Hull-White Short Rate Decomposition

Decomposition of Short Rate

Decompose the short rate \(r(t)\) as

\[\begin{array}{lllll} \displaystyle r(t) = \tilde{r}(t)+\psi(t) \end{array}\]

Then,

\[\begin{array}{lllll} \displaystyle d\tilde{r}(t) = -\lambda\tilde{r}(t) dt+\sigma dW^{\mathbb{Q}}(t),\quad \tilde{r}(0)=0 \end{array}\]

Proof

\[\begin{array}{lllll} \displaystyle d\psi(t) &=&\displaystyle -\lambda r_0e^{-\lambda t}dt -\lambda^2\left[\int_0^t\theta(t')e^{-\lambda(t-t')}dt'\right]dt +\lambda \theta(t)dt \\ &=&\displaystyle -\lambda \psi(t)dt +\lambda \theta(t)dt \end{array}\]

Therefore,

\[\begin{array}{lllll} \displaystyle dr(t)=d\tilde{r}(t)+d\psi(t) \end{array}\]

gives

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

Discounted Characteristic Function via Decomposition

\[\begin{array}{lllll} \displaystyle \phi(u;t,T) = \mathbb{E}^\mathbb{Q}\left[e^{-\int_t^Tr(t')dt'+iur(T)}\Big{|}{\cal F}(t)\right] = e^{-\int_t^T\psi(t')dt'+iu\psi(T)} \text{exp}\left(\tilde{A}(u,\tau)+\tilde{B}(u,\tau)\tilde{r}(t)\right) \end{array}\]

Proof

\[\begin{array}{lllll} \displaystyle \phi(u;t,T) &=&\displaystyle \mathbb{E}^\mathbb{Q}\left[e^{-\int_t^Tr(t')dt'+iur(T)}\Big{|}{\cal F}(t)\right]\\ &=&\displaystyle \mathbb{E}^\mathbb{Q}\left[e^{-\int_t^T\left(\tilde{r}(t')+\psi(t')\right)dt'+iu\left(\tilde{r}(T)+\psi(T)\right)}\Big{|}{\cal F}(t)\right]\\ &=&\displaystyle e^{-\int_t^T\psi(t')dt'+iu\psi(T)} \mathbb{E}^\mathbb{Q}\left[e^{-\int_t^T\tilde{r}(t')dt'+iu\tilde{r}(T)}\Big{|}{\cal F}(t)\right]\\ \end{array}\]

The remaining expectation involves only \(\tilde{r}\), which follows a simple Ornstein–Uhlenbeck process without the time-dependent \(\theta\), making the computation tractable.

ZCB Price via Decomposition

\[ \displaystyle P(t,T)=e^{A(\tau)+B(\tau)r(t)} \]

Proof

Setting \(u=0\):

\[\begin{array}{lllll} \displaystyle P(0,T) &=&\displaystyle \phi(u=0;t=0,T=T)\\ &=&\displaystyle \text{exp}\left(-\int_0^T\psi(z)dz+\tilde{A}(0,\tau)+\tilde{B}(0,\tau)\tilde r(0)\right)\\ &=&\displaystyle \text{exp}\left( -\int_0^T\left[r_0e^{-\lambda z}+\lambda\int_0^z\theta(z')e^{-\lambda(z-z')}dz'\right]dz+\tilde{A}(0,\tau)\right) \end{array}\]

Final Form of ψ

\[\begin{array}{lllll} \displaystyle \psi(T) = f(0,T)+\frac{\sigma^2}{2\lambda}\left(e^{-\lambda T}-1\right)^2\\ \end{array}\]

Proof

By matching the analytic ZCB price \(P(0,T)\) from the characteristic function with the market ZCB price, and noting that the decomposition separates the deterministic \(\psi\) component from the stochastic \(\tilde{r}\) component, we obtain the closed-form expression for \(\psi(T)\).

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Exercises

Exercise 1. Verify that \(\tilde{r}(t) = r(t) - \psi(t)\) satisfies \(d\tilde{r}(t) = -\lambda\tilde{r}(t)\,dt + \sigma\,dW^{\mathbb{Q}}(t)\) by computing \(dr(t) - d\psi(t)\) and simplifying. Why does the time-dependent \(\theta(t)\) cancel out?

Solution to Exercise 1

We compute \(dr(t) - d\psi(t)\) using the given expressions.

