Ho & Lee yield curve fitting with zero coupon bond market prices

Question

The Ho & Lee model for interest rates is given by the SDE: $$ \mathrm d r = \eta(t) \mathrm d t + c\,\mathrm d X $$ The calibration function for $\eta(t)$ is given by $$ \eta^*(t)=c^2(t-t^*)-\frac{\partial^2}{\partial t^2}\operatorname{log}(Z_M(t^*;t)) $$ where $Z_M(t^*, t)$ are the discount factors in the market from today $= t^*$ to maturity $t$ (Source: Paul Wilmott on Quantitative Finance, p. 526).

The term $\frac{\partial^2}{\partial t^2}\operatorname{log}(Z_M(t^*;t))$ confuses me.

I have a set of discount factors $Z_M$, which are numbers (e.g. $Z_M(0;\,0.5)=0.99750, Z_M(0;\,1)=0.989060)$.

So, the $\operatorname{log}Z_M$ is also a number.

How can I compute the partial derivative $\frac{\partial^2}{\partial t^2}\operatorname{log}(Z_M(t^*;t))$ of a number?

EDIT: My current understanding is that I have to use some interpolation method which is twice differentiable (for example spline interpolation) using the discount factors as support points. Would this be correct?

Answer 1

When no functional form is available in differential analysis then one should use a computational method. As Daneel comments a common computational approximation of the second order derivative can be obtained using finite differences.

For example if we assume the points you have available for your discount factors $Z_M$ are equally spaced with gap $\Delta t$ then you get the following approximation via the second order central finite difference method:

$ \eta^*(t) = c^2(t) - \frac{\partial^2}{\partial t^2}\log(Z_M(0;t))$

$ \approx c^2(t) - \frac{\log(Z_M(0;t+\Delta t))-2\log(Z_M(0;t))+\log(Z_M(0;t-\Delta t))}{(\Delta t)^2}$

I wouldn't first fit an interpolation and then differentiate as I have not seen that used in practice. I would assess which finite difference method is the most appropriate and use that.