Solve the stochastic differential equation $dY_t = rdt + \alpha Y_tdB_t$ where $B_t$ is a Brownian motion.

Question

I need some help me with a stochastic differential equation that I am stuck on, since my professor loves to not give us any suggested solution.

I got a hint first to calculate the dynamics of the process $Y_{}tF_{t}$ where $$F_t = exp(-\alpha B_t+0.5\alpha^2t)$$

My solution is as follows:

$$d(F_tY_t) = F_t r dt + \alpha F_t Y_t dB_t + 0.5 \alpha^2 F_t Y_t dt - \alpha F_t Y_t dB_t $$

$$ d(F_t Y_t) = 0.5 \alpha^2 F_t Y_t dt + F_t r dt$$

Take integral of both sides

$$F_t Y_t = \int ^{t}_{0}r F_s ds + 0.5 \alpha^2 \int ^{t}_{0} F_s Y_s ds$$

So I am stuck at this point, any advice?

Answer 1

Since $F_t = \exp(-\alpha B_t + \alpha^2 t / 2)$, we can write $F_t = f(t,B_t)$, where $f(t,x) = \exp(-\alpha x + \alpha^2 t/2)$. Thus, by Itô's Lemma, $$ \mathrm d F_t = \frac 12 \alpha^2 F_t \,\mathrm d t -\alpha F_t \, \mathrm d B_t + \frac 12 \alpha^2 F_t \,\mathrm d t=\alpha^2 F_t \,\mathrm d t -\alpha F_t\, \mathrm d B_t.$$

Then, by the product rule, \begin{align*} \mathrm d(FY)_t &= F_t\,\mathrm d Y_t + Y_t\, \mathrm dF_t + \mathrm d [F,Y]_t \\ &= r F_t\,\mathrm d t + \alpha F_t Y_t \,\mathrm d B_t +\alpha^2 F_t Y_t \,\mathrm dt -\alpha F_t Y_t\,\mathrm d B_t -\alpha^2 F_t Y_t \,\mathrm d t \\ &=r F_t \,\mathrm dt. \end{align*}

This means that \begin{align*} F_tY_t &= F_0 Y_0 + r\int_{[0,t]} F_s \,\mathrm ds \\ &= Y_0 +r\int_{[0,t]} \exp\left( -\alpha B_s + \frac 12 \alpha^2 s \right) \,\mathrm d s. \end{align*}

Since $F_t \neq 0$ for all $t\ge 0$, we have that $$ Y_t = \exp\left( \alpha B_t - \frac 12 \alpha^2 t \right)\left(Y_0 + r\int_{[0,t]} \exp\left( -\alpha B_s + \frac 12 \alpha^2 s \right) \,\mathrm d s\right). $$

Let's check that this satisfies the SDE $$ \mathrm d Y_t = r\,\mathrm d t + \alpha Y_t \,\mathrm d B_t.\label{*}\tag{*} $$

To this end, define the processes $$ X_t = \exp\left( \alpha B_t - \frac 12 \alpha^2 t \right) $$ and $$ Z_t = Y_0 + r\int_{[0,t]} \exp\left( -\alpha B_s + \frac 12 \alpha^2 s \right) \,\mathrm d s. $$ It's easy to check via Itô's lemma that $X$ satisfies $$ \mathrm d X_t = \alpha X_t \,\mathrm d B_t. $$ Moreover, $$ \mathrm d Z_t = rF_t \,\mathrm dt. $$ Thus, as $Z$ has finite variation, and $FX = 1$, the product rule gives us that \begin{align*} \mathrm d (XZ)_t &= \alpha X_t Z_t \,\mathrm d B_t + r \,\mathrm d t \end{align*} which means that $Y = XZ$ satisfies the SDE \eqref{*}.

Answer 2

You had done the solution, there isn't any thing left to solve. It seemed wierd since unlike ODE, solution to SDE seldom have explicit form, after integration, you've got an adapted representation for $Y_t$, that's engouh.