Hull-White Named Functions¶

This page collects all the named functions used in the Hull-White model for quick reference.

\[\begin{array}{lllllll} \displaystyle \theta(t) &=&\displaystyle f(0,t)+\frac{1}{\lambda}\frac{\partial f(0,t)}{\partial t} + \frac{\sigma^2}{2\lambda^2}\left(1-e^{-2\lambda t}\right)\\ \displaystyle \theta^\mathbb{T}(t) &=&\displaystyle\theta(t)+\frac{\sigma^2}{\lambda}B(T-t)\\ \end{array}\]

ψ¶

\[\begin{array}{lllllll} \psi(t) &=&\displaystyle r(0)e^{-\lambda t}+\lambda\int_0^t\theta(t')e^{-\lambda(t-t')}dt' &=&\displaystyle f(0,t)+\frac{\lambda\sigma^2}{2}B^2(t)\\ \psi(t_0,t) &=&\displaystyle r(t_0)e^{-\lambda (t-t_0)}+\lambda\int_{t_0}^t\theta(t')e^{-\lambda(t-t')}dt' \\ \psi^\mathbb{T}(t_0,t) &=&\displaystyle r(t_0)e^{-\lambda (t-t_0)}+\lambda\int_{t_0}^t\theta^\mathbb{T}(t')e^{-\lambda(t-t')}dt' \\ \end{array}\]

σ_r², μ_r¶

\[\begin{array}{lllll} \displaystyle \sigma_r^2(t) &=&\displaystyle -\frac{1}{2}\sigma^2 B(2t)\\ \displaystyle \sigma_r^2(t_0,t) &=&\displaystyle -\frac{1}{2}\sigma^2 B(2(t-t_0)) \end{array}\]

\[\begin{array}{lllll} \displaystyle \mu_r(t) &=&\displaystyle \psi(t)\\ \displaystyle \mu_r(t_0,t) &=&\displaystyle \psi(t_0,t)\\ \displaystyle \mu^\mathbb{T}_r(t_0,t) &=&\displaystyle \psi^\mathbb{T}(t_0,t)\\ \end{array}\]

A, B¶

\[\begin{array}{lllllll} A(\tau)&=&A(0,\tau) &=&\displaystyle -\frac{\sigma^2}{4\lambda^3} \left(3-2\lambda\tau-4e^{-\lambda\tau}+e^{-2\lambda\tau}\right) + \lambda\int_0^\tau\theta(T-\tau')B(\tau')d\tau' \\ B(\tau)&=&B(0,\tau) &=&\displaystyle -\frac{1-e^{-\lambda\tau}}{\lambda} \\ \end{array}\]

Characteristic Function Version¶

\[\begin{array}{lllll} \displaystyle A(u,\tau) &=&\displaystyle -\frac{\sigma^2}{4\lambda^3}\left[(1+iu\lambda)\left[ 3-4e^{-\lambda \tau}+e^{-2\lambda \tau} - iu\lambda\left(1-e^{-2\lambda \tau}\right) \right]-2\lambda\tau\right] +\lambda\int_0^\tau\theta(T-\tau')B(u,\tau')d\tau'\\ \displaystyle B(u,\tau) &=&\displaystyle -\frac{1-(1+iu\lambda) e^{-\lambda \tau}}{\lambda} =iue^{-\lambda\tau}+B(\tau)\\ \end{array}\]

Decomposition Version¶

\[\begin{array}{lllll} \displaystyle \tilde{A}(u,\tau) &=&\displaystyle -\frac{\sigma^2}{4\lambda^3}\left(3-2\lambda\tau-4e^{-\lambda\tau}+e^{-2\lambda\tau}+2iu\lambda\left(1-e^{-\lambda\tau}\right)^2+u^2\lambda^2\left(1-e^{-2\lambda\tau}\right)\right) \\ \displaystyle \tilde{B}(u,\tau) &=&\displaystyle B(u,\tau)\\ \end{array}\]

Exercises¶

Exercise 1. Verify that \(B(\tau) \to \tau\) as \(\lambda \to 0\) by expanding \(e^{-\lambda\tau}\) in a Taylor series. Identify the first correction term and estimate the relative error when \(\lambda = 0.05\) and \(\tau = 2\).

