Geometric Brownian motion probability question

Question

Let $S(t)$, $t \geq 0$ be a geometric Brownian motion with drift parameter $\mu = 0.3$ and volatility parameter $\sigma = 0.3$. Find

$(a) P(S(1) > S(0))$

$(b) P(S(2)) > S(0))$

I think I did $(a)$ right as follows:

$P(S(1) > S(0)) = P(S(1)/S(0) > 1) = P(\log(S(1)/S(2)) > 0) = P(Z > -2/3) = 0.2525.$

I don't know how to do $b$ though. Can someone please help me?

Answer 1

I think you can solve $(b)$ using the same way as you did in $(a)$. \begin{align*}P(S(2)>S(0))&=P(S(2)/S(0)>1)\\&=P(\log(S(2)/S(1))>0)\\&=P(Y>0)\end{align*} where $Y$ is also Gaussian, which can be derived from the property of the brown motion. Be careful about the expectation and variance of $Y$, which are different with the expectation and variance of $Z$.

Answer 2

Let $(B(t))_{t\ge 0}$ be the standard brownian motion.

The individual random variable $B(t)$ has thus mean $0$ and variance $\sigma_{B(t)}^2=t$.

Then we have en.wikipedia.org/wiki/Geometric_Brownian_motion $$ S(t) = S(0)\exp\left(\left(\mu-\frac 12\sigma^2\right)t+\sigma B(t)\right)\ . $$ The probabilities to be computed are then (for a more general time $t$): $$ \begin{aligned} \Bbb P(\ S(t)>S(0)\ ) &= \Bbb P\left(\ \frac{S(t)}{S(0)}>1\ \right) \\ &= \Bbb P\left(\ \exp\left(\left(\mu-\frac 12\sigma^2\right)t+\sigma B(t)\right)>1\ \right) \\ &= \Bbb P\left(\ \left(\mu-\frac 12\sigma^2\right)t+\sigma B(t)>0\ \right) \\ &= \Bbb P\left(\ B(t)>-\frac 1\sigma\left(\mu-\frac 12\sigma^2\right)t\ \right) \\ &= \Bbb P\left(\ \frac{B(t)}{\sqrt t} < + \frac 1\sigma\left(\mu-\frac 12\sigma^2\right)\sqrt t\ \right) \\ & =\Phi\left(\ \frac 1\sigma\left(\mu-\frac 12\sigma^2\right)\sqrt t\ \right) \ . \end{aligned} $$ Here, $\Phi$ is the repartition of an $N(0,1^2)$ random variable. The computer gives now (via sage).

sage: si, mu = 0.3, 0.3
sage: for t in (1, 2):
....:     print "t=%s probability=%f" % (t, T.cum_distribution_function( (mu-si^2/2)/si*sqrt(t) ))
....: 
t=1 probability=0.802337
t=2 probability=0.885334