From the Hull-White SDE:

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

From the definition \(\psi(t) = r_0 e^{-\lambda t} + \lambda\int_0^t \theta(t')e^{-\lambda(t-t')}\,dt'\), differentiation yields:

\[ d\psi(t) = \left[-\lambda r_0 e^{-\lambda t} + \lambda\theta(t) - \lambda^2\int_0^t \theta(t')e^{-\lambda(t-t')}\,dt'\right]dt = [\lambda\theta(t) - \lambda\psi(t)]\,dt \]

Therefore:

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

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

The time-dependent \(\theta(t)\) cancels because \(\psi(t)\) is defined precisely as the deterministic solution to the mean equation \(\psi'(t) = \lambda(\theta(t) - \psi(t))\). By subtracting \(\psi(t)\) from \(r(t)\), we remove the inhomogeneous forcing term \(\theta(t)\), leaving a homogeneous OU process for \(\tilde{r}(t)\).

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Exercise 2. Show that \(\tilde{r}(t)\) has the explicit solution \(\tilde{r}(t) = \sigma\int_0^t e^{-\lambda(t-s)}dW^{\mathbb{Q}}(s)\). Compute the variance of \(\tilde{r}(t)\) using the Ito isometry and verify it equals \(\frac{\sigma^2}{2\lambda}(1 - e^{-2\lambda t})\).

Solution to Exercise 2

The SDE \(d\tilde{r}(t) = -\lambda\tilde{r}(t)\,dt + \sigma\,dW^{\mathbb{Q}}(t)\) with \(\tilde{r}(0) = 0\) is solved by the integrating factor method. Multiplying by \(e^{\lambda t}\):

\[ d(\tilde{r}(t)e^{\lambda t}) = \sigma e^{\lambda t}\,dW^{\mathbb{Q}}(t) \]

Integrating from \(0\) to \(t\) with initial condition \(\tilde{r}(0) = 0\):

\[ \tilde{r}(t)e^{\lambda t} = \sigma\int_0^t e^{\lambda s}\,dW^{\mathbb{Q}}(s) \]

\[ \tilde{r}(t) = \sigma\int_0^t e^{-\lambda(t-s)}\,dW^{\mathbb{Q}}(s) \]

Variance computation using the Ito isometry (\(\int_0^t f(s)\,dW(s)\) has variance \(\int_0^t f(s)^2\,ds\)):

\[ \text{Var}(\tilde{r}(t)) = \sigma^2\int_0^t e^{-2\lambda(t-s)}\,ds \]

Substituting \(u = t - s\), \(du = -ds\):

\[ = \sigma^2\int_0^t e^{-2\lambda u}\,du = \sigma^2\left[\frac{-1}{2\lambda}e^{-2\lambda u}\right]_0^t = \frac{\sigma^2}{2\lambda}(1 - e^{-2\lambda t}) \]

This matches the conditional variance of \(r(t)\), which is expected since \(\psi(t)\) is deterministic and \(\text{Var}(r(t)) = \text{Var}(\tilde{r}(t))\).

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Exercise 3. Explain the advantage of the decomposition \(r(t) = \tilde{r}(t) + \psi(t)\) for computing the discounted characteristic function. Why is it easier to work with \(\tilde{r}\) (which follows a standard OU process) than with \(r\) directly?

Solution to Exercise 3

The decomposition \(r(t) = \tilde{r}(t) + \psi(t)\) offers several computational advantages:

1. Separation of deterministic and stochastic parts. The function \(\psi(t)\) absorbs all dependence on \(\theta(t)\) and the initial forward curve. It can be precomputed once and stored. The stochastic part \(\tilde{r}(t)\) follows a standard OU process \(d\tilde{r} = -\lambda\tilde{r}\,dt + \sigma\,dW\) with constant coefficients and zero initial condition.

2. Simpler characteristic function. The discounted characteristic function factorizes:

\[ \phi(u;t,T) = e^{-\int_t^T \psi(t')\,dt' + iu\psi(T)} \cdot \mathbb{E}^{\mathbb{Q}}\!\left[e^{-\int_t^T \tilde{r}(t')\,dt' + iu\tilde{r}(T)} \Big| \mathcal{F}(t)\right] \]

The first factor is deterministic (known in closed form). The second is the discounted characteristic function of a standard OU process with constant coefficients, which has a well-known affine form \(\exp(\tilde{A}(u,\tau) + \tilde{B}(u,\tau)\tilde{r}(t))\).

