Hull-White Model Class Guide

This guide describes the design and implementation of the HullWhite class, the central object for one-factor Hull-White model computations. The class encapsulates the model parameters \((\sigma, \lambda)\) and the market yield curve \(P^M(0,T)\), and provides methods for computing the named functions, bond prices, short rate simulation, and derivative pricing. The companion .py file contains the full implementation; this page explains the design choices and maps each method to its mathematical formula.

Class Overview

The HullWhite class is initialized with three inputs:

• sigma: the short rate volatility \(\sigma > 0\)
• lambd: the mean-reversion speed \(\lambda > 0\)
• P: a callable representing the market discount curve \(P^M(0, T)\)

All other quantities (\(\theta(t)\), \(A(t,T)\), \(B(t,T)\), the short rate distribution parameters) are derived from these three inputs.

Constructor and Parameters

python class HullWhite: def __init__(self, sigma, lambd, P): self.sigma = sigma self.lambd = lambd self.P = P

The market curve P is a function \(P^M: [0, \infty) \to (0, 1]\) satisfying \(P^M(0) = 1\). In practice, this is obtained by interpolating market bond prices or by fitting a parametric model (e.g., Nelson-Siegel) to swap rates.

Named Function Methods

compute_B(t, T)

Computes the duration-like function:

\[ B(t, T) = \frac{e^{-\lambda(T-t)} - 1}{\lambda} \]

This function appears in the bond price formula, the bond volatility, and the measure change.

compute_A(t, T)

Computes the bond price intercept function by numerical integration:

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

The integral is evaluated using the trapezoidal rule on a grid of 250 points. The function compute_theta provides the integrand.

compute_theta(t)

Computes the time-dependent drift:

\[ \theta(t) = f^M(0, t) + \frac{1}{\lambda}\frac{\partial f^M(0, t)}{\partial t} + \frac{\sigma^2}{2\lambda^2}\left(1 - e^{-2\lambda t}\right) \]

where \(f^M(0, t)\) is the instantaneous forward rate, obtained by numerical differentiation of \(\ln P^M(0, t)\).

compute_theta_T(t, T)

Computes the drift under the \(T\)-forward measure:

\[ \theta^{\mathbb{T}}(t) = \theta(t) + \frac{\sigma^2}{\lambda}B(T - t) \]

Bond Pricing Methods

compute_ZCB(t, T, r_t)

Computes the zero-coupon bond price:

\[ P(t, T) = \exp\!\bigl(A(t, T) + B(t, T)\,r_t\bigr) \]

compute_sigma_P(t, T)

Computes the bond price volatility:

\[ \sigma_P(t, T) = \sigma\,B(t, T) \]

Short Rate Distribution Methods

compute_mu_r_T(T)

Computes the conditional mean of \(r_T\) under \(\mathbb{Q}\):

\[ \mu_r(T) = r_0\,e^{-\lambda T} + \lambda\int_0^T \theta(s)\,e^{-\lambda(T-s)}\,ds \]

compute_mu_r_T_ForwardMeasure(T)

Computes the conditional mean of \(r_T\) under \(\mathbb{Q}^T\):

\[ \mu_r^{\mathbb{T}}(T) = r_0\,e^{-\lambda T} + \lambda\int_0^T \theta^{\mathbb{T}}(s)\,e^{-\lambda(T-s)}\,ds \]

compute_sigma_square_r_T(T)

Computes the conditional variance of \(r_T\):

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

This variance is the same under both \(\mathbb{Q}\) and \(\mathbb{Q}^T\) because the Girsanov transformation only changes the drift, not the diffusion.

Simulation Methods

generate_sample_paths(num_paths, num_steps, T, seed)

Generates short rate paths using Euler-Maruyama discretization:

\[ r_{t_{i+1}} = r_{t_i} + \lambda\left(\theta(t_i) - r_{t_i}\right)\Delta t + \sigma\,\sqrt{\Delta t}\,Z_{i+1} \]

where \(Z_{i+1} \sim N(0, 1)\). The method also computes the money market account \(M(t_i) = \prod_{k=0}^{i-1}e^{r_{t_k}\Delta t}\).

Returns: time grid t, short rate paths R, money market account M.

Moment Matching

When num_paths > 1, the normal draws are centered and standardized: \(Z_i \leftarrow (Z_i - \bar{Z}) / \text{std}(Z)\). This ensures that the sample mean and variance match the theoretical values exactly, reducing Monte Carlo bias for finite samples.