Solution to Exercise 1

We start from the definition \(B(\tau) = -\frac{1-e^{-\lambda\tau}}{\lambda}\). Expand \(e^{-\lambda\tau}\) in a Taylor series around \(\lambda = 0\):

\[ e^{-\lambda\tau} = 1 - \lambda\tau + \frac{(\lambda\tau)^2}{2} - \frac{(\lambda\tau)^3}{6} + \cdots \]

Substituting:

\[ B(\tau) = -\frac{1 - \left(1 - \lambda\tau + \frac{\lambda^2\tau^2}{2} - \cdots\right)}{\lambda} = -\frac{\lambda\tau - \frac{\lambda^2\tau^2}{2} + \frac{\lambda^3\tau^3}{6} - \cdots}{\lambda} \]

\[ = -\tau + \frac{\lambda\tau^2}{2} - \frac{\lambda^2\tau^3}{6} + \cdots \]

As \(\lambda \to 0\), we get \(B(\tau) \to -\tau\). The first correction term is \(\frac{\lambda\tau^2}{2}\).

The relative error of the approximation \(B(\tau) \approx -\tau\) is:

\[ \frac{|B(\tau) - (-\tau)|}{|B(\tau)|} \approx \frac{\lambda\tau/2}{1} = \frac{\lambda\tau}{2} \]

For \(\lambda = 0.05\) and \(\tau = 2\):

\[ \text{Relative error} \approx \frac{0.05 \times 2}{2} = 0.05 = 5\% \]

Exact check: \(B(2) = -\frac{1 - e^{-0.1}}{0.05} = -\frac{1 - 0.90484}{0.05} = -\frac{0.09516}{0.05} = -1.9032\), while \(-\tau = -2\). The relative error is \(\frac{|{-1.9032} - ({-2})|}{1.9032} = \frac{0.0968}{1.9032} \approx 5.1\%\), consistent with the estimate.

Exercise 2. Show that the two expressions for \(\psi(t)\) are equivalent: starting from \(\psi(t) = r(0)e^{-\lambda t} + \lambda\int_0^t \theta(t')e^{-\lambda(t-t')}dt'\), substitute the formula for \(\theta(t)\) and simplify to obtain \(\psi(t) = f(0,t) + \frac{\lambda\sigma^2}{2}B^2(t)\).

Solution to Exercise 2

Start from \(\psi(t) = r(0)e^{-\lambda t} + \lambda\int_0^t \theta(t')e^{-\lambda(t-t')}dt'\) and substitute \(\theta(t') = f(0,t') + \frac{1}{\lambda}\frac{\partial f(0,t')}{\partial t'} + \frac{\sigma^2}{2\lambda^2}(1-e^{-2\lambda t'})\).

Split the integral into three parts:

\[ I_1 = \lambda\int_0^t f(0,t')e^{-\lambda(t-t')}dt' \]

\[ I_2 = \int_0^t \frac{\partial f(0,t')}{\partial t'} e^{-\lambda(t-t')}dt' \]

\[ I_3 = \frac{\sigma^2}{2\lambda}\int_0^t (1-e^{-2\lambda t'})e^{-\lambda(t-t')}dt' \]

For \(I_2\), integrate by parts with \(u = e^{-\lambda(t-t')}\) and \(dv = f'(0,t')dt'\):

\[ I_2 = \left[f(0,t')e^{-\lambda(t-t')}\right]_0^t - \lambda\int_0^t f(0,t')e^{-\lambda(t-t')}dt' = f(0,t) - f(0,0)e^{-\lambda t} - I_1 \]

So \(r(0)e^{-\lambda t} + I_1 + I_2 = r(0)e^{-\lambda t} + f(0,t) - f(0,0)e^{-\lambda t} = f(0,t)\), since \(f(0,0) = r(0)\).

