Calibrate Hull-white one factor model with swaption in analytical formula

Question

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 1

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

1. function $f$ is monotonically decreasing ($B_i$'s and $\tilde{c}_i$ are positive),

$$f(x) = \sum_{i=1}^n \tilde{c}_i {\rm e}^{A_i-B_ix}, $$

2. its derivative is analytical,

$$f'(x) = \sum_{i=1}^n -B_i\tilde{c}_i {\rm e}^{A_i-B_ix}, $$ and

3. 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 2

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.