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Hull-White Short Rate

Hull-White Model

\[\begin{array}{lllll} \displaystyle dr(t)=\lambda\left(\theta^\mathbb{Q}(t)-r(t)\right) dt+\sigma dW^{\mathbb{Q}}(t) \end{array}\]

where

\[ \displaystyle \theta^\mathbb{Q}(t) = f^M(0,t) +\frac{1}{\lambda}\frac{\partial f^M(0,t)}{\partial t} +\frac{\sigma^2}{2\lambda^2}\left(1-e^{-2\lambda t}\right) \]
Proof
\[\begin{array}{lllll} \displaystyle \sigma(t)=\sigma(t,t)=\sigma \end{array}\]
\[\begin{array}{lllll} \displaystyle \mu(t) &=&\displaystyle \mu(t,t)+ \frac{\partial f(0,t)}{\partial t} +\int_0^t\frac{\partial \mu(t',t)}{\partial t}dt' +\int_0^t\frac{\partial \sigma(t',t)}{\partial t}dW^{\mathbb{Q}}(t')\\ &=&\displaystyle \frac{\partial f(0,t)}{\partial t} +\int_0^t\left[2\sigma^2e^{-2\lambda(t-t')} -\sigma^2e^{-\lambda(t-t')}\right]dt' -\lambda \int_0^t\sigma(t',t)dW^{\mathbb{Q}}(t')\\ &=&\displaystyle \frac{\partial f(0,t)}{\partial t} +\frac{\sigma^2}{\lambda}e^{-2\lambda t}\left(e^{\lambda t}-1\right)-\lambda \int_0^t\sigma(t',t)dW^{\mathbb{Q}}(t')\\ \end{array}\]

Handling the stochastic integral term:

\[\begin{array}{l} \displaystyle r(t) = f(t,t) = f(0,t)+\int_0^t\mu(t',t)dt'+\int_0^t\sigma(t',t)dW^{\mathbb{Q}}(t') \end{array}\]

implies

\[\begin{array}{llllllll} \displaystyle \int_0^t\sigma(t',t)dW^{\mathbb{Q}}(t') &=&\displaystyle r(t)-f(0,t)-\frac{\sigma^2}{2\lambda^2}e^{-2\lambda t}\left(e^{\lambda t}-1\right)^2 \end{array}\]

Substituting back:

\[\begin{array}{lllll} \displaystyle \mu(t) &=&\displaystyle \lambda\left( f(0,t) +\frac{1}{\lambda}\frac{\partial f(0,t)}{\partial t} +\frac{\sigma^2}{2\lambda^2}\left(1-e^{-2\lambda t}\right) -r(t) \right)\\ &:=&\displaystyle \lambda\left( \theta(t) -r(t) \right)\\ \end{array}\]

Hull-White Model Solution

\[\begin{array}{lllll} \displaystyle r(t) &=&\displaystyle r(s)e^{-\lambda (t-s)}+ \lambda\int_{s}^t\theta^\mathbb{Q}(t')e^{-\lambda (t-t')} dt'+\sigma \int_{s}^t e^{-\lambda (t-t')} dW^{\mathbb{Q}}(t')\\ &=&\displaystyle r(s)e^{-\lambda (t-s)}+ \alpha(t)-\alpha(s)e^{-\lambda(t-s)}+\sigma \int_{s}^t e^{-\lambda (t-t')} dW^{\mathbb{Q}}(t')\\ \end{array}\]

where

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

Therefore, \(r(t)\) conditional on \({\cal F}(s)\) is normally distributed with mean and variance:

