Deriving the Hull–White Short Rate from the HJM Forward-Rate Model

Summary

The document shows how the one-factor Hull–White short-rate model follows from a Heath–Jarrow–Morton specification for forward rates. It starts with the HJM volatility that decays exponentially with time to maturity and applies the risk-neutral drift condition, in which forward-rate drift is determined by the volatility and its maturity integral.

After integrating the forward-rate dynamics, the derivation sets maturity equal to the current time to identify the instantaneous forward rate with the short rate. The resulting expression for the short rate contains the initial curve, the mean-reversion factor, the time-dependent input, and a stochastic integral. Differentiating it recovers the Hull–White drift and diffusion terms. The answer also supplies an initial forward curve consistent with the chosen short-rate specification. This is an algebraic model-consistency derivation under the stated one-factor setup; it does not discuss calibration, empirical fit, or extensions such as time-varying volatility or mean reversion.

Key ideas

• The HJM drift condition links forward-rate drift to the volatility across maturities.
• With exponentially decaying volatility, the model’s short rate can be obtained by evaluating the forward rate at maturity equal to the current time.
• The resulting short-rate expression differentiates to the Hull–White mean-reverting process.
• The initial forward curve must be consistent with the short-rate specification and its starting value.
• The derivation covers a one-factor setup and does not address calibration or empirical performance.

Tags

# Hull-White model: match between HJM framework and short model formulation


# Hull-White model: match between HJM framework and short model formulation












I need to show that the Hull-White model $$dr=(\theta(t)-ar)dt+\sigma dW^Q$$ corresponds to the Heath-Jarrow-Morton formulation $$df(t,T)=\alpha(t,T)dt+\sigma e^{-a(T-t)}dW^Q.$$ I obtained the drift by with drift condition $$\alpha(t,T)=\sigma(t,T)\int_t^T\sigma(t,s)ds,$$ where $\sigma(t,T)=\sigma e^{-a(T-t)}.$ Then, I integrated the resulting $df(t,T)$ and set $T=t\rightarrow f(t,t)=r(t)$. Finally, I looked for the differential $dr(t)$, but the resulting expression looks completely diffeent form the Hull-White formulation.

Could you show me how to perform needed calculations?

Thank you, Giulio

## Answer by Gordon (score 3)

https://quant.stackexchange.com/a/53769

Note that \begin{align*} f(t, T) = f(0, T) + \int_0^t\alpha(u,T)du+\int_0^t\sigma e^{-a(T-u)}dW_u, \end{align*} where, based on this question, \begin{align*} f(0, T) = \int_0^T \theta(u) e^{-a(T-u)} du - \frac{\sigma^2}{2a^2}\big(e^{-a T} -1\big)^2 + e^{-a T} r_0. \end{align*} Note also that \begin{align*} \int_0^t\alpha(u,T)du &= \int_0^t\sigma(u,T)\int_u^T\sigma(u,s)dsdu\\ &=\int_0^t\sigma e^{-a(T-u)}\int_u^T\sigma e^{-a(s-u)}dsdu\\ &=-\frac{\sigma^2}{2a^2}\Big[\big(e^{-a(T-t)}-1\big)^2 - \big(e^{-aT}-1 \big)^2 \Big]. \end{align*} Therefore, \begin{align*} r_t &= f(t, t)\\ &=e^{-at}\int_0^t \theta(u) e^{au} du + e^{-a t} r_0+e^{-at}\int_0^t\sigma e^{au}dW_u. \end{align*} Then, it is clear that \begin{align*} dr_t &= -ar_t dt + \theta(t) dt + \sigma dW_t\\ &= (\theta(t) - a r) dt + \sigma dW_t. \end{align*}

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.