Relationships Between Named Functions¶
The Hull-White model involves a collection of named functions -- \(B(\tau)\), \(A(t,T)\), \(\theta(t)\), \(\psi(t)\), \(\alpha(t)\), \(\sigma_r^2(t)\), and \(V(t,T)\) -- that appear in different contexts (bond pricing, option pricing, simulation, calibration). These functions are not independent: they are connected by a web of algebraic and differential relationships that reflect the internal consistency of the model. Understanding these relationships is essential for verifying implementations, simplifying derivations, and recognizing when different-looking formulas are in fact equivalent. This section catalogs the key relationships and proves the most important ones.
Prerequisites
- Named functions definition (sibling section)
- Derivation via Riccati equations (sibling section)
- Hull-White SDE, solution, and conditional distribution
- Affine bond price formula
Learning Objectives
By the end of this section, you will be able to:
- State all major named functions and their definitions
- Derive relationships between \(B\), \(A\), \(\theta\), \(\psi\), \(\alpha\), and \(V\)
- Express bond prices, yields, and forward rates using named functions
- Verify internal consistency of the Hull-White formulas
- Translate between different notational conventions in the literature
Catalog of Named Functions¶
For reference, the complete set of named functions with \(\tau = T - t\):
| Function | Definition | Role |
|---|---|---|
| \(B(\tau)\) | \(\dfrac{1-e^{-a\tau}}{a}\) | Duration-like function (note: positive convention) |
| \(\theta(t)\) | \(f'(0,t) + af(0,t) + \dfrac{\sigma^2}{2a}(1-e^{-2at})\) | Time-dependent drift |
| \(\alpha(t)\) | \(f(0,t) + \dfrac{\sigma^2}{2a^2}(1-e^{-at})^2\) | Deterministic mean of \(r_t\) |
| \(\psi(t)\) | \(r_0 e^{-at} + a\displaystyle\int_0^t \theta(u) e^{-a(t-u)}\, du\) | Expected short rate given \(r_0\) |
| \(\sigma_r^2(t)\) | \(\dfrac{\sigma^2}{2a}(1-e^{-2at})\) | Variance of \(r_t\) given \(r_0\) |
| \(V(t,T)\) | \(\dfrac{\sigma^2}{a^2}\!\left[\tau - 2B(\tau) + \dfrac{1-e^{-2a\tau}}{2a}\right]\) | Variance of \(\int_t^T r_s\, ds\) |
| \(A(t,T)\) | \(\ln\dfrac{P(0,T)}{P(0,t)} + B(\tau) f(0,t) + \dfrac{\sigma^2}{4a} B(\tau)^2(1-e^{-2at})\) | Bond price intercept |
Here \(B(\tau)\) uses the positive convention \(B(\tau) = (1-e^{-a\tau})/a > 0\), so the bond price is \(P(t,T) = e^{A(t,T) - B(\tau) r_t}\).
Fundamental Relationships¶
B and Its Derivatives¶
The function \(B(\tau) = (1-e^{-a\tau})/a\) satisfies the ODE
with \(B(0) = 0\). This can also be written as \(aB(\tau) = 1 - e^{-a\tau}\), or equivalently \(e^{-a\tau} = 1 - aB(\tau)\).
Proposition: Powers of \(B\)
Useful algebraic identities:
The last identity follows from \(1 - e^{-2a\tau} = (1-e^{-a\tau})(1+e^{-a\tau})\).
theta, alpha, and psi¶
The three drift-related functions are connected by differential relationships.
Theorem: \(\alpha\) Satisfies the Mean ODE
The deterministic mean \(\alpha(t)\) satisfies
Proof
Differentiate \(\alpha(t) = f(0,t) + \frac{\sigma^2}{2a^2}(1-e^{-at})^2\):
Now compute \(\theta(t) - a\alpha(t)\):
Expanding: \((1-e^{-2at}) - (1-e^{-at})^2 = 1 - e^{-2at} - 1 + 2e^{-at} - e^{-2at} = 2e^{-at}(1 - e^{-at})\).
