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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 $f(t,t)$.

Xander Henderson
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epsilon
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    I don't know if this is still relevant, but: In fixed income modelling by Claus Munk page 57-58 he proves Leibniz rule for stochastic integrals and on page 282 he does the step you dislike (in the HJM) model. The proof of Leibniz rule is really ugly and only uses stochastic fubini, but when you got the result pretty much follows. I think it will be clear to you if you find the book. – htd Dec 21 '12 at 16:28
  • Related: https://quant.stackexchange.com/q/30484 (Differential of integral of a stochastic process), https://math.stackexchange.com/q/1298990/532409 (Stochastic Differential Equation for Time Integral of Stochastic Process), https://math.stackexchange.com/q/4431021/532409 (Properties of the time integral of Ito process) – Quillo Apr 19 '22 at 10:12

1 Answers1

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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.

Diego F Medina
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