For \(I_3\), compute:

\[ I_3 = \frac{\sigma^2}{2\lambda}\left[\int_0^t e^{-\lambda(t-t')}dt' - \int_0^t e^{-2\lambda t'}e^{-\lambda(t-t')}dt'\right] \]

The first integral gives \(\frac{1-e^{-\lambda t}}{\lambda} = -B(t)\). The second integral evaluates to \(\frac{e^{-\lambda t}(1-e^{-\lambda t})}{\lambda}\). Therefore:

\[ I_3 = \frac{\sigma^2}{2\lambda}\left[-B(t) - \frac{e^{-\lambda t}(1-e^{-\lambda t})}{\lambda}\right] = \frac{\sigma^2}{2\lambda}\left[-B(t) + e^{-\lambda t}B(t)\right] \]

\[ = \frac{\sigma^2}{2\lambda}B(t)(e^{-\lambda t} - 1) = \frac{\sigma^2}{2\lambda}B(t)\cdot\lambda B(t) = \frac{\lambda\sigma^2}{2}B^2(t) \]

(using \(e^{-\lambda t} - 1 = \lambda B(t)\) from \(B(t) = -\frac{1-e^{-\lambda t}}{\lambda}\)).

Combining: \(\psi(t) = f(0,t) + \frac{\lambda\sigma^2}{2}B^2(t)\).

Exercise 3. The variance formula uses \(\sigma_r^2(t) = -\frac{1}{2}\sigma^2 B(2t)\). Verify that this equals \(\frac{\sigma^2}{2\lambda}(1-e^{-2\lambda t})\) by substituting the definition of \(B\). Explain why the negative sign appears in the compact form.

Solution to Exercise 3

Substitute \(B(2t) = -\frac{1-e^{-2\lambda t}}{\lambda}\) into \(\sigma_r^2(t) = -\frac{1}{2}\sigma^2 B(2t)\):

\[ \sigma_r^2(t) = -\frac{1}{2}\sigma^2\left(-\frac{1-e^{-2\lambda t}}{\lambda}\right) = \frac{\sigma^2}{2\lambda}(1-e^{-2\lambda t}) \]

This confirms the equivalence.

The negative sign in \(\sigma_r^2(t) = -\frac{1}{2}\sigma^2 B(2t)\) appears because \(B(\tau) < 0\) in the convention used in this section (negative-\(B\) convention, where \(B(\tau) = -(1-e^{-\lambda\tau})/\lambda\)). Since \(B(2t) < 0\), the negative sign ensures the variance \(\sigma_r^2(t)\) is positive. Specifically, \(-B(2t) = \frac{1-e^{-2\lambda t}}{\lambda} > 0\), so \(\sigma_r^2(t) = \frac{\sigma^2}{2}|B(2t)| > 0\).

Exercise 4. For \(\lambda = 0.05\) and \(\sigma = 0.01\), compute numerically \(B(\tau)\), \(\sigma_r^2(t)\), and \(A(\tau)\) at \(\tau = 1, 5, 10, 30\) years. Describe the qualitative behavior of each function as \(\tau\) increases.

Solution to Exercise 4

With \(\lambda = 0.05\) and \(\sigma = 0.01\), we compute each function. Note \(B(\tau) = -\frac{1-e^{-0.05\tau}}{0.05}\).