Therefore \(\theta(t) - a\alpha(t) = f'(0,t) + \frac{\sigma^2}{a} e^{-at}(1-e^{-at}) = \alpha'(t)\). \(\square\)
Corollary: \(\psi(t) = \alpha(t)\)
The expected short rate \(\psi(t) = \mathbb{E}[r_t \mid r_0]\) equals \(\alpha(t)\):
Proof
Both \(\psi(t)\) and \(\alpha(t)\) satisfy the same ODE \(y' = \theta(t) - ay\) with \(y(0) = r_0\) (since \(\alpha(0) = f(0,0) + 0 = r_0\)). By uniqueness of ODE solutions, \(\psi(t) = \alpha(t)\). \(\square\)
Variance Functions¶
Theorem: Relationship Between \(\sigma_r^2\) and \(B\)
The short rate variance can be expressed as
More generally, the conditional variance starting from time \(t_0\) is
The variance of the integrated rate \(V(t,T)\) also relates to \(B\):
Proposition: \(V\) in Terms of \(B\)
where \(\tau = T - t\).
Proof
From the definition:
using \(B(\tau) = (1-e^{-a\tau})/a\) and \(B(2\tau) = (1-e^{-2a\tau})/a\). \(\square\)
Bond Price in Named Function Notation¶
Using the named functions with the positive-\(B\) convention:
Theorem: Bond Price Formula
Equivalently, using \(\alpha(t)\):
The second form is useful because \(\alpha(t) - r_t = -\tilde{r}_t\) is the zero-mean stochastic part of the short rate from the decomposition \(r_t = \alpha(t) + \tilde{r}_t\).
Yield and Forward Rate Formulas¶
Proposition: Yield in Terms of Named Functions
The continuously compounded yield for maturity \(\tau = T - t\) is
Since \(B(\tau)/\tau \to 1\) as \(\tau \to 0\) and \(B(\tau)/\tau \to 0\) as \(\tau \to \infty\), the yield is sensitive to \(r_t\) at the short end and nearly independent of \(r_t\) at the long end.
Proposition: Forward Rate in Terms of Named Functions
The instantaneous forward rate is
The loading on \(r_t\) is \(e^{-a\tau} = B'(\tau) = 1 - aB(\tau)\).
Consistency Relations¶
The named functions satisfy several cross-consistency checks that are useful for verifying implementations.
Proposition: Consistency Checks
The following identities hold:
1. Bond price at \(t = 0\):
2. Variance decomposition:
(The cross-covariance does not vanish because \(r_s\) for \(s \leq t\) affects \(r_u\) for \(u > t\) through mean reversion.)
3. Short-rate variance limit:
The \(\tau^3\) scaling for small \(\tau\) matches the Ho-Lee limit.
4. Duration identity:
so \(B(\tau)\) is the "dollar duration per unit notional" divided by the bond price.
Translation Between Conventions¶
Different references use different sign conventions. The two main variants:
| Convention | Bond price | \(B\) function | Used by |
|---|---|---|---|
| Positive \(B\) | \(P = e^{A - Br}\) | \(B = (1-e^{-a\tau})/a > 0\) | Brigo-Mercurio, this text |
| Negative \(B\) | \(P = e^{A + Br}\) | \(B = -(1-e^{-a\tau})/a < 0\) | Some academic papers |
The relationship is simply \(B_{\text{neg}} = -B_{\text{pos}}\) and \(A\) is the same in both conventions. The positive convention is more natural because \(B(\tau) > 0\) has the interpretation of a modified duration.
