Numerical Root Finding for Hull–White Swaption Calibration
Summary
The document addresses a calibration step in the one-factor Hull–White interest-rate model: finding the critical rate that satisfies the equation used in an analytical swaption pricing formula. It explains that this root generally has no closed-form solution, but that this does not require an impractically slow calibration process.
The proposed method is to solve the scalar equation numerically with bisection or Newton–Raphson. The function is monotonic and has an analytical derivative, and the answer provides bounds for the root based on the model coefficients, which can make the search stable. A second answer notes that the broader parameter calibration may use simulated annealing and may take longer in a higher-dimensional parameter space. These are distinct computational steps: a fast one-dimensional root solve does not by itself guarantee that the full model calibration will be fast. The document provides algorithmic guidance but no benchmark or validation results.
Key ideas
- The critical rate in the described swaption equation has no closed-form solution.
- Bisection or Newton–Raphson can solve the scalar root numerically.
- Monotonicity and an analytical derivative support a stable and efficient root search.
- Bounds derived from the coefficients can constrain the root search interval.
- Full parameter calibration may still require a separate optimization procedure such as simulated annealing.
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# Calibrate Hull-white one factor model with swaption in analytical formula
# Calibrate Hull-white one factor model with swaption in analytical formula
I've been trying to calibrate Hull-white one factor model with swaption but I have a trouble making closed form solution of swaption
Below is the part of paper I've been referencing to
https://people.kth.se/~aaurell/Teaching/SF2975_HT17/calibration-hull-white.pdf
The problem is r* part.
In order to calculate the price of swaption following the instruction of the paper, I need to solve the equation (16) to come up with r*.
But it seems that there is no closed-form solution to this equation finding r*.
However, if no closed-form solution exists for pricing swaption, the whole calibration process takes too long. I think it is not what the author intended.
Is there any closed-form solution for finding r* in this equation?
Many thanks in advance for helping me.
## Answer by ir7 (score 2, accepted)
https://quant.stackexchange.com/a/63175
There is no closed-form solution, but solving for $r^\star$ such that
$$f(r^\star) = \tilde{c}^{-1} $$
should be fast and safe with a standard single dimension solver, bisection or Newton-Raphson, as
- function $f$ is monotonically decreasing ($B_i$'s and $\tilde{c}_i$ are positive),
$$f(x) = \sum_{i=1}^n \tilde{c}_i {\rm e}^{A_i-B_ix}, $$
- its derivative is analytical,
$$f'(x) = \sum_{i=1}^n -B_i\tilde{c}_i {\rm e}^{A_i-B_ix}, $$ and
- we know the solution $r^\star$ belongs to the interval
$$ \left[ \frac{\ln (\tilde{c}A_d)}{B_u}, \frac{\ln (\tilde{c}A_u)}{B_d} \right],$$
where
$$ \tilde{c} = \sum_{i=1}^n c_i, \: \: \tilde{c}_i = c_i/\tilde{c},$$
$$ A_u = \max_{i=1,...,n} {\rm e}^{A_i}, \: \: A_d = \min_{i=1,...,n} {\rm e}^{A_i}, $$
$$ B_u = \max_{i=1,...,n} {B_i}, \: \: B_d = \min_{i=1,...,n} {B_i}. $$
## Answer by wgajate (score 1)
https://quant.stackexchange.com/a/63172
Given the non-linear nature of the constrained optimization problem ie. $exp(A(T0,Ti)-B(T0,Ti)*r)$, you will need to employ numerical solvers.
The authors of the document used Simulated Annealing (shown in Appendix B) for fast convergence. They note that it could take up to 10 seconds to solve a 10-dimensional parameter space.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.