Assume that $B(t)$ is a Brownian motion and that $S(t)$ is defined as $S(t)=A\cdot e^{B(t)}$ for some positive constant $A$. Calculate the probability of the event ${S(3)>2S(1)}$.
How could I go about calculating such a probability? I believe it'll involving finding the probability that $S(3) - 2S(1) > 0$
This is what I've done so far:
$B(T) = \sqrt{t} * N$ where $N$ is the standard normal.
With a bit of algebra,
$B(3) > 2*B(1)$
I am stuck trying to find the probability that:
$P[B(3) - 2*B(1) > 0]$
I am omitting any algebra details or anything from probability that you should know by now. Observe: $$\newcommand{\P}[1]{\mathbb{P}\left[#1 \right]} \begin{align*} \P{S(3) > 2S(1)} &= \P{Ae^{B(3)}>2Ae^{B(1)}} \\ &= \P{e^{B(3)}>2e^{B(1)}} \text{ since }A > 0 \\ &= \P{\dfrac{e^{B(3)}}{e^{B(1)}} > 2} \\ &= \P{e^{B(3)-B(1)} > 2}\text{.} \end{align*}$$ Notice that $B(3)-B(1) \sim \mathcal{N}(0, 2)$ (see here for example). Hence, $$\dfrac{B(3)-B(1)}{\sqrt{2}} \sim \mathcal{N}(0, 1)\text{.}$$ Thus, $$\begin{align*} \P{S(3) > 2S(1)} &= \P{B(3)-B(1)>\ln(2)} \\ &= \P{\dfrac{B(3)-B(1)}{\sqrt{2}} > \dfrac{\ln(2)}{\sqrt{2}}} \\ &= 1 - \Phi\left(\dfrac{\ln(2)}{2} \right) \end{align*}$$ where $\Phi$ is the standard normal CDF. Note that some people use $N$ instead of $\Phi$, especially when using Brownian motion or similar concepts.
We can calculate $\Phi\left(\dfrac{\ln(2)}{2} \right)$ using a table, or using
R:so our final answer is $1-0.6355442 \approx 0.364$.