Summary¶
The Hull-White named functions form a tightly connected system. The core relationships are: \(B(\tau) = (1-e^{-a\tau})/a\) with \(B'(\tau) = 1 - aB(\tau)\); \(\alpha(t) = \psi(t)\) both satisfying \(y' = \theta(t) - ay\); \(\sigma_r^2(t) = \frac{\sigma^2}{2} B(2t)\); \(V(t,T) = \frac{\sigma^2}{a^2}[\tau - 2B(\tau) + B(2\tau)/2]\); and the bond price \(P(t,T) = \frac{P(0,T)}{P(0,t)} \exp(B(\tau)[f(0,t) - r_t] + \frac{\sigma^2}{4a} B(\tau)^2(1-e^{-2at}))\). These identities enable cross-checking of implementations and simplification of complex pricing formulas, and the translation table between sign conventions resolves the most common source of confusion in the literature.
Exercises¶
Exercise 1. Starting from \(\alpha(t) = f(0,t) + \frac{\sigma^2}{2a^2}(1-e^{-at})^2\), verify that \(\alpha'(t) = \theta(t) - a\alpha(t)\) by computing \(\alpha'(t)\) directly and substituting the formula for \(\theta(t)\). Show all algebraic steps.
Solution to Exercise 1
Compute \(\alpha'(t)\) directly from \(\alpha(t) = f(0,t) + \frac{\sigma^2}{2a^2}(1-e^{-at})^2\):
Now compute \(\theta(t) - a\alpha(t)\) using \(\theta(t) = f'(0,t) + af(0,t) + \frac{\sigma^2}{2a}(1-e^{-2at})\):
Expand the bracket:
Therefore:
This confirms \(\alpha'(t) = \theta(t) - a\alpha(t)\).
Exercise 2. Prove the identity \(\sigma_r^2(t) = \frac{\sigma^2}{2}B(2t)\) by substituting the definition of \(B\). Then use the doubling formula \(B(2\tau) = B(\tau)(2 - aB(\tau))\) to express \(\sigma_r^2(t)\) in terms of \(B(t)\) alone.
Solution to Exercise 2
First part: Substitute \(B(2t) = (1-e^{-2at})/a\) into \(\sigma_r^2(t) = \frac{\sigma^2}{2}B(2t)\):
This matches the definition.
Second part: Use the doubling formula \(B(2\tau) = B(\tau)(2 - aB(\tau))\). Setting \(\tau = t\):
Therefore:
This expresses the short rate variance purely in terms of \(B(t)\), without any exponentials.
Exercise 3. The duration identity states \(\frac{\partial P(t,T)}{\partial r_t} = -B(\tau)P(t,T)\). Derive this from \(P(t,T) = e^{A(t,T) - B(\tau)r_t}\). Explain why \(B(\tau)\) is called a "duration-like" function and how it differs from Macaulay duration.
Solution to Exercise 3
From \(P(t,T) = e^{A(t,T) - B(\tau)r_t}\) (positive-\(B\) convention), differentiate with respect to \(r_t\):
This is the duration identity.
Why \(B(\tau)\) is called "duration-like": The modified duration \(D\) of a bond is defined by \(\frac{\partial P}{\partial y} = -D\cdot P\), where \(y\) is the yield. In the Hull-White model, \(B(\tau)\) plays the analogous role with \(r_t\) replacing \(y\). For small \(\tau\), \(B(\tau) \approx \tau\), which matches the Macaulay duration of a zero-coupon bond.
How it differs from Macaulay duration: Macaulay duration is always equal to \(\tau\) for a zero-coupon bond, regardless of the interest rate model. In contrast, \(B(\tau) = (1-e^{-a\tau})/a < \tau\) for \(a > 0\), reflecting the mean-reverting nature of the Hull-White model. Mean reversion dampens the long-run impact of rate changes, so the effective duration saturates at \(1/a\) rather than growing linearly with maturity. Only in the \(a \to 0\) limit does \(B(\tau) \to \tau\), recovering Macaulay duration.
Exercise 4. Verify the consistency check \(A(0,T) - B(T)r_0 = \ln P(0,T)\) numerically for \(a = 0.05\), \(\sigma = 0.01\), \(r_0 = 0.03\), and a flat market curve \(P^M(0,T) = e^{-0.03T}\) at \(T = 1, 5, 10\).
