SDE Solution: Hull-White extension of Vasicek model I am trying to figure out the particular ansatz (if that's all there is) for the solution to the SDE:
$
dr_t = [v_t - ar_t]dt + \sigma dW_t,
$
where $a$ is constant and $v,t$ are, potentially, time-dependent. Note the difference to the standard Vasicek mean-reverting SDE.
Any help in solving this SDE is appreciated.
Thanks,
 A: The SDE can be solved similarly as in the Vasicek model. Define $F(t,r(t)) = e^{\alpha t}r(t)$, then
\begin{cases}
\displaystyle \frac{\partial F}{\partial t} &= \alpha e^{\alpha t} r(t) \\
\displaystyle \frac{\partial F}{\partial r(t)} &= e^{\alpha t} \\
\displaystyle \frac{\partial^2 F}{\partial r(t)^2} &= 0.
\end{cases}
Applying Itô's lemma yields
\begin{align}
dY(t,r(t)) &= \left(\alpha e^{\alpha t} r(t)+(v(t)-\alpha r(t))e^{\alpha t} \right)+\sigma e^{\alpha t} dW(t) \\
&=v(t)e^{\alpha t} dt+\sigma e^{\alpha t}dW(t),
\end{align}
or in integral form
$$Y(t,r(t)) = \underbrace{Y(0,r(0))}_{=r(0)}+\int_{0}^{t} e^{\alpha s}v(s)ds + \sigma \int_{0}^{t} e^{\alpha s} dW(s).$$
Making use of the definition of $F$, we finally obtain
$$r(t) = e^{-\alpha t} r(0)+\int_{0}^{t} e^{\alpha(s-t)}v(s)ds+\sigma e^{-\alpha t} \int_{0}^{t} e^{\alpha s}dW(s).$$
Note that if $v(t)$ is constant, i.e $v(t)=v$, then you obtain the solution $r(t)$ in the Vasicek model. Moreover, one can do a similar reasoning for a time dependent $\sigma(t)$. 