\[\begin{array}{lllll} \displaystyle \mathbb{E}\left(r(t)|{\cal F}(s)\right) &=&\displaystyle r(s)e^{-\lambda (t-s)}+ \alpha(t)-\alpha(s)e^{-\lambda(t-s)}\\ \mathbb{Var}\left(r(t)|{\cal F}(s)\right) &=&\displaystyle \frac{\sigma^2}{2\lambda}\left(1-e^{-2\lambda(t-s)}\right) \end{array}\]
Proof
\[\begin{array}{lllll} \displaystyle &&dr(t)=\lambda\left(\theta(t)-r(t)\right) dt+\sigma dW^{\mathbb{Q}}(t)\\ \\ &\Rightarrow& \displaystyle e^{\lambda t}dr(t) + \lambda r(t)e^{\lambda t} dt=\lambda\theta(t)e^{\lambda t} dt+\sigma e^{\lambda t} dW^{\mathbb{Q}}(t)\quad(y(t)=r(t)e^{\lambda t})\\ \\ &\Rightarrow& \displaystyle dy(t)=\lambda\theta(t)e^{\lambda t} dt+\sigma e^{\lambda t} dW^{\mathbb{Q}}(t)\\ \\ &\Rightarrow& \displaystyle r(t) = r(0)e^{-\lambda t}+ \lambda\int_0^t\theta(t')e^{-\lambda (t-t')} dt'+\sigma \int_0^t e^{-\lambda (t-t')} dW^{\mathbb{Q}}(t')\\ \end{array}\]

Therefore,

\[\begin{array}{lllll} \displaystyle \mathbb{E}^\mathbb{Q}\left(r(t)\big|{\cal F}(0)\right) = r(0)e^{-\lambda t}+ \lambda\int_0^t\theta(t')e^{-\lambda (t-t')} dt' = \psi(t)\\ \end{array}\]
\[\begin{array}{lllll} \displaystyle \mathbb{Var}^\mathbb{Q}\left(r(t)\big|{\cal F}(0)\right) = \sigma^2 \int_0^t e^{-2\lambda (t-t')} dt' = \frac{\sigma^2}{2\lambda} \left(1-e^{-2\lambda t}\right) =-\frac{1}{2}\sigma^2 B(2t) \end{array}\]

Hull-White Short Rate Sample Path

```python import matplotlib.pyplot as plt

def main(): hw = HullWhite(sigma=0.01, lambd=0.01, P=P_market) t, R, M = hw.generate_sample_paths( num_paths=20, num_steps=100, T=30, seed=42 )

plt.figure(figsize=(10, 4))
plt.title("Hull-White Short Rate Sample Paths")
for i in range(20):
    plt.plot(t, R[i, :], alpha=0.3, lw=0.5)
plt.plot(t, R.mean(axis=0), 'k--', lw=2, label='Mean')
plt.xlabel("Time")
plt.ylabel("r(t)")
plt.legend()
plt.show()

```

Exercises

Exercise 1. Starting from the Hull-White SDE \(dr(t) = \lambda(\theta^{\mathbb{Q}}(t) - r(t))\,dt + \sigma\,dW^{\mathbb{Q}}(t)\), derive the explicit solution using the integrating factor \(e^{\lambda t}\). Show every step of the derivation.

Solution to Exercise 1

Starting from the Hull-White SDE:

\[ dr(t) = \lambda(\theta^{\mathbb{Q}}(t) - r(t))\,dt + \sigma\,dW^{\mathbb{Q}}(t) \]

Step 1: Apply the integrating factor. Multiply both sides by \(e^{\lambda t}\):

\[ e^{\lambda t}\,dr(t) + \lambda r(t) e^{\lambda t}\,dt = \lambda\theta^{\mathbb{Q}}(t) e^{\lambda t}\,dt + \sigma e^{\lambda t}\,dW^{\mathbb{Q}}(t) \]

Step 2: Recognize the product rule. The left side is \(d(r(t)e^{\lambda t})\) by the product rule for Ito processes (since \(e^{\lambda t}\) is deterministic, there is no Ito correction). Define \(y(t) = r(t)e^{\lambda t}\), so:

\[ dy(t) = \lambda\theta^{\mathbb{Q}}(t)e^{\lambda t}\,dt + \sigma e^{\lambda t}\,dW^{\mathbb{Q}}(t) \]

