Suppose I have a model for the short rate $r$ as ($W(t)$ is standard Brownian motion)

$r(t) = c+ \int_0^t \sigma (s) ^2 (t-s) ds+ \int_0^t \sigma (s) dW(s)$

I then want to find the dynamics of $r$, but how do I do that when the process itself contains integrals w.r.t Brownian motion? I get stuck when using Ito's formula and trying to calculate the integral $ \frac{\partial}{\partial t} \int_0^t \sigma (s) dW(s)$

  • $\begingroup$ This partial derivative of an Ito stochastic integral wrt to time does not really make sense, you should see $Y_t=\int_0^t \sigma (s) dW(s)$ as what it is, i.e. a stochastic process, and apply Ito's lemma to $r_t = f(t,Y_t)$. Have you had a look at this question where the poster has a similar issue than yours: quant.stackexchange.com/questions/7726/… $\endgroup$
    – Quantuple
    Commented Oct 20, 2016 at 18:54
  • $\begingroup$ You don't need to use the partial derivative for extracting dynamic of $r_t$. $\endgroup$
    – user16651
    Commented Oct 20, 2016 at 19:11

1 Answer 1


Let $$X_t = \int_0^t \sigma(s) dW_s$$ denote a stochastic integral in the Itô sense. In that case one can write $r_t = f(t,X_t)$ where $$f:(t,x) \to c + \int_0^t \sigma^2(s)(t-s) ds + x \tag{1}$$ and use Itô's lemma to compute the differential $$ dr_t = \partial_t f(t,X_t) dt + \partial_x f(t,X_t) dX_t + \partial_{xx} f(t,X_t) d\langle X \rangle_t $$ where from $(1)$ \begin{align} \partial_t f(t,X_t) &= \partial_t \int_0^t \sigma^2(s)(t-s) ds \\ &= \int_0^t \sigma^2(s) ds + 1 (\sigma^2(t)(t-t)) - 0 (\sigma^2(0)(0-s)) \\ &= \int_0^t \sigma^2(s) ds \end{align} from Leibniz integral rule and $$ \partial_x f(t,X_t) = 1,\quad \partial_{xx} f(t,X_t) = 0 $$ along with, by definition of the Itô integral: $$ dX_t = \sigma(t) dW_t,\quad d\langle X \rangle_t = \sigma^2(t) dt $$ such that finally: $$ dr_t = \left(\int_0^t \sigma^2(s) ds\right) dt + \sigma(t) dW_t $$

And indeed by integrating this last equation from $0$ to $t$ one gets: $$ r_t - r_0 = \int_0^t \left( \int_0^u \sigma^2(s) ds\right) du + \int_0^t \sigma(u) dW_u $$ and noting that \begin{align} \int_0^t \int_0^u \sigma^2(s) ds du &= \int_0^t \int_s^t \sigma^2(s) du ds\\ &= \int_0^t \sigma^2(s) (t-s) ds \end{align} by Fubini theorem, one gets $$ r_t = r_0 + \int_0^t \sigma^2(s) (t-s) ds + \int_0^t \sigma(s) dW_s $$


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