and Gaussian increments; so $B(t)-B(s)$ is independent of $\mathcal F(s)$ and is a Gaussian random variable with zero mean and variance equal to the increment $t-s$.

The first fact allows you to move out the first part from the expectation; the second fact allows you to write that $$\mathbf E[\mathrm e^{\sigma (B(t) - B(s))}| \mathcal F(s)] = \mathbf E[\mathrm e^{\sigma Y}]$$ where $Y\sim N(0,t-s)$; then, using the moment generating formula, you have the value of the expectation.

Answer 2

$\mathbb{E}[S ( t )|\mathcal { F } ( s )] = e ^ { \mu t + \sigma B ( s )}\mathbb{E}[e^{\sigma\left( B ( t ) - B ( s ) \right) }]=e ^ { \mu t + \sigma B ( s )}e^{\sigma^{2}(t-s)/2}$