# Details of calibration of Hull-White model

Consider the one-factor Hull-White model

$$\mathrm{d}r(t) = (\theta(t)-\kappa r(t))\mathrm{d}t + \sigma\mathrm{d}W(t)$$

When one calibrates the model to market data one chooses

$$\theta(t) = \frac{\partial f^M}{\partial T}(0,t) + \kappa f^M(0,t) + \frac{\sigma^2}{2\kappa}\left(1-\mathrm{e}^{-2\kappa t}\right)$$

where $f^M(0,T) = -\frac{\partial}{\partial T}\log(P^M(0,T))$ with the observed bond term structure $P^M(0,T)$ at the time of calibration.

I have several questions regarding this calibration:

• How do I come up with this formula for $\theta(t)$? I always read that this aligns the model with the market's zero curve. How can you derive that this formula in fact establishes the desired consistency?

• How do I come up with $P^M(0,T)$ and $f^M(0,T)$? Am I right that the following approach is taken?

1. First the zero curve $y^M(t)$ is bootstrapped using coupon bearing instruments.
2. The bootstrapped curve is just given at a finite number of points given by the maturities of the considered instruments. We interpolate these points (e.g. via spline interpolation) to obtain the function $y^M(t)$ on a continuous domain.
3. Now we obtain $P^M(0,T)$ by setting $P^M(0,T) = \exp(-y^M(T)T)$.
4. In a final step we compute the derivative $f^M(0,T):=-\frac{\partial}{\partial T}\log (P^M(0,T)) = -\frac{\partial}{\partial T}\left(y^M(T)T\right) = -\left(\frac{\partial}{\partial T}y^M(T)\right)T - y^M(T)$

Regarding your first question: the equation for $\theta(t)$ is obtained from the consistency condition $$\forall T, \;\; E\left[e^{-\int_0^T r(t) dt} \right] = P^M(0,T)$$ after a somewhat involved calculation using the integrated version of the SDE for $r$ $$r(t)=e^{-\kappa t}r(0) + \int_0^t e^{-\kappa (t-u)} \theta(u) du + \int_0^t e^{-\kappa (t-u)} \sigma dW(u)$$
As a sidenote if you define $x(t) = r(t) - f^M(0,t)$ then the SDE for $x$ is $$dx(t) = \left(-\kappa x(t) + \frac{\sigma^2}{2 \kappa}(1- e^{-2 \kappa t})\right) dt + \sigma dW(t)$$ Thus when doing MC simulation or finite differences schemes you can use $x(t)$ as the state variable, and then simply add $f^M(0,t)$ to obtain $r(t)$, so that in fact you do need to compute $\frac{\partial f^M(0,t)}{\partial t}$, which means that zero curve interpolation methods that are not twice differentiable but only once differentiable (such as linear on yield or linear on log discounts) will produce $P^M(0,T)$ and $f^M(0,t)$ that can still be used with the model.
• Thanks for your answer! Yet, there are some open questions remaining: You just said that the first point in the approach described is correct. What about the other steps? Secondly, regarding your sidenote: Where do I need the function to be twice differentiable? There is only one derivative appearing? And what is the advantage of using $x(t)$? I don't get your point there. Mar 27, 2018 at 12:20
• 1) all the steps in your approach are correct. 2) For computing $\theta(t)$ you need $\frac{\partial f^M(0,t)}{\partial t} = \frac{\partial}{\partial t}(-\frac{\partial}{\partial t} \ln(P^M(0,t))) = -\frac{\partial^2}{\partial t^2} \ln(P^M(0,t))$. This is why working with $x(t)$ is generally better, as there is no need to compute $\frac{\partial f^M(0,t)}{\partial t}$. Mar 27, 2018 at 12:29
• Sorry, I’m new to interest models and still not sure if I get your second comment so let me summarize what I understand: You want to compute $\frac{\partial }{\partial t}f^M(0,t)$. One approach would be to calculate $\frac{\partial }{\partial t}f^M(0,t) = -t\frac{\partial^2}{\partial t^2}y^M(t)-2\frac{\partial}{\partial t}y^M(t)$ following the forth step in my approach. This would require our interpolation $y(t)$ to be twice differentiable. Another approach now would be: Mar 27, 2018 at 13:20
• $\frac{\partial }{\partial t}f^M(0,t) = -\frac{\partial^2 }{\partial t^2}\log\left(\mathbb{E}\left(\exp(\int_0^t r(s)\mathrm{d}s)\right)\right)\approx -\frac{\partial^2 }{\partial t^2}\log\left(\sum_{i=1}^n\exp\left(\int_0^t r_i(s)\mathrm{d}s \right)\right)\ = \frac{\partial}{\partial t}\sum_{i=1}^n r_i(t) =\frac{\partial}{\partial t} \sum_{i=1}^n (x_i(t)+f^M(0,t)) \approx \frac{\partial}{\partial t}\mathbb{E}(x(t)+f^M(0,t))$ Mar 27, 2018 at 13:21
• Where $r_i(t)$ and $x_i(t)$ are sample paths generated using MC. The last derivative can then be computed using MC sensitivity schemes like pathwise derivatives of likelihood ratios. Am I right there? Mar 27, 2018 at 13:21