Integrating the Hull–White Short-Rate SDE with an Integrating Factor
Summary
The document explains how to solve the time-dependent Hull–White short-rate stochastic differential equation using an integrating factor. Applying Itô’s lemma to the short rate multiplied by the exponential of the integrated mean-reversion coefficient cancels the rate-dependent drift term. Integrating the resulting differential from one time to another yields the difference between the transformed rate at the endpoints, alongside an ordinary integral and a stochastic integral.
The endpoint identity follows from the definition of integrating a differential: increments telescope, so integrating dY over an interval gives Y at the end minus Y at the start. The answer also sketches this fact through the partition-sum definition of the Itô integral with a constant integrand. This clarifies why evaluating the transformed process at the interval boundaries is justified, rather than an informal substitution. The exposition gives the solution method but does not discuss parameter estimation, bond pricing, or assumptions needed for the integrals to exist.
Key ideas
- An integrating factor removes the short-rate-dependent drift from the Hull–White SDE.
- Itô’s lemma derives the differential of the transformed rate process.
- Integrating a process differential over an interval gives the difference between its endpoint values.
- The transformed equation retains both an ordinary time integral and a stochastic integral.
- The derivation assumes suitable regularity for the coefficients and stochastic integrals.
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# Integration in the Hull-White SDE
# Integration in the Hull-White SDE
I'm stuck in solving the SDE in Hull-White interest rate model. I do not have a thorough background in math (only Real Analysis during my blissful undergrad years), so I am having trouble understanding the integration process in explicitly solving the Hull-White SDE.
So, the Hull-White interest model follows the SDE $$ dR(u) = (a(u) - b(u) R(u)) du + \sigma(u) d\tilde{W}(u) $$ It says the explicit solution can be obtained by applying Ito's Lemma to $$ e^{\int_0^u b(v) dv} R(u) $$ and integrating both sides.
This is where I am having trouble understanding. $$ \int_t^T d\left(e^{\int_0^u b(v) dv} R(u)\right) = e^{\int_0^T b(v) dv} R(T) - e^{\int_0^t b(v) dv} R(t) $$ It seems that we are naively replacing $u$ with $T$ in the first term and with $t$ in the second term. Could we simply do this due to the fundamental theorem of Calculus? Or is there some other working mechanism behind the scenes?
## Answer by Quantuple (score 2, accepted)
https://quant.stackexchange.com/a/25663
Applying Itô's lemma to $$ Y_t := e^{\int_0^t b(v) dv} r_t $$
You get \begin{align} dY_t &= b(t) e^{\int_0^t b(v) dv} r_t dt + e^{\int_0^t b(v) dv} dr_t + 0\\ &= e^{\int_0^t b(v) dv} (b(t) r_t dt + dr_t) \\ &= e^{\int_0^t b(v) dv} (a(t) dt + \sigma(t) dW_t) \end{align} where the last line is obtained by using the fact that $$ dr_t = (a(t)-b(t)r_t) dt + \sigma(t) dW_t $$
Your question regards the integration of the LHS, well it is simply given by \begin{align} \int_t^T dY_u &= Y_T - Y_t \\ &= e^{\int_0^T b(v) dv} r_T - e^{\int_0^t b(v) dv} r_t \end{align}
For the RHS, you get $$ \int_t^T e^{\int_0^u b(v) dv} (a(u) du + \sigma(u) dW_u) = \int_t^T a(u) e^{\int_0^u b(v) dv} du + \int_t^T \sigma(u) e^{\int_0^u b(v) dv} dW_u $$ which you cannot really simplify much more.
To obtain the solution to the original equation, work from here using the fact that $$r_t := e^{-\int_0^t b(v) dv} Y_t $$
[Remark]
$$ \int_0^t dY_u = (Y_t - Y_0) $$
is merely a consequence of how we define stochastic integrals to begin with (wheter Ito or Stratonovich). Assuming Itô formalism, for a sufficiently well-behaved integrand $\psi_t$ and a semi-martingale $X_t$, the stochastic integral writes $$ I_t := \int_0^t \psi_u dX_u = \lim_{\Vert P \Vert \rightarrow 0} \sum_{i=1}^N \psi_{t_{i-1}} (X_{t_i}-X_{t_{i-1}}) $$ where the limit (when it exists) is taken in the mean square sense, as the partition $P = \{ 0=t_0 < \dots < t_N=t \}$ is refined more and more (the maximum interval $t_{i}-t_{i-1}$ tends towards 0).
In your particular case, just replace $\psi_t$ by $1$ to see that: $$ I_t = \int_0^t dX_u = \lim_{\Vert P \Vert \rightarrow 0} \sum_{i=1}^N X_{t_i}-X_{t_{i-1}} = X_t - X_0 $$ due to the telescopic sum, as in ordinary calculus.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.