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?
Best Answer
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{*}.