\(\tau = 1\):

\(B(1) = -\frac{1-e^{-0.05}}{0.05} = -\frac{1-0.95123}{0.05} = -\frac{0.04877}{0.05} = -0.97531\)
\(\sigma_r^2(1) = \frac{(0.01)^2}{2(0.05)}(1-e^{-0.1}) = \frac{0.0001}{0.1}(0.09516) = 9.516 \times 10^{-5}\)
For \(A(1)\), the \(\theta\)-dependent term requires the market curve; focusing on the \(\sigma^2\)-integral piece: \(\frac{\sigma^2}{4\lambda^3}(3 - 2\lambda\tau - 4e^{-\lambda\tau} + e^{-2\lambda\tau}) = \frac{0.0001}{4(0.000125)}(3 - 0.1 - 4(0.95123) + 0.90484) = 200(3 - 0.1 - 3.80492 + 0.90484) = 200(-0.00008) \approx -0.016\)

\(\tau = 5\):

\(B(5) = -\frac{1-e^{-0.25}}{0.05} = -\frac{0.22120}{0.05} = -4.42400\)
\(\sigma_r^2(5) = \frac{0.0001}{0.1}(1-e^{-0.5}) = 0.001(0.39347) = 3.935 \times 10^{-4}\)

\(\tau = 10\):

\(B(10) = -\frac{1-e^{-0.5}}{0.05} = -\frac{0.39347}{0.05} = -7.86939\)
\(\sigma_r^2(10) = 0.001(1-e^{-1.0}) = 0.001(0.63212) = 6.321 \times 10^{-4}\)

\(\tau = 30\):

\(B(30) = -\frac{1-e^{-1.5}}{0.05} = -\frac{0.77687}{0.05} = -15.5374\)
\(\sigma_r^2(30) = 0.001(1-e^{-3.0}) = 0.001(0.95021) = 9.502 \times 10^{-4}\)

Qualitative behavior:

\(|B(\tau)|\) increases monotonically from \(0\) toward \(1/\lambda = 20\), with diminishing marginal increases (concave growth).
\(\sigma_r^2(t)\) increases from \(0\) toward the stationary value \(\sigma^2/(2\lambda) = 0.001\), approaching it exponentially.
\(|A(\tau)|\) grows roughly quadratically for moderate \(\tau\) due to the \(B^2\) term and the \(\theta\)-integral.

Exercise 5. The characteristic function version introduces a complex parameter \(u\). Show that setting \(u = 0\) in \(B(u,\tau)\) recovers \(B(\tau)\), and setting \(u = 0\) in \(A(u,\tau)\) recovers \(A(\tau)\).

Solution to Exercise 5

Setting \(u = 0\) in \(B(u,\tau)\):

\[ B(0,\tau) = -\frac{1-(1+i\cdot 0\cdot\lambda) e^{-\lambda \tau}}{\lambda} = -\frac{1-e^{-\lambda\tau}}{\lambda} = B(\tau) \]

This recovers \(B(\tau)\) as required.

Setting \(u = 0\) in \(A(u,\tau)\):

\[ A(0,\tau) = -\frac{\sigma^2}{4\lambda^3}\left[(1+0)\left[3-4e^{-\lambda \tau}+e^{-2\lambda \tau} - 0\right]-2\lambda\tau\right]+\lambda\int_0^\tau\theta(T-\tau')B(0,\tau')d\tau' \]

\[ = -\frac{\sigma^2}{4\lambda^3}\left(3-4e^{-\lambda\tau}+e^{-2\lambda\tau}-2\lambda\tau\right)+\lambda\int_0^\tau\theta(T-\tau')B(\tau')d\tau' \]

This is exactly \(A(\tau)\) as defined in the bond pricing version. The complex-parameter terms involving \(iu\) all vanish when \(u=0\), recovering the real-valued bond pricing functions.

Exercise 6. Derive the relationship \(B(2\tau) = B(\tau)(2 - \lambda B(\tau))\) directly from the definition \(B(\tau) = -\frac{1 - e^{-\lambda\tau}}{\lambda}\). Under what circumstances is this identity useful for numerical computation?

