Deriving Short-Rate Dynamics from a Brownian Integral
Summary
The document derives the dynamics of a short-rate process whose definition includes both a deterministic time integral and an Itô integral. The key step is to name the stochastic integral as a process, then express the rate as a function of time and that process. Itô's lemma can then be applied without attempting to take an ordinary time derivative of the Brownian integral.
The time derivative of the deterministic term follows from Leibniz's rule, while the stochastic process contributes its defining Itô differential. Combining these terms gives the rate's drift and diffusion, and integrating the resulting dynamics recovers the original representation by changing the order of integration. The derivation assumes the integrability and regularity needed for these operations; it offers a worked mathematical argument rather than empirical evidence or a calibrated financial model.
Key ideas
- Represent the Itô integral as a stochastic process before applying Itô's lemma.
- Use the defining differential of that process instead of taking an ordinary time derivative of the stochastic integral.
- Leibniz's rule gives the time derivative of the deterministic integral term.
- Fubini's theorem verifies that integrating the derived dynamics recovers the stated rate process.
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Full text
# Partial derivative of an integral
# Partial derivative of an integral
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)$
## Answer by Quantuple (score 3, accepted)
https://quant.stackexchange.com/a/30656
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 $$ r_t = r_0 + \int_0^t \sigma^2(s) (t-s) ds + \int_0^t \sigma(s) dW_s $$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.