Stochastic process, stochastic differential equation A stochastic process $\{X_t
, t ≥ 0\}$ satisfies stochastic differential equation
$$\frac{dX_t}{X_t} = 3\mu\ dt + 2\sigma dB_t.$$
where $-\infty<\mu<\infty$ and $\sigma>0$ are given constants, and $\{B_t
, t \geq 0\}$ is the standard Brownian motion.
(a) Show that the solution of the stochastic differential equation is given by
$$X_t = X_0e^{(3\mu − 2\sigma^2)t + 2\sigma B_t}.$$
(b) Assume $X_0 = 1$. For a given time $t> 0$, find the probability density function of $X_t$.
.
 A: a) 
We use Ito's formula on the function $f(t, S_t) = \ln(S_t)$. 
This gives 
\begin{align}
df(t, S_t) &= \frac{\partial f}{\partial t}dt + \frac{\partial f}{\partial S_t}dS_t +\frac{1}{2} \frac{\partial ^2f}{\partial S_t^2}(dS_t)^2\\
&= \frac{1}{S_t}dS_t - \frac{1}{2S_t^2}(dS_t)^2.  
\end{align} 
Using that
\begin{align}(dS_t)^2 &= (3\mu S_t)^2(dt)^2 + 3\mu  2\sigma S_t^2 dB_tdt +(2\sigma)^2S_t^2 (dB_t)^2\\ 
&= (2\sigma)^2S_t^2 dt, 
\end{align}
we find 
\begin{align}
\frac{dS_t} { S_t} &= dln(S_t) + 2\sigma^2 dt.
\end{align}
We recoognize the LHS. as $3\mu + 2\sigma dBt$
Integrating both sides:
\begin{align}
3\mu t +2\sigma Bt &= ln(S_t) - ln(S_0) + 2\sigma^2t\\
ln\left(\frac{S_t}{S_0}\right) &= 3\mu t + 2\sigma Bt - 2\sigma^2 t\\
S_t &= S_0 \exp\left((3\mu  - 2\sigma^2)t  + 2\sigma Bt \right)
\end{align}
b) For a fixed t, only $Bt$ contributes to the randomness. 
\begin{align}
S_t &= S_0\exp(3\mu -2\sigma^2t)\exp(2\sigma Bt).
\end{align}
We know that in distribution,  \begin{align}
2\sigma Bt = 2\sigma \sqrt{t} Z, 
\end{align} 
for $Z \sim \mathcal{N}(0, 1)$. Hence 
\begin{align}
S_t \sim \exp(3\mu -2\sigma^2t + 2\sigma \sqrt{t}Z).
\end{align}