Derivative Pricing Methods

compute_ZCB_Option_Price(K, T1, T2, CP)

Prices a European call or put on a ZCB \(P(T_1, T_2)\) with strike \(K\):

\[ V = P(0, T_1)\,e^{A(\tau)}\left[e^{\frac{1}{2}B^2(\tau)\sigma_r^2 + B(\tau)\mu_r^{\mathbb{T}}}\,N(d_1) - \hat{K}\,N(d_2)\right] \]

The put price uses put-call parity: \(\text{Put} = \text{Call} - P(0, T_2) + K\,P(0, T_1)\).

compute_Caplet_Floorlet_Price(N, K, T1, T2, CP)

Prices a caplet or floorlet using the caplet-as-ZCB-put relationship:

\[ \text{Caplet}(K) = (1 + \delta K)\,\text{Put}\!\left(\frac{1}{1+\delta K}\right) \]

where \(\delta = T_2 - T_1\).

compute_SwapPrice(t, r_t, notional, K, Ti, Tm, n, CP)

Computes the swap value at time \(t\) given \(r_t\):

\[ V_{\text{swap}} = N\left[P(t, T_i) - P(t, T_m) - K\sum_{k} \delta_k\,P(t, T_k)\right] \]

Design Principles

1. Single responsibility: each method computes one named function or formula
2. Composability: complex computations (e.g., swaption pricing via Jamshidian) compose simpler methods
3. Market curve as input: the class takes \(P^M(0,T)\) as a callable, allowing any curve representation (interpolation, parametric, bootstrap)
4. Numerical differentiation: forward rates and their derivatives are computed by central differences with step size \(10^{-4}\), avoiding the need for an explicit forward curve representation

Summary

The HullWhite class implements the complete one-factor Hull-White model computation chain: from model parameters and market curve through named functions (\(\theta\), \(A\), \(B\)), bond pricing, short rate simulation, and derivative pricing (ZCB options, caplets, swaps). Each method maps directly to a mathematical formula from the theory sections, with numerical integration and differentiation used where closed-form expressions require the initial forward curve. The class is designed for composability, enabling the construction of higher-level pricing and calibration routines.

---

Exercises

Exercise 1. Using a flat market curve at 4\% and Hull-White parameters \(\sigma = 0.01\), \(\lambda = 0.10\), compute \(\theta(t)\) for \(t = 0.5, 1, 5, 10\) years. Verify that in the flat-curve case, \(\theta(t)\) converges to the long-run level \(r + \sigma^2/(2\lambda^2)\) as \(t \to \infty\). What is the numerical value of this limit?

Solution to Exercise 1

For a flat market curve at 4%, we have \(P^M(0, t) = e^{-0.04\,t}\), so the instantaneous forward rate is \(f^M(0, t) = 0.04\) for all \(t\), and \(\partial f^M / \partial t = 0\). The \(\theta(t)\) formula simplifies to:

\[ \theta(t) = 0.04 + 0 + \frac{(0.01)^2}{2(0.10)^2}(1 - e^{-2 \cdot 0.10 \cdot t}) = 0.04 + 0.005(1 - e^{-0.2t}) \]

Computing at the specified times:

• \(t = 0.5\): \(\theta(0.5) = 0.04 + 0.005(1 - e^{-0.1}) = 0.04 + 0.005 \times 0.09516 = 0.04048\)
• \(t = 1\): \(\theta(1) = 0.04 + 0.005(1 - e^{-0.2}) = 0.04 + 0.005 \times 0.18127 = 0.04091\)
• \(t = 5\): \(\theta(5) = 0.04 + 0.005(1 - e^{-1.0}) = 0.04 + 0.005 \times 0.63212 = 0.04316\)
• \(t = 10\): \(\theta(10) = 0.04 + 0.005(1 - e^{-2.0}) = 0.04 + 0.005 \times 0.86466 = 0.04432\)

As \(t \to \infty\), \(e^{-2\lambda t} \to 0\), so:

\[ \lim_{t \to \infty} \theta(t) = r + \frac{\sigma^2}{2\lambda^2} = 0.04 + \frac{(0.01)^2}{2(0.10)^2} = 0.04 + 0.005 = 0.045 \]

The long-run level is \(0.045\) or 4.5%. The convexity adjustment \(\sigma^2/(2\lambda^2)\) raises the drift target above the flat rate to compensate for the Jensen inequality effect in bond pricing.

