Let $\{ X_t \}_{t \ge 0}$ defined by $ X_t := 10+3t+3B_t$ where $\{ B_t: t \ge 0 \}$ is a standard Brownian Motion. Let $\{ S_t \}_{t \ge 0}$ defined by $S_t := e^{X_t}$
a) Write explicitly the density of $S_5$
b) Compute $P(S_1 \gt S_0)$
c) Calculate $P(S_1 \gt S_3, S_1 \gt0)$
What I've done:
S is a Geometric Brownian Motion which starts on 10, with drift coefficient 3 and variance 9.
a) $P(S_5 \le x)=P(e^{10+15+3B_5}\le x)=P(B_5\le \frac{lnx-25}{3})=P(Z\le \frac{lnx-25}{3 \sqrt 5})$ where $Z \sim N(0,1)$ since $B_5 \sim N(0,5)$ then $P(S_5 \le x)={\int}_{- \infty}^{\frac{lnx-25}{3}} \frac {1}{\sqrt{2 \pi}} e^{-y^2/2} dy$ In order to find the distribution I should derivate but I'm not sure how to do it and I'm not sure if what I did is correct.
b) $P(S_1 \gt S_0)= P(\frac {S_1} {S_0} \gt 1)= P(\frac{e^{10+3+3B_1}}{e^{10}}\gt 1) =P(3+3B_1>ln(1)) =P(B_1 \gt -1)$ Due to $B_1 \sim N(0,1)$ it is easy to get the value.
c) $P(S_1 \gt S_3, S_1 \gt0)=P(0 \gt S_3 - S_1, S_1\gt 0) =P(S_3 - S_1 \lt 0)P(S_1\gt 0)$ . How can I compute $P(S_3 - S_1 \lt 0)$?
a) $P(S_5\leq x)=P(Z\leq \frac{lnx-25}{3\sqrt{5}})=\int_{-\infty}^{\frac{lnx-25}{3\sqrt{5}}}\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}dy=\int_{-\infty}^{\frac{lnx-25}{3\sqrt{5}}}I(y)dy$.
So, thanks to fundamental theorem of calculus $f_{S_5}(x)=I(\frac{lnx-25}{3\sqrt{5}})\frac{\partial}{\partial x}(\frac{lnx-25}{3\sqrt{5}})$ is the density of $S_5$.
c) $P(S_3-S_1<0) = P(e^{3(B_3-B_1)} < e^{-6}) = P(B_3-B_1<-3)$
But $(B_t)_t$ is a Brownian Motion so $B_t-B_s \sim N(0,t-s)$ (for $0\leq s\leq t$), so $Y=B_3-B_1 \sim N(0,2)$.
Then, $P(Y<-3)=P(Z<-\frac{3}{\sqrt{2}})$ with $Z\sim N(0,1)$.