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.


  • $\begingroup$ better asking on quant.stackexchange.com $\endgroup$ – Mark Joshi Apr 18 '16 at 21:52

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 $\sigma(t)$.

  • $\begingroup$ Perfect. Thanks. I suppose that solving most variants of the Vasicek model follow the same approach. $\endgroup$ – user5619709 Apr 19 '16 at 13:55

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