---

Exercise 2. The compute_B(t, T) method returns \((e^{-\lambda(T-t)} - 1)/\lambda\). Show that \(B(t, T) \to -(T-t)\) as \(\lambda \to 0\) and \(B(t, T) \to -1/\lambda\) as \(T - t \to \infty\). Interpret each limit financially in terms of the bond price sensitivity to the short rate.

Solution to Exercise 2

Limit as \(\lambda \to 0\): Apply L'Hopital's rule or a Taylor expansion. Write \(e^{-\lambda\tau} \approx 1 - \lambda\tau + \frac{1}{2}\lambda^2\tau^2 - \cdots\), so:

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

As \(\lambda \to 0\), \(B(t, T) \to -(T - t)\). This is the Vasicek limit with zero mean reversion: the bond price has log-linear sensitivity to the short rate with coefficient \(-(T-t)\), identical to the duration of a zero-coupon bond under flat rates.

Limit as \(T - t \to \infty\): As \(\tau \to \infty\), \(e^{-\lambda\tau} \to 0\), so:

\[ B(t, T) = \frac{e^{-\lambda\tau} - 1}{\lambda} \to \frac{-1}{\lambda} \]

This means the sensitivity of the bond price to the short rate saturates at \(-1/\lambda\). Financially, mean reversion implies that the short rate will eventually revert to its long-run level regardless of its current value. Therefore, very long-dated bonds are insensitive to the current short rate beyond the mean-reversion horizon \(1/\lambda\). A rate shock today has an effect that decays exponentially, so the bond price sensitivity cannot grow without bound.

---

Exercise 3. The variance of the short rate is \(\sigma_r^2(T) = \frac{\sigma^2}{2\lambda}(1 - e^{-2\lambda T})\). This is the same under both \(\mathbb{Q}\) and \(\mathbb{Q}^T\). Explain why the Girsanov theorem guarantees this invariance. Compute \(\sigma_r(T)\) for \(T = 1\) and \(T = 10\) with \(\sigma = 0.01\), \(\lambda = 0.05\).

Solution to Exercise 3

The Girsanov theorem states that the change of measure from \(\mathbb{Q}\) to \(\mathbb{Q}^T\) transforms the drift of the Brownian motion but does not alter the diffusion coefficient. In the Hull-White SDE:

\[ dr_t = \lambda(\theta(t) - r_t)\,dt + \sigma\,dW_t^{\mathbb{Q}} \]

Under \(\mathbb{Q}^T\), the SDE becomes:

\[ dr_t = \lambda(\theta^{\mathbb{T}}(t) - r_t)\,dt + \sigma\,dW_t^{\mathbb{Q}^T} \]

The volatility \(\sigma\) is unchanged because the Girsanov density only introduces a drift adjustment to the Brownian motion. Since the conditional variance depends only on \(\sigma\) and \(\lambda\) (not on the drift), it is invariant under the measure change.

Computing \(\sigma_r(T)\) with \(\sigma = 0.01\), \(\lambda = 0.05\):

\[ \sigma_r^2(T) = \frac{(0.01)^2}{2 \times 0.05}(1 - e^{-2 \times 0.05 \times T}) = 0.001(1 - e^{-0.1T}) \]

For \(T = 1\):

\[ \sigma_r^2(1) = 0.001(1 - e^{-0.1}) = 0.001 \times 0.09516 = 9.516 \times 10^{-5} \]

\[ \sigma_r(1) = \sqrt{9.516 \times 10^{-5}} \approx 0.00976 = 0.976\% \]

For \(T = 10\):

\[ \sigma_r^2(10) = 0.001(1 - e^{-1.0}) = 0.001 \times 0.63212 = 6.321 \times 10^{-4} \]

\[ \sigma_r(10) = \sqrt{6.321 \times 10^{-4}} \approx 0.02514 = 2.514\% \]

The standard deviation grows from 0.976% at 1 year to 2.514% at 10 years, approaching the asymptotic limit \(\sigma/\sqrt{2\lambda} = 0.01/\sqrt{0.1} \approx 3.162\%\).

---

Exercise 4. The generate_sample_paths method uses moment matching: \(Z_i \leftarrow (Z_i - \bar{Z})/\text{std}(Z)\). Explain why this reduces Monte Carlo bias. Generate 10,000 paths for \(T = 5\) years with \(\sigma = 0.01\), \(\lambda = 0.05\), flat curve at 3\%, and compare the sample mean and variance of \(r_5\) with the analytical values from compute_mu_r_T(5) and compute_sigma_square_r_T(5).

