Ito's Lemma applied to the Solution to the General Linear Equation

Question

When I apply Ito's lemma to the solution of the general linear equation I don't get the correct result.

Starting with the general linear equation,

$dX_t=\left( a(t)X+c(t)\right) dt + \left( b(t)X+d(t)\right)dW_t$

The solution is

$X(t,W_t)=\Phi(t,W_t) I(t,W_t)$

$\Phi(t,W_t)=\exp\left(\int_{0}^{t}{\left( a(s) - \frac{1}{2}b^2(s)\right) ds}+\int_{0}^{t}{b(s)dW_s}\right)$

$I(t,W_t)=X_0 + \int_{0}^{t}{\Phi^{-1}\left( c(t)-b(t)d(t)\right)ds}+\int_{0}^{t}{\Phi^{-1}d(s)dW_s}$

Applying Ito's lemma to the solution,

$d\left(\Phi I\right)=Id\Phi+\Phi dI+dI d\Phi$

When I compute each term I get

$d\Phi=a(t) \Phi dt + b(t) \Phi dW_t$

$dI=\Phi^{-1}\left( c(t) - b(t)d(t)\right) dt + \Phi^{-1}d(t) dW_t -\frac{1}{2}\Phi^{-1}b(t)d(t) dt $

$=\Phi^{-1}\left( c(t) - \frac{3}{2}b(t)d(t)\right) dt + \Phi^{-1}d(t) dW_t $

Putting these into Ito's lemma,

$dX_t=\left( a(t)X+c(t)-\frac{1}{2}b(t)d(t)\right) dt + \left( b(t)X+d(t)\right)dW_t$

This is off by the term $-\frac{1}{2}b(t)d(t) dt$. I believe that my expression for dI is off but I'm not certain where the mistake is.

Thanks for the help!

Answer 1

With your finally correct definitions but with better notation $$ \Phi_t=\textstyle\exp\left(\int_0^t( a - \frac{1}{2}b^2)\,\mathrm{d}s+\int_0^tb\,\mathrm{d}W_s\right)\,, $$ $$ I_t=X_0 + \int_0^t\frac{(c-b\,d)\,\mathrm ds+d\,\mathrm dW_s}{\Phi_s} $$ we have

$$\mathrm{d}\Phi_t=a\,\Phi_t\,\mathrm{d}t+b\,\Phi_t\,\mathrm{d}W_t\,,\quad\mathrm{d}I_t=\frac{(c - b\,d)\,\mathrm{d}t + d\,\mathrm{d}W_t}{\Phi_t}\,,\quad \mathrm{d}\langle I,\Phi\rangle_t=b\,d\,\mathrm{d}t\,,\quad X_t=\Phi_t\,I_t\,.$$\begin{align}\mathrm{d}X_t&=I_t\,\mathrm{d}\Phi_t+\Phi_t\,\mathrm{d}I_t+\mathrm{d}\langle I,\Phi\rangle_t\\[2mm]&=X_t(a\,\mathrm{d}t+b\,\mathrm{d}W_t)+(c-b\,d\,\mathrm{d}t)\,\mathrm{d}t+d\,\mathrm{d}W_t+b\,d\,\mathrm dt\\[2mm]&=(aX_t+c)\,\mathrm{d}t+(bX_t+d)\,\mathrm{d}W_t\end{align}

as expeted.