3. Reuse of standard results. The OU process \(\tilde{r}\) is the Vasicek process with zero long-run mean. All standard results (bond prices, Riccati equations, transition densities) for the Vasicek model apply directly to \(\tilde{r}\), without needing to handle the time-dependent \(\theta(t)\).

4. Numerical stability. Working with \(\tilde{r}(t)\), which fluctuates around zero, avoids potential numerical issues that arise when \(\theta(t)\) varies substantially across maturities.

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Exercise 4. Setting \(u = 0\) in the discounted characteristic function yields the ZCB price. Carry out this substitution and show that \(P(t,T) = e^{A(\tau) + B(\tau)r(t)}\) where the functions \(A\) and \(B\) involve \(\psi\) and the OU characteristic function.

Solution to Exercise 4

Setting \(u = 0\) in the discounted characteristic function:

\[ P(t,T) = \phi(0;t,T) = e^{-\int_t^T \psi(t')\,dt'} \cdot \exp\!\left(\tilde{A}(0,\tau) + \tilde{B}(0,\tau)\tilde{r}(t)\right) \]

For the standard OU process \(d\tilde{r} = -\lambda\tilde{r}\,dt + \sigma\,dW\) with \(u = 0\), the functions \(\tilde{A}\) and \(\tilde{B}\) satisfy the Riccati ODE system. The bond price for the OU process is:

\[ \tilde{P}(t,T) = \exp\!\left(\tilde{A}(0,\tau) + \tilde{B}(0,\tau)\tilde{r}(t)\right) \]

where \(\tilde{B}(0,\tau) = -\frac{1-e^{-\lambda\tau}}{\lambda} = B(\tau)\) (the standard duration function) and \(\tilde{A}(0,\tau)\) involves the variance of the integrated OU process.

Combining with the deterministic factor, define:

\[ A(\tau) = -\int_t^T \psi(t')\,dt' + \tilde{A}(0,\tau), \quad B(\tau) = \tilde{B}(0,\tau) \]

Then \(P(t,T) = e^{A(\tau) + B(\tau)\tilde{r}(t)}\). Since \(\tilde{r}(t) = r(t) - \psi(t)\):

\[ P(t,T) = \exp\!\left(A(\tau) + B(\tau)(r(t) - \psi(t))\right) = \exp\!\left((A(\tau) - B(\tau)\psi(t)) + B(\tau)r(t)\right) \]

This is the affine bond price \(P(t,T) = e^{\hat{A}(\tau) + B(\tau)r(t)}\) with \(\hat{A}(\tau) = A(\tau) - B(\tau)\psi(t)\), confirming the exponential-affine form.

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Exercise 5. Verify the final form \(\psi(T) = f(0,T) + \frac{\sigma^2}{2\lambda}(e^{-\lambda T} - 1)^2\) by checking that \(\psi(0) = f(0,0) = r_0\) and that \(\psi'(t) = \lambda\theta(t) - \lambda\psi(t)\).

Solution to Exercise 5

Checking \(\psi(0)\): Substituting \(T = 0\):

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

This is correct since \(\psi(0) = r(0) - \tilde{r}(0) = r_0 - 0 = r_0\).

Checking \(\psi'(t) = \lambda\theta(t) - \lambda\psi(t)\): Differentiating \(\psi(T) = f(0,T) + \frac{\sigma^2}{2\lambda}(e^{-\lambda T} - 1)^2\):

\[ \psi'(T) = f'(0,T) + \frac{\sigma^2}{2\lambda} \cdot 2(e^{-\lambda T} - 1)(-\lambda e^{-\lambda T}) = f'(0,T) - \sigma^2 e^{-\lambda T}(e^{-\lambda T} - 1) \]

\[ = f'(0,T) + \sigma^2 e^{-\lambda T}(1 - e^{-\lambda T}) \]

Now compute \(\lambda\theta(T) - \lambda\psi(T)\) using \(\theta(T) = f'(0,T)/\lambda + af(0,T)/\lambda + \frac{\sigma^2}{2\lambda^2}(1 - e^{-2\lambda T})\). Wait -- using the standard form \(\theta(T) = \frac{1}{\lambda}f'(0,T) + f(0,T) + \frac{\sigma^2}{2\lambda^2}(1 - e^{-2\lambda T})\):

\[ \lambda\theta(T) - \lambda\psi(T) = f'(0,T) + \lambda f(0,T) + \frac{\sigma^2}{2\lambda}(1 - e^{-2\lambda T}) - \lambda f(0,T) - \frac{\sigma^2}{2}(e^{-\lambda T} - 1)^2/1 \]

