I am attempting to solve the Vasicek model SDE (using Wikipedia parametrisation):

$$ dr_t = a(b-r_t)dt + \sigma dW_t $$

Every solution is proceeding to multiply both sides of the equation by the integrating factor $e^{at}$ (akin to solving linear ODEs). After multiplication and rearrangement we get the following equation:

$$ e^{at}dr_t + e^{at}ar_tdt= e^{at}(abdt + \sigma dW_t) $$ Now the left hand side is apparently equal to $d(e^{at}r_t)$. How is that exactly the case?

Is it by Ito product rule? If so what is $X(t)$ and $Y(t)$?

Is it by Ito's lemma but then what is the $f(x,t)$


1 Answer 1


Apply the Ito product rule, noting the cov of a deterministic and stochastic term is zero:

$$\begin{align} d\left(e^{at}r_t\right)&=e^{at} dr_t+r_t de^{at} \\[6pt] &=e^{at} dr_t+r_t e^{at} d(at) \\[6pt] &=e^{at} dr_t+r_t e^{at} a dt \end{align}$$

  • $\begingroup$ By the "cov of a deterministic and stochastic term" do you mean $dr_tde^{at}$ from Ito's Product rule which when we multiply out and substitute for $dr_t$ we get $[a(b-r_t)dt + \sigma dW_t]de^{at}$ and hence we only get the $dtdt$ and $dtdW_t$ terms which are all 0? $\endgroup$ Commented Oct 24, 2018 at 18:39
  • $\begingroup$ That’s right! Though this result can be generalised $\endgroup$ Commented Oct 24, 2018 at 18:47

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