# Question on swaption price in quant research paper "Hull-White one factor model: results and implementation"

#1

I would like to ask a question on Marc Henrard’s quantitative research paper, specifically about Theorem 2 (the exact European swaption price).

As far as I can tell, the exact European swaption price relies on the same Jamshidian-style decomposition of the payoff as in the single-curve case.
However, as Marc states directly after the theorem, “the original proof required that d_i > 0, while here we have coupon equivalent like 1 - \beta with usually \beta > 1.”

Why does this theorem not require that all d_i > 0?
If we value a swaption in OG with a few negative coupons (which would usually be the case), is this exact formula still exact?

I wasn’t able to find an explicit derivation of the theorem for the dual-curve case, and it seems to me that payoff decomposition is not valid when not all of the coupons are positive.
Does it become an approximation in the presence of negative coupons?

#2

You are right, the document is not very clear around the values d_i and their sign.

The condition d_i>0 in the original paper is a sufficient condition which was natural in the single curve framework (coupons are positive). Even in that framework the condition is not necessary.

What we are trying to prove in the theorem is that there is a unique \kappa solution of the equation (9)(and that it is easy to solve the equation numerically). The dominant term is the one corresponding to the final notional (d_n=1+c_n). What may happen if the coupons are becoming to large and negative (with respect to the notional) is that the function may not be decreasing everywhere and there may be several solution to the equation (9). If that was to happen, the exercise domain of the option would not be a single interval but a union of disjoint intervals.

The formula given is always exact if the equation (9) for kappa has a unique solution. The 1-\beta_i can be negative but are small with respect to the notional 1; for “reasonably” small values, the decreasing property of the function from which kappa is the root is still satisfied. I have not quantified the “reasonably”; that would be an interesting exercise to see if it is possible to give a numerical lower bound on the “d_i” for which the result is still valid even if they are negative.

To answer more directly to your question, the result is exact in most cases, even if the d_i>0 condition is not satisfied strictly; very large spreads would be required to be outside the domain of applicability of the result.

#3

Marc,
Thank you very much for your detailed answer, that makes perfect sense.