Step 3: Integrate from \(s\) to \(t\). Integrating both sides:

\[ y(t) - y(s) = \lambda\int_s^t \theta^{\mathbb{Q}}(t')e^{\lambda t'}\,dt' + \sigma\int_s^t e^{\lambda t'}\,dW^{\mathbb{Q}}(t') \]

Step 4: Substitute back. Since \(y(t) = r(t)e^{\lambda t}\) and \(y(s) = r(s)e^{\lambda s}\):

\[ r(t)e^{\lambda t} = r(s)e^{\lambda s} + \lambda\int_s^t \theta^{\mathbb{Q}}(t')e^{\lambda t'}\,dt' + \sigma\int_s^t e^{\lambda t'}\,dW^{\mathbb{Q}}(t') \]

Step 5: Divide by \(e^{\lambda t}\):

\[ r(t) = r(s)e^{-\lambda(t-s)} + \lambda\int_s^t \theta^{\mathbb{Q}}(t')e^{-\lambda(t-t')}\,dt' + \sigma\int_s^t e^{-\lambda(t-t')}\,dW^{\mathbb{Q}}(t') \]

This is the explicit solution. The first term shows exponential decay of the initial condition, the second is the deterministic mean driven by \(\theta^{\mathbb{Q}}\), and the third is the stochastic contribution filtered by the mean-reversion kernel.

Exercise 2. Verify that \(\alpha(t) = f^M(0,t) + \frac{\sigma^2}{2\lambda^2}(1 - e^{-\lambda t})^2\) satisfies \(\alpha(0) = r_0 = f^M(0,0)\). Then show that \(\alpha(t)\) is the conditional mean \(\mathbb{E}^{\mathbb{Q}}[r(t) | \mathcal{F}(0)]\).

Solution to Exercise 2

Checking \(\alpha(0)\): Substituting \(t = 0\):

\[ \alpha(0) = f^M(0,0) + \frac{\sigma^2}{2\lambda^2}(1 - e^0)^2 = f^M(0,0) + 0 = f^M(0,0) = r_0 \]

since the instantaneous forward rate at time zero equals the spot rate.

Showing \(\alpha(t) = \mathbb{E}^{\mathbb{Q}}[r(t) | \mathcal{F}(0)]\): From the explicit solution with \(s = 0\):

\[ \mathbb{E}^{\mathbb{Q}}[r(t) | \mathcal{F}(0)] = r(0)e^{-\lambda t} + \lambda\int_0^t \theta^{\mathbb{Q}}(t')e^{-\lambda(t-t')}\,dt' \]

Using \(\theta^{\mathbb{Q}}(t) = f^M(0,t) + \frac{1}{\lambda}\frac{\partial f^M(0,t)}{\partial t} + \frac{\sigma^2}{2\lambda^2}(1 - e^{-2\lambda t})\), one can verify by direct integration (using integration by parts on the \(f^M\) terms) that:

\[ r(0)e^{-\lambda t} + \lambda\int_0^t \theta^{\mathbb{Q}}(t')e^{-\lambda(t-t')}\,dt' = f^M(0,t) + \frac{\sigma^2}{2\lambda^2}(1 - e^{-\lambda t})^2 = \alpha(t) \]

Alternatively, one can verify that \(\alpha(t)\) satisfies the ODE \(\alpha'(t) = \lambda(\theta^{\mathbb{Q}}(t) - \alpha(t))\) with \(\alpha(0) = r_0\), which is exactly the equation for the conditional mean (obtained by taking expectations in the SDE). Since the ODE has a unique solution, \(\alpha(t) = \mathbb{E}^{\mathbb{Q}}[r(t) | \mathcal{F}(0)]\).

Exercise 3. For \(\lambda = 0.05\), \(\sigma = 0.01\), \(r(0) = 0.03\), and a flat forward curve \(f^M(0,t) = 0.03\), compute \(\mathbb{E}[r(t) | \mathcal{F}(0)]\) and \(\text{Var}[r(t) | \mathcal{F}(0)]\) at \(t = 1, 10, 50\). Plot the mean and the 95% confidence band.

