Conditional expectation of geometric brownian motion

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 $$\mathbb{E}[e^{\sigma B_t}]=e^{\frac{1}{2}\sigma^2t},\qquad \sigma\in\mathbb{R}.$$

$$\mathcal F(s)$$ is the filtration of $$B(t)$$ for $$t, hence, $$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.

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

• This is what the question states; I don't see how you are answering any of the points raised by the poster. – Riccardo Sven Risuleo Mar 5 '19 at 16:27
• @RiccardoSvenRisuleo I made an edit to my problem after his answer. – Blade Mar 5 '19 at 16:57
• I apologize for that then! – Riccardo Sven Risuleo Mar 5 '19 at 16:58