Solution to Exercise 4

Why moment matching reduces bias: In a standard Monte Carlo simulation, finite-sample draws from \(N(0,1)\) will have sample mean \(\bar{Z} \neq 0\) and sample variance \(s^2 \neq 1\). These sampling errors propagate through the Euler-Maruyama scheme, biasing the terminal distribution of \(r_T\). By applying \(Z_i \leftarrow (Z_i - \bar{Z})/\text{std}(Z)\), we enforce \(\bar{Z} = 0\) and \(s^2 = 1\) exactly at each time step, eliminating the first two sources of finite-sample bias. This ensures the simulated paths have the correct drift and diffusion properties at every step.

For the numerical comparison with \(\sigma = 0.01\), \(\lambda = 0.05\), flat curve at 3%, \(T = 5\):

Analytical values:

\[ \mu_r(5) = r_0 e^{-\lambda T} + \lambda \int_0^T \theta(s) e^{-\lambda(T-s)}\,ds \]

With a flat curve, \(\theta(s) = r + \frac{\sigma^2}{2\lambda^2}(1 - e^{-2\lambda s})\) where \(r = 0.03\). The analytical conditional mean evaluates to approximately \(\mu_r(5) \approx 0.03 + \frac{\sigma^2}{2\lambda^2}(1 - e^{-2\lambda \cdot 5}) \times (\text{integrated contribution}) \approx 0.03079\).

\[ \sigma_r^2(5) = \frac{(0.01)^2}{2 \times 0.05}(1 - e^{-0.5}) = 0.001 \times 0.39347 = 3.935 \times 10^{-4} \]

With moment matching and 10,000 paths, the sample mean and variance of \(r_5\) should match these analytical values to within \(O(1/\sqrt{N})\) statistical error, which for \(N = 10{,}000\) is of order \(10^{-4}\) for the mean.

---

Exercise 5. The compute_theta_T method adjusts \(\theta(t)\) for the \(T\)-forward measure. Show that the adjustment \(\sigma^2 B(T-t)/\lambda\) has the correct sign to produce a lower expected short rate under the \(T\)-forward measure compared to \(\mathbb{Q}\) (when \(T > t\)). Why does a lower expected rate under the \(T\)-forward measure make economic sense for bond pricing?

Solution to Exercise 5

The adjustment to the drift under the \(T\)-forward measure is:

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

Since \(B(\tau) = (e^{-\lambda\tau} - 1)/\lambda < 0\) for all \(\tau > 0\) (because \(e^{-\lambda\tau} < 1\)), the adjustment \(\frac{\sigma^2}{\lambda}B(T - t)\) is negative when \(T > t\).

Therefore \(\theta^{\mathbb{T}}(t) < \theta(t)\), which means the drift of the short rate is lower under the \(T\)-forward measure than under \(\mathbb{Q}\). This produces a lower expected short rate under \(\mathbb{Q}^T\).

Economic interpretation: Under the \(T\)-forward measure \(\mathbb{Q}^T\), the numeraire is the zero-coupon bond \(P(t, T)\). When the short rate is lower, bond prices are higher. The \(T\)-forward measure "tilts" the distribution toward states where the bond (the numeraire) has high value, i.e., low-rate states. This is the Radon-Nikodym effect: the measure change reweights paths by \(P(t, T)/P(0, T)\), giving more probability weight to paths with low rates. This tilt is essential for the forward-measure pricing formula \(\mathbb{E}^{\mathbb{Q}^T}[V(T)] = V(0)/P(0, T)\) to hold.

---

Exercise 6. Using the compute_SwapPrice method, price a 5-year payer swap (annual payments) with fixed rate \(K = 3\%\) at \(t = 0\) with \(r_0 = 3\%\) and a flat market curve at 3\%. Verify that the swap value is approximately zero (since \(K\) equals the par rate). Then compute the swap value at \(t = 0\) for \(K = 4\%\) and interpret the sign.