Solution to Exercise 3

With a flat forward curve \(f^M(0,t) = 0.03\), we have \(\alpha(t) = 0.03 + \frac{(0.01)^2}{2(0.05)^2}(1 - e^{-0.05t})^2 = 0.03 + 0.02(1 - e^{-0.05t})^2\).

The conditional mean and variance are:

\[ \mathbb{E}[r(t) | \mathcal{F}(0)] = \alpha(t) = 0.03 + 0.02(1 - e^{-0.05t})^2 \]
\[ \text{Var}[r(t) | \mathcal{F}(0)] = \frac{\sigma^2}{2\lambda}(1 - e^{-2\lambda t}) = \frac{(0.01)^2}{0.10}(1 - e^{-0.10t}) = 0.001(1 - e^{-0.10t}) \]

At \(t = 1\):

\[ \mu_1 = 0.03 + 0.02(1 - e^{-0.05})^2 = 0.03 + 0.02(0.04877)^2 \approx 0.03005 \]
\[ \sigma_1^2 = 0.001(1 - e^{-0.10}) = 0.001 \times 0.09516 \approx 9.516 \times 10^{-5}, \quad \sigma_1 \approx 0.00976 \]

At \(t = 10\):

\[ \mu_{10} = 0.03 + 0.02(1 - e^{-0.5})^2 = 0.03 + 0.02(0.3935)^2 \approx 0.03309 \]
\[ \sigma_{10}^2 = 0.001(1 - e^{-1.0}) = 0.001 \times 0.6321 \approx 6.321 \times 10^{-4}, \quad \sigma_{10} \approx 0.02514 \]

At \(t = 50\):

\[ \mu_{50} = 0.03 + 0.02(1 - e^{-2.5})^2 = 0.03 + 0.02(0.9179)^2 \approx 0.04685 \]
\[ \sigma_{50}^2 = 0.001(1 - e^{-5.0}) = 0.001 \times 0.99326 \approx 9.933 \times 10^{-4}, \quad \sigma_{50} \approx 0.03152 \]

The 95% confidence band is \(\mu_t \pm 1.96\sigma_t\). Note that the mean drifts upward due to the convexity correction, and the variance saturates at \(\sigma^2/(2\lambda) = 0.001\) as \(t \to \infty\).

Exercise 4. Show that the conditional variance \(\frac{\sigma^2}{2\lambda}(1 - e^{-2\lambda(t-s)})\) can also be written as \(-\frac{1}{2}\sigma^2 B(2(t-s))\) where \(B(\tau) = (e^{-\lambda\tau} - 1)/\lambda\). Interpret the relationship between the variance and the bond price function \(B\).

Solution to Exercise 4

Recall \(B(\tau) = \frac{e^{-\lambda\tau} - 1}{\lambda} = -\frac{1 - e^{-\lambda\tau}}{\lambda}\). Substituting \(\tau = 2(t-s)\):

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

This is exactly the conditional variance \(\text{Var}(r(t) | \mathcal{F}(s))\).

Interpretation: The function \(B(\tau)\) appears in the affine bond price formula \(P(t,T) = e^{A(\tau) + B(\tau)r(t)}\) where it measures the duration-like sensitivity of the bond price to \(r(t)\). The variance of \(r(t)\) can be expressed through \(B\) evaluated at twice the time horizon because:

  • The variance involves \(\int_s^t e^{-2\lambda(t-t')}\,dt'\), which has the same structure as \(B\) but with \(2\lambda\) replacing \(\lambda\)
  • The factor of \(-\frac{1}{2}\sigma^2\) is a scaling that connects the diffusion coefficient \(\sigma\) to the second moment
  • This relationship reflects the deep connection between the affine structure of the model (which produces exponential bond price functions) and the Gaussian distribution of the short rate

Exercise 5. The sample path code uses hw.generate_sample_paths to simulate 20 paths over 30 years. Describe the Euler-Maruyama discretization used for this simulation and explain why exact simulation (using the transition density) would be preferable.

