Deriving Ho–Lee Zero-Coupon Bond Dynamics with Itô's Lemma
Summary
The document works through a derivation of the price dynamics of a zero-coupon bond in the Ho–Lee short-rate model. The question starts from the model's short-rate process and a conditional expectation representation of the bond price, then applies Itô's lemma to a price function of time and the current short rate. A drift term appears that seems inconsistent with the expected bond dynamics.
The answer explains that differentiating a conditional expectation with respect to time must account for the changing information set, not just the explicit time argument. It supplies an exponential expression for the bond price and differentiates that expression; the resulting time derivative includes terms involving the short-rate drift and the time to maturity. These terms allow the remaining Itô calculation to be completed. The exchange is a derivation aid, not a broad treatment of the model: it does not discuss calibration, empirical performance, or model limitations. The lesson concerns careful differentiation under a time-dependent conditional expectation.
Key ideas
- In the Ho–Lee model, the bond price depends on both time and the current short rate.
- Applying Itô's lemma requires the correct partial derivative with respect to time.
- A conditional expectation's information set changes with time and must be considered in differentiation.
- An explicit exponential bond-price expression helps recover the missing drift terms.
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Full text
# Bond dynamics in Ho Lee model
# Bond dynamics in Ho Lee model
The short rate in the Ho-Lee model is given by :
$$dr_t=\left( \frac{df(0,t)}{dt} +\sigma^2t\right)dt + \sigma dW_t$$
I'm trying to find the bond dynamics given by :
$$dP(t,T)/P(t,T)=r_tdt-\sigma(T-t)dW_t$$
I started from :
$$P(t,T)=E_t[e^{-\int_t^T r_sds}]$$
and I applied Itô to the function $P(t,T)=\phi(t,r)$:
$$d\phi(t,r) = \frac{\partial \phi(t,r)}{\partial t}dt+\frac{\partial \phi(t,r)}{\partial r} dr_t+ \frac{1}{2} \frac{\partial^2\phi(t,r)}{\partial r^2}(dr_t)^2$$
I computed the derivatives :
$$\frac{\partial \phi(t,r)}{\partial t}=r_tP(t,T)$$
$$\frac{\partial \phi(t,r)}{\partial r} = -(T-t)P(t,T)$$
$$\frac{1}{2} \frac{\partial^2\phi(t,r)}{\partial r^2} = (T-t)^2P(t,T)$$
Assembling everything I get :
$$dP(t,T)/P(t,T) = r_tdt-(T-t)\sigma dW_t +\left[ \frac{1}{2}(T-t)^2\sigma^2-(T-t)\left( \frac{df(0,t)}{dt}+\sigma^2t \right) \right] dt $$
I don't know how to get rid of the last $dt$ term. Any Help? Or did I get the derivatives wrong? I checked them several times but I don't see where the probem comes from. Thank you
## Answer by Gordon (score 3)
https://quant.stackexchange.com/a/44031
When taking the partial derivative $\frac{\partial}{\partial t}$ in a conditional expectation, not only the parameter $t$ within the expectation needs to be considered, the information set $\mathscr{F}_t$ should also be considered.
For this particular question, based on an answer to this question, \begin{align*} P(t, T) = e^{-(T-t)r_t - \int_t^T (T-u)\theta_u du + \frac{\sigma^2}{6}(T-t)^3}, \end{align*} where \begin{align*} \theta_t &= \frac{df(0,t)}{dt} +\sigma^2t. \end{align*} Then, \begin{align*} \frac{\partial P(t, T)}{\partial t} &= P(t, T)\Big(r_t + (T-t) \theta_t -\frac{\sigma^2}{2}(T-t)^2\Big)\\ &= P(t, T)\Big(r_t + (T-t) \Big(\frac{df(0,t)}{dt} +\sigma^2t\Big) -\frac{\sigma^2}{2}(T-t)^2\Big). \end{align*} The remaining is now straightforward.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.