Suppose $$f(t,u)=f(0,u)+\int_0^t{\mu (w,u)dw}+\int_0^t{\sigma(w,u)dB_w},$$ where $B_w$ is a standard Brownian motion. I would like to calculus the drift and diffusion of $Y_t=-\int_t^s{f(t,u)du}$ (under sufficient conditions that guarantee all the regularities).
The problem comes from the HJM model in finance, and the answer is $$dY_t=\left(f(t,t)-\int_t^s{\mu(t,u)du}\right)dt+\left(-\int^s_t{\sigma(t,u)du}\right)dB_t.$$ I am really confused about where the term $f(t,t)$ in the drift comes from. Formally, I can calculate this using Leibniz rule and get $$dY_t=-\int_t^s{d_t f(t,u)du}+f(t,t)dt$$ which is equivalent to the answer, but this calculation is not justified. And the notes I have says the answer is due to Fubini's theorem for stochastic integrals. I understand that I can use Fubini's theorem to get
$$Y_t=-\int_t^sf(0,u)du-\int^t_0\int^s_t\mu(w,u)dudw-\int_0^t\int_t^s\sigma(w,u)dudB_w.$$
But I don't know how this could lead to the answer, since the "drift" $\int^s_t\mu(w,u)du$ also depends on $t$, and how I could get the term $f(t,t)$.
Best Answer
I am probably very late in answering this, but maybe the proof is useful for someone like me coming just now to this question.
Using the integral form of $f$: \begin{align*} Y_t = & - \int_t^T \left[ f(0, u) + \int_0^t \alpha(s, u)ds + \int_0^t \sigma(s, u) dw_s \right] du \,, \end{align*} Using Fubini's theorem (twice): \begin{align*} Y_t = & - \int_t^T f(0, u) du - \int_0^t \int_t^T \alpha(s, u) du ds - \int_0^t \int_t^T \sigma(s, u) du dw_s \,, \\ = & - \int_t^T f(0, u) du - \int_0^t \int_s^T \alpha(s, u) du ds - \int_0^t \int_s^T \sigma(s, u) du dw_s \,, \\ & + \int_0^t \int_s^t \alpha(s, u) du ds + \int_0^t \int_s^t \sigma(s, u) du dw_s \,, \\ = & - \int_0^T f(0, u) du - \int_0^t \int_s^T \alpha(s, u) du ds - \int_0^t \int_s^T \sigma(s, u) du dw_s \,, \\ & + \int_0^t f(0, u) du + \int_0^t \int_0^u \alpha(s, u) ds du + \int_0^t \int_0^u \sigma(s, u) dw_s du \,, \end{align*}
Given that $r(u) = f(u, u)$ and using the definition for $Y_0$ we have that: \begin{align*} Y_t = & Y_0 - \int_0^t \int_s^T \alpha(s, u) du\,ds - \int_0^t \int_s^T \sigma(s, u) du\,dw_s + \int_0^t r(u) du \,. \end{align*} Which is the desired result in the integral form.