Solution to Exercise 6

For a 5-year payer swap with annual payments, \(K = 3\%\), \(t = 0\), \(r_0 = 3\%\), flat curve at 3%:

\[ V_{\text{swap}} = N\left[P(0, T_0) - P(0, T_5) - K\sum_{k=1}^{5}\delta_k P(0, T_k)\right] \]

With \(\delta_k = 1\) (annual), \(T_k = k\), and \(P(0, k) = e^{-0.03k}\):

\[ \sum_{k=1}^{5} P(0, k) = e^{-0.03} + e^{-0.06} + e^{-0.09} + e^{-0.12} + e^{-0.15} \]

\[ = 0.97045 + 0.94176 + 0.91393 + 0.88692 + 0.86071 = 4.57377 \]

The par rate for this swap satisfies \(P(0, 0) - P(0, 5) = K_{\text{par}} \sum P(0, T_k)\), giving \(K_{\text{par}} = (1 - e^{-0.15})/4.57377 = 0.13929/4.57377 \approx 0.03046\) or about 3.046% (slightly above 3% due to continuous vs. discrete compounding effects).

Since the par rate is approximately 3.046%, a swap with \(K = 3\%\) has a small positive value for the payer (receiving floating, paying fixed at a rate slightly below par): \(V \approx N \times (K_{\text{par}} - K) \times \text{annuity} \approx N \times 0.00046 \times 4.574 \approx 0.0021 \times N\).

For \(K = 4\%\), the payer swap has a negative value because the fixed rate paid (4%) exceeds the par rate (3.046%). The payer is locked into paying above-market fixed rates:

\[ V_{\text{swap}}(K=4\%) = N\left[(1 - 0.86071) - 0.04 \times 4.57377\right] = N[0.13929 - 0.18295] = -0.04366 \times N \]

The negative sign confirms the payer is at a disadvantage.

---

Exercise 7. The compute_ZCB_Option_Price method prices European options on zero-coupon bonds. Price a call on \(P(2, 5)\) with strike \(K = 0.92\), using \(\sigma = 0.01\), \(\lambda = 0.05\), and a flat curve at 3\%. Then use put-call parity to obtain the put price. Verify by calling the method directly with CP=OptionType.PUT.

Solution to Exercise 7

With \(\sigma = 0.01\), \(\lambda = 0.05\), flat curve at 3%, we price a call on \(P(2, 5)\) with strike \(K = 0.92\).

First, compute the bond price volatility:

\[ \sigma_P = \frac{\sigma}{\lambda}(1 - e^{-\lambda(T_2 - T_1)})\sqrt{\frac{1 - e^{-2\lambda T_1}}{2\lambda}} \]

\[ = \frac{0.01}{0.05}(1 - e^{-0.05 \times 3})\sqrt{\frac{1 - e^{-0.1 \times 2}}{0.1}} \]

\[ = 0.2 \times (1 - e^{-0.15})\sqrt{\frac{1 - e^{-0.2}}{0.1}} \]

\[ = 0.2 \times 0.13929 \times \sqrt{\frac{0.18127}{0.1}} = 0.02786 \times \sqrt{1.8127} = 0.02786 \times 1.3464 = 0.03751 \]

Next, compute the relevant bond prices:

\[ P(0, 2) = e^{-0.06} = 0.94176, \qquad P(0, 5) = e^{-0.15} = 0.86071 \]

Compute \(d_1\) and \(d_2\):

\[ d_1 = \frac{\ln(P(0,5)/(K \cdot P(0,2)))}{\sigma_P} + \frac{\sigma_P}{2} = \frac{\ln(0.86071/(0.92 \times 0.94176))}{0.03751} + \frac{0.03751}{2} \]

\[ = \frac{\ln(0.86071/0.86642)}{0.03751} + 0.01876 = \frac{\ln(0.99341)}{0.03751} + 0.01876 = \frac{-0.00661}{0.03751} + 0.01876 \]

\[ = -0.17622 + 0.01876 = -0.15746 \]

\[ d_2 = d_1 - \sigma_P = -0.15746 - 0.03751 = -0.19497 \]

The call price:

\[ C = P(0,5)\,\Phi(d_1) - K\,P(0,2)\,\Phi(d_2) = 0.86071 \times \Phi(-0.15746) - 0.92 \times 0.94176 \times \Phi(-0.19497) \]

\[ = 0.86071 \times 0.43743 - 0.86642 \times 0.42272 = 0.37647 - 0.36622 = 0.01025 \]

Using put-call parity:

\[ \text{Put} = C - P(0, 5) + K \cdot P(0, 2) = 0.01025 - 0.86071 + 0.92 \times 0.94176 = 0.01025 - 0.86071 + 0.86642 = 0.01596 \]

Calling the method with CP=OptionType.PUT directly should return the same value of approximately \(0.01596\).