Super basic question. I think I am doing this correctly, but just want a sanity check.

Say I have a stochastic process $r(t)$.

Say I have an equation

$$d(e^{\beta (t-s)}r(s))=\dots$$

where the $e^{\beta (t-s)}$ term is deterministic and $t\geq s > 0$.

Then say we want to integrate both sides from $s$ to $t$

$$\int_s^t{d(e^{\beta (t-u)}r(u))}=\dots$$

we then have

$$e^{\beta (t-t)}r(t)-e^{\beta (t-s)}r(s)=\dots$$

$$r(t)-e^{\beta (t-s)}r(s)=\dots$$

Please set me straight if I have this wrong. Thanks.


I think there is a typo in your first equation. The running variable should be $s$, as in $d\left( e^{\beta(t-s)} r(s) \right)$.

Let's start with your integral. Let $R_u = e^{-\beta u} r_u$. Your integral becomes $$ e^{\beta t} \int_s^t d R_u \, . $$ Recall that $dR_u = R_{u+du} - R_u$. The integral evaluates to $e^{\beta t}(R_t -R_s)$, which simplifies to $r_t - e^{\beta(t-s)} r_s$, which is what you got.

  • 1
    $\begingroup$ Thanks William! It was indeed a typo, thanks for catching that and thanks for the elaboration. Helpful indeed. $\endgroup$ – Joe Oct 4 '13 at 4:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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