Solution to Exercise 5

Euler-Maruyama discretization: Partition \([0, T]\) into \(N\) steps with \(\Delta t = T/N\). The scheme is:

\[ r(t_{k+1}) = r(t_k) + \lambda(\theta^{\mathbb{Q}}(t_k) - r(t_k))\Delta t + \sigma\sqrt{\Delta t}\,Z_k, \quad Z_k \sim \mathcal{N}(0,1) \]

This introduces discretization error of order \(O(\Delta t)\) in the weak sense and \(O(\sqrt{\Delta t})\) in the strong sense.

Exact simulation: Since the transition density is known in closed form:

\[ r(t_{k+1}) | r(t_k) \sim \mathcal{N}\!\left(r(t_k)e^{-\lambda\Delta t} + \alpha(t_{k+1}) - \alpha(t_k)e^{-\lambda\Delta t},\; \frac{\sigma^2}{2\lambda}(1 - e^{-2\lambda\Delta t})\right) \]

one can simulate exactly:

\[ r(t_{k+1}) = \mu(t_k, t_{k+1}) + \Sigma(t_k, t_{k+1})\,Z_k \]

Why exact simulation is preferable:

  • No discretization error: Euler-Maruyama accumulates error over each step; exact simulation gives the correct distribution regardless of step size
  • Larger time steps: With exact simulation, one can use \(\Delta t = 1\) year (or even jump directly to maturity) without loss of accuracy, whereas Euler-Maruyama requires small \(\Delta t\) for accuracy
  • Computational savings: Fewer steps means fewer random number generations and arithmetic operations
  • Exact variance: The Euler scheme slightly distorts the variance for finite \(\Delta t\), while exact simulation preserves it exactly

Exercise 6. Derive the autocovariance function \(\text{Cov}(r(t), r(t+h))\) for \(h > 0\) in the Hull-White model. Show that it decays exponentially with lag \(h\).

Solution to Exercise 6

From the explicit solution starting at \(s = 0\):

\[ r(t) = r(0)e^{-\lambda t} + \lambda\int_0^t \theta^{\mathbb{Q}}(t')e^{-\lambda(t-t')}\,dt' + \sigma\int_0^t e^{-\lambda(t-t')}\,dW^{\mathbb{Q}}(t') \]

The stochastic part is \(X(t) = \sigma\int_0^t e^{-\lambda(t-t')}\,dW^{\mathbb{Q}}(t')\). Since the deterministic parts do not contribute to the covariance:

\[ \text{Cov}(r(t), r(t+h)) = \text{Cov}(X(t), X(t+h)) \]

Expanding:

\[ X(t+h) = \sigma\int_0^{t+h} e^{-\lambda(t+h-t')}\,dW^{\mathbb{Q}}(t') = \sigma\int_0^t e^{-\lambda(t+h-t')}\,dW^{\mathbb{Q}}(t') + \sigma\int_t^{t+h} e^{-\lambda(t+h-t')}\,dW^{\mathbb{Q}}(t') \]

The second integral is independent of \(\mathcal{F}(t)\), hence independent of \(X(t)\). For the first integral, note \(e^{-\lambda(t+h-t')} = e^{-\lambda h}e^{-\lambda(t-t')}\). Therefore:

\[ \text{Cov}(X(t), X(t+h)) = e^{-\lambda h}\,\text{Cov}\!\left(X(t),\, \sigma\int_0^t e^{-\lambda(t-t')}\,dW^{\mathbb{Q}}(t')\right) = e^{-\lambda h}\,\text{Var}(X(t)) \]

Using the Ito isometry:

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

Therefore:

\[ \text{Cov}(r(t), r(t+h)) = \frac{\sigma^2}{2\lambda}(1 - e^{-2\lambda t})\,e^{-\lambda h} \]

The autocovariance decays exponentially with lag \(h\) at rate \(\lambda\). As \(t \to \infty\), the stationary autocovariance is \(\frac{\sigma^2}{2\lambda}e^{-\lambda h}\), which is the autocovariance of the stationary Ornstein-Uhlenbeck process.