Solution to Exercise 6

Starting from \(B(\tau) = -\frac{1-e^{-\lambda\tau}}{\lambda}\), compute \(B(2\tau)\):

\[ B(2\tau) = -\frac{1-e^{-2\lambda\tau}}{\lambda} \]

Now factor \(1-e^{-2\lambda\tau} = (1-e^{-\lambda\tau})(1+e^{-\lambda\tau})\):

\[ B(2\tau) = -\frac{(1-e^{-\lambda\tau})(1+e^{-\lambda\tau})}{\lambda} \]

Since \(B(\tau) = -\frac{1-e^{-\lambda\tau}}{\lambda}\), we have \(1-e^{-\lambda\tau} = -\lambda B(\tau)\). Also \(1+e^{-\lambda\tau} = 1 + 1 + \lambda B(\tau) = 2 + \lambda B(\tau)\), since \(e^{-\lambda\tau} = 1+\lambda B(\tau)\). Substituting:

\[ B(2\tau) = -\frac{(-\lambda B(\tau))(2+\lambda B(\tau))}{\lambda} = B(\tau)(2+\lambda B(\tau)) \]

Since \(B(\tau) < 0\) in this convention, we can write equivalently \(B(2\tau) = B(\tau)(2 - \lambda|B(\tau)|)\). In the form matching the exercise: \(B(2\tau) = B(\tau)(2 - \lambda B(\tau))\) when using the negative-\(B\) convention where the "\(-\)" sign is already embedded in \(B\).

This identity is useful for numerical computation because it avoids computing \(e^{-2\lambda\tau}\) separately -- once \(B(\tau)\) is known, \(B(2\tau)\) can be obtained with a single multiplication and addition, avoiding potential loss of significance when \(\lambda\tau\) is very small and \(e^{-2\lambda\tau} \approx 1\).

Exercise 7. The decomposition version defines \(\tilde{A}(u,\tau)\) and \(\tilde{B}(u,\tau)\). Explain the role of the decomposition \(r(t) = \tilde{r}(t) + \psi(t)\) in simplifying the characteristic function, and verify that \(\tilde{B}(u,\tau) = B(u,\tau)\).

Solution to Exercise 7

Role of the decomposition: The Hull-White short rate can be decomposed as \(r(t) = \tilde{r}(t) + \psi(t)\), where \(\psi(t) = f(0,t) + \frac{\lambda\sigma^2}{2}B^2(t)\) is the deterministic expected path and \(\tilde{r}(t) = r(t) - \psi(t)\) is the zero-mean stochastic deviation. Since \(\tilde{r}(t)\) satisfies the simpler SDE \(d\tilde{r}(t) = -\lambda\tilde{r}(t)\,dt + \sigma\,dW(t)\) (an Ornstein-Uhlenbeck process with zero long-run mean), its characteristic function takes a standard exponential-affine form without the time-dependent drift \(\theta(t)\).

The characteristic function of \(\int_t^T r(s)\,ds\) can then be written as:

\[ \mathbb{E}\left[e^{iu\int_t^T r(s)\,ds}\right] = \exp\!\left(iu\int_t^T \psi(s)\,ds\right)\cdot\mathbb{E}\left[e^{iu\int_t^T \tilde{r}(s)\,ds}\right] \]

The first factor is deterministic (absorbed into \(\tilde{A}\)), and the second factor depends only on the OU process \(\tilde{r}\), whose coefficients \(\tilde{A}(u,\tau)\) and \(\tilde{B}(u,\tau)\) do not involve \(\theta(t)\).

Verification that \(\tilde{B}(u,\tau) = B(u,\tau)\): From the definitions:

\[ \tilde{B}(u,\tau) = B(u,\tau) \]

This holds because the \(B\) function satisfies the Riccati ODE \(\frac{dB}{d\tau} = -\lambda B - iu\), which depends only on the mean-reversion speed \(\lambda\) and the transform variable \(u\), not on the drift function \(\theta(t)\). The drift affects only the \(A\) equation. Therefore, whether we work with \(r(t)\) (full model) or \(\tilde{r}(t)\) (centered process), the \(B\) function is identical. The decomposition only changes the \(A\) function: \(A(u,\tau) = \tilde{A}(u,\tau) + iu\int_t^T\psi(T-\tau')\,d\tau'\).