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 $X_t$. .

  • $\begingroup$ You need to integrate both sides of your equation. The right hand side is easy. You might guess that the left hand side will evaluate to something like $\log X_t$ use Ito's lemma on this guess to check for any additional quadratic variation terms. $\endgroup$ Jun 3, 2020 at 0:19

1 Answer 1



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}


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