2
$\begingroup$

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 $f(x,t)$

$\endgroup$

1 Answer 1

5
$\begingroup$

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

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

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$ Oct 24, 2018 at 18:39
  • $\begingroup$ That’s right! Though this result can be generalised $\endgroup$ Oct 24, 2018 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.