Ho–Lee Short-Rate Integration and Zero-Coupon Bond Pricing
Summary
The document derives the time integral of the Ho–Lee short rate, where the short rate has constant volatility and a deterministic drift adjustment under the risk-neutral measure. Rewriting the rate from its value at the start of the pricing interval separates the known starting rate from Brownian and drift increments. Changing the order of integration gives each increment a weight equal to the remaining time until maturity.
The Brownian integral is normally distributed, so its exponential expectation follows from the Gaussian moment-generating function. This yields the conditional bond-pricing expression: the starting rate contributes over the full interval, the deterministic drift contributes through a weighted integral, and volatility contributes a cubic maturity term. The post also clarifies that integrating the drift from time zero requires a separate accumulated term up to the current time. The result relies on deterministic bounded drift and constant volatility; it does not address calibration or extensions with time-varying or stochastic parameters.
Key ideas
- Changing the order of integration turns the accumulated short-rate drift into a maturity-weighted integral.
- The Brownian contribution to integrated rates is a Gaussian stochastic integral.
- The conditional exponential expectation produces a volatility adjustment proportional to the cube of the pricing interval.
- When the drift integral begins before the current time, its pre-existing accumulation must be included separately.
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# Ho and lee derivation for short rates model
# Ho and lee derivation for short rates model
A silly question that is bugging me. I am working my way through Baxter and Rennie (again) and I am getting my wires crossed on the short rate models in particular the straight forward Ho and Lee analysis.
So given the SDE (under the $\mathbb{Q}$ measure) $$ dr_t = \sigma dW_t +\theta_t dt $$ where $\theta_t$ is both deterministic and bounded and $\sigma$ is constant. This becomes $$ r_t = f(0,t) + \sigma W_t +\int_0^t \theta_s ds $$ (I hope).
How do I compute the integral $$ \int_t^T r_sds $$
Basically it comes down to computing $$ \int_t^T W_sds $$ and $$ \int_t^T \int_0^s \theta_k dk. $$ From first integral is just book work (though it be nice to see a derivation here other than by parts?) It is the later which I am not sure about, as the result is apparently $$ \int_t^T (T-s)\theta_s ds $$ Which I am puzzled by?
$\textbf{edit}$ Actually an important piece of information is that I am trying to compute $$ -\log\mathbb{E}_{\mathbb{Q}}\left(\mathrm{e}^{-\int_t^T r_sds}\vert r_t = x\right)=x(T-t) -\frac{1}{6}\sigma^2 (T-t)^3 + \int_0^T (T-s)\theta_sds $$
## Answer by Gordon (score 11, accepted)
https://quant.stackexchange.com/a/15962
For any $s \geq t$, note that \begin{align*} r_s = r_t + \sigma\int_t^s dW_u + \int_t^s \theta_u du. \end{align*} Then, \begin{align*} \int_t^T r_s ds &= (T-t)r_t + \sigma\int_t^T\int_t^s dW_u ds + \int_t^T \int_t^s\theta_u du ds\\ &=(T-t)r_t + \sigma\int_t^T\int_u^T ds\, dW_u +\int_t^T\int_u^T\theta_u ds du\\ &=(T-t)r_t + \sigma\int_t^T (T-u)dW_u +\int_t^T (T-u) \theta_u du. \end{align*} Moreover, \begin{align*} E_Q\Big(e^{-\int_t^T r_s ds} \mid r_t \Big) &= e^{-(T-t)r_t - \int_t^T (T-u) \theta_u du}E_Q\Big(e^{-\sigma\int_t^T (T-u)dW_u} \mid r_t\Big)\\ &=e^{-(T-t)r_t - \int_t^T (T-u) \theta_u du}e^{\frac{\sigma^2}{2}\int_t^T(T-u)^2 du} \\ &=e^{-(T-t)r_t - \int_t^T (T-u)\theta_u du + \frac{\sigma^2}{6}(T-t)^3}. \end{align*} That is, \begin{align*} -\ln E_Q\Big(e^{-\int_t^T r_s ds} \mid r_t \Big) = (T-t)r_t + \int_t^T (T-u)\theta_u du - \frac{\sigma^2}{6}(T-t)^3. \end{align*} If you really need the integral $\int_t^T\int_0^s \theta_u du ds$, you can proceed as follows: \begin{align*} \int_t^T\int_0^s \theta_u du ds &= \int_t^T\int_0^t \theta_u du ds + \int_t^T\int_t^s \theta_u du ds \\ &=(T-t)\int_0^t \theta_u ds + \int_t^T\int_u^T \theta_u ds du\\ &=(T-t)\int_0^t \theta_u ds + \int_t^T (T-u)\theta_u du. \end{align*}
## Answer by Giogre (score 1)
https://quant.stackexchange.com/a/70233
It seems this specific passage of the Ho-Lee short rate model has left many readers puzzled, so the authors themselves have expanded on this derivation with a pdf add-on that can be found at the book website.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.