I have been trying to learn HW1F on my own, out of nothing more than genuine curiosity during my twilight years, and I'm confused on the issue of calibrating. I don't know why, but all the research papers seem to be missing something, or there's some crucial detail I just don't understand regarding the calculus.

I am using Brigo and Mercurio as a guide.

Suppose we are given a the process as

$dr(t) = [v(t) - ar(t)]dt + \sigma dW(t)$

with the market instantaneous forward $f^M(0,T)$ as

$f^M(0,T) = -\frac{\partial lnP^M(0,T)}{\partial T}$

where $P^M(0,t)$ is the market discount factor. Then we are given

$v(t) = \frac{\partial f^M(0,T)}{\partial T} + af^M(0,T) + \frac{\sigma^2}{2a}(1 - e^{-2at})$

where $a$ is an input to the model where practitioners typically use 5% (this point also making me uncomfortable, since 5% seems very arbitrary).

At first I was thinking there was something magical I didn't understand about the partial derivative in $f^M(0,T)$. But then I suppose if $P^M(0,t)$ is observable by the market, or rather, $f^M(0,T)$ is the z-curve, is it not?

From there, I could just fit a cubic spline through a few points on the curve to get the $f^M(0,T)$ for any $T$.

And here is where my 40 years post-calculus begins to show... is there some simple trick for understanding the partial derivative $\frac{ \partial f^M(0,T) }{\partial T}$? I believe $-\frac{\partial lnP^M(0,T)}{\partial T}$ is just $\frac{-ln(P(0,T))}{\Delta T}$ from my understanding of continuous compounding, but I have long since forgotten why.

Then I supposed, once I calculated a few $v(t)$ from the observable market, then I could fit a spline the $v(t)$ to get a $v(t)$ for any $t$ within the range of observed $f^M(0,T)$. Does this sound reasonable?

Coming to the $\sigma$, I have seen several comments alluding to calibrating the volatility to a set of swaptions or caplets, depending on the types of things one is trying to model. Why would you not just take a treasury option Black-Scholes implied vol for calibrating $\sigma$? I guess this is because the rate being modeled is the instantaneous short rate and not the yield-to-maturity on the zero-coupon bond associated with $P^M(0,T)$? Beyond this, I get completely lost. I have seen some commentary on using the diagonal of the volatility surface for calibration, but I'm not sure why.

So my question(s) then is... is the approach for the $v(t)$ correct? Should I question further where $a=$ 5% comes from, or take it as a given? Why is the diagonal of the swaption surface used in calibrating $\sigma$? Lastly, I notice that Hull and White advocate a lattice for calibration. But then some approaches I believe calibration is done without a lattice using possibly a root-solver. Is this my ignorance? Does (mostly) everyone use a lattice to calibrate?


1 Answer 1


I remember having the same confusion about that when I started to learn the theory, especially about those instantaneous forward rates and their derivatives. Calculating instantaneous forward rates would mean taking derivatives of the yield curve. The result would then heavily depend on the interpolation on the yield curve.

The trick is that you never have to calculate the mentioned derivatives, or in fact any instantaneous forward rates, by doing a clever substitution of variables before you start.

Basically you reformulate all equations in terms of the variable $$ x(t) = r(t) - f(0,t) $$ instead of $r(t)$.

It is a bit too much to write the entire theory here, but an excellent reference is Part II of the series "Interest rate modeling" of L. Andersen and V.Piterbarg. All details are written out clearly there.

Regarding other questions: Often the value of the mean reversion is pre-chosen. 0.03 seems to be a common choice. Then the sigma is calibrated. sigma is typically chosen either to be a constant or a piecewise constant time dependent function.

I did not use a grid when I calibrated the Hull-White model but rather some semi-analytical formulas. See for example this article of M.Henrard : https://arxiv.org/pdf/0901.1776.pdf

Which options you choose for your calibration is much a matter of choice and how you want to use your calibrated model.


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