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 $f^M(0,T):=-\frac{\partial}{\partial T}\log (P^M(0,T)) = -\frac{\partial}{\partial T}\left(y^M(T)T\right) = -\left(\frac{\partial}{\partial T}y^M(T)\right)T - y^M(T)$

1 Answer 1


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 $r$ $$ r(t)=e^{-\kappa t}r(0) + \int_0^t e^{-\kappa (t-u)} \theta(u) du + \int_0^t e^{-\kappa (t-u)} \sigma dW(u) $$

Regarding your second question yes you bootstrap the zero curve on the choosen instruments, bonds or swaps depending on the market you are modeling. You may choose splines, or any other type of interpolation as long as the required derivatives can be computed.

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.

  • $\begingroup$ 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. $\endgroup$
    – lbf_1994
    Commented Mar 27, 2018 at 12:20
  • 1
    $\begingroup$ 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 $\frac{\partial f^M(0,t)}{\partial t}$. $\endgroup$ Commented Mar 27, 2018 at 12:29
  • $\begingroup$ 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: $\endgroup$
    – lbf_1994
    Commented Mar 27, 2018 at 13:20
  • $\begingroup$ $\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)) $ $\endgroup$
    – lbf_1994
    Commented Mar 27, 2018 at 13:21
  • $\begingroup$ 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? $\endgroup$
    – lbf_1994
    Commented Mar 27, 2018 at 13:21

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