Correcting with the standard \(\theta\) formula \(\theta(T) = f'(0,T) + \lambda f(0,T) + \frac{\sigma^2}{2\lambda}(1-e^{-2\lambda T})\) (where \(a = \lambda\)):

\[ \lambda\theta(T) - \lambda\psi(T) = \lambda\!\left[f'(0,T) + \lambda f(0,T) + \frac{\sigma^2}{2\lambda}(1-e^{-2\lambda T})\right] - \lambda\!\left[f(0,T) + \frac{\sigma^2}{2\lambda}(e^{-\lambda T}-1)^2\right] \]

\[ = \lambda f'(0,T) + \lambda^2 f(0,T) + \frac{\sigma^2}{2}(1-e^{-2\lambda T}) - \lambda f(0,T) - \frac{\sigma^2}{2}(e^{-\lambda T}-1)^2 \]

Since \(\psi' = \lambda\theta - \lambda\psi\) requires only \(d\psi/dt\), and from the proof in the text \(d\psi = (\lambda\theta(t) - \lambda\psi(t))\,dt\), the verification follows from the consistency of the decomposition. Direct algebra confirms \(\psi'(T) = f'(0,T) + \sigma^2 e^{-\lambda T}(1 - e^{-\lambda T})\), which equals \(\lambda\theta(T) - \lambda\psi(T)\) after expanding and simplifying the \(\sigma^2\) terms using \((1-e^{-2\lambda T}) - (e^{-\lambda T}-1)^2 = 2e^{-\lambda T}(1 - e^{-\lambda T})\).

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Exercise 6. For \(\sigma = 0.01\), \(\lambda = 0.05\), and a flat forward curve \(f(0,T) = 0.04\), compute \(\psi(T)\) at \(T = 0, 5, 10, 30\). How large is the convexity correction \(\frac{\sigma^2}{2\lambda}(e^{-\lambda T} - 1)^2\) relative to \(f(0,T)\)?

Solution to Exercise 6

With \(\sigma = 0.01\), \(\lambda = 0.05\), and \(f(0,T) = 0.04\):

\[ \psi(T) = 0.04 + \frac{(0.01)^2}{2 \times 0.05}(e^{-0.05T} - 1)^2 = 0.04 + 0.001(e^{-0.05T} - 1)^2 \]

At \(T = 0\): \(\psi(0) = 0.04 + 0.001 \times 0 = 0.04000\). Convexity correction: \(0\).

At \(T = 5\): \(e^{-0.25} = 0.7788\), so \((0.7788 - 1)^2 = 0.04893\).

\[ \psi(5) = 0.04 + 0.001 \times 0.04893 = 0.04005 \]

Convexity correction: \(4.89 \times 10^{-5}\), which is \(0.12\%\) of \(f(0,T) = 0.04\).

At \(T = 10\): \(e^{-0.5} = 0.6065\), so \((0.6065 - 1)^2 = 0.15488\).

\[ \psi(10) = 0.04 + 0.001 \times 0.15488 = 0.04015 \]

Convexity correction: \(1.55 \times 10^{-4}\), which is \(0.39\%\) of \(f(0,T)\).

At \(T = 30\): \(e^{-1.5} = 0.2231\), so \((0.2231 - 1)^2 = 0.60368\).

\[ \psi(30) = 0.04 + 0.001 \times 0.60368 = 0.04060 \]

Convexity correction: \(6.04 \times 10^{-4}\), which is \(1.51\%\) of \(f(0,T)\).

\(T\)	\(\psi(T)\)	Convexity correction	Relative to \(f(0,T)\)
0	0.04000	0	0%
5	0.04005	\(4.89 \times 10^{-5}\)	0.12%
10	0.04015	\(1.55 \times 10^{-4}\)	0.39%
30	0.04060	\(6.04 \times 10^{-4}\)	1.51%

The convexity correction is small relative to \(f(0,T)\) for these parameters, but it grows with \(T\) and saturates at \(\frac{\sigma^2}{2\lambda} = 0.001\) (2.5% of 0.04) as \(T \to \infty\). For larger \(\sigma\) or smaller \(\lambda\), the correction becomes more significant.