Solution to Exercise 4
With \(a = 0.05\), \(\sigma = 0.01\), \(r_0 = 0.03\), and a flat market curve \(P^M(0,T) = e^{-0.03T}\), we have \(f(0,t) = 0.03\) for all \(t\).
The consistency check requires \(A(0,T) - B(T)r_0 = \ln P(0,T) = -0.03T\).
Compute \(A(0,T)\) from the formula:
Since \(P^M(0,0) = 1\), \(f(0,0) = 0.03\), and \(1 - e^0 = 0\):
Then:
This holds for all \(T\), not just specific values. Let us verify numerically at each \(T\):
\(T = 1\): \(B(1) = (1-e^{-0.05})/0.05 = 0.97531\). \(A(0,1) = -0.03 + 0.03(0.97531) = -0.03 + 0.02926 = -0.00074\). Check: \(A(0,1) - B(1)\cdot 0.03 = -0.00074 - 0.02926 = -0.03 = \ln P^M(0,1)\). \(\checkmark\)
\(T = 5\): \(B(5) = (1-e^{-0.25})/0.05 = 4.42400\). \(A(0,5) = -0.15 + 0.03(4.42400) = -0.15 + 0.13272 = -0.01728\). Check: \(-0.01728 - 4.42400(0.03) = -0.01728 - 0.13272 = -0.15\). \(\checkmark\)
\(T = 10\): \(B(10) = (1-e^{-0.5})/0.05 = 7.86939\). \(A(0,10) = -0.30 + 0.03(7.86939) = -0.30 + 0.23608 = -0.06392\). Check: \(-0.06392 - 7.86939(0.03) = -0.06392 - 0.23608 = -0.30\). \(\checkmark\)
Exercise 5. The variance decomposition identity involves a cross-covariance term between \(\int_0^t r_s\,ds\) and \(\int_t^T r_s\,ds\). Explain why this cross-covariance is nonzero in the Hull-White model. Compute it explicitly using the covariance function \(\text{Cov}(r_s, r_u) = \frac{\sigma^2}{2a}e^{-a|u-s|}(1-e^{-2a\min(s,u)})\).
Solution to Exercise 5
In the Hull-White model, \(r_t\) is an Ornstein-Uhlenbeck process: \(r_u = r_t e^{-a(u-t)} + \text{deterministic} + \sigma\int_t^u e^{-a(u-s)}dW_s\) for \(u > t\). Therefore \(r_u\) depends on \(r_s\) for \(s \leq t\) through the dependence on \(r_t\), and \(r_t\) in turn depends on \(\{r_s\}_{s \leq t}\) through the SDE dynamics.
The cross-covariance is nonzero because, for \(s \leq t < u\):
Since \(r_t\) carries information about \(r_s\) through the mean-reversion dynamics, \(\text{Cov}(r_s, r_t) > 0\), and this nonzero covariance propagates to future times. Specifically:
for \(s \leq u\).
The cross-covariance between the integrals is:
Evaluating the inner integral:
Then the outer integral becomes:
This integral is nonzero and can be evaluated by expanding the product and integrating each exponential term, confirming that the variance decomposition includes a genuine cross-covariance contribution from the mean-reverting dynamics.
Exercise 6. A reference uses the negative-\(B\) convention \(P = e^{A + Br}\) with \(B = -(1-e^{-a\tau})/a < 0\). Translate the bond price formula, the duration identity, and the yield formula from this section into the negative-\(B\) convention. Verify that all pricing results are numerically identical.
Solution to Exercise 6
In the negative-\(B\) convention, \(B_{\text{neg}} = -B_{\text{pos}} = -(1-e^{-a\tau})/a < 0\) and \(P = e^{A + B_{\text{neg}}\, r}\).
Bond price formula: Starting from \(P(t,T) = \frac{P(0,T)}{P(0,t)}\exp\left(B_{\text{pos}}(\tau)[f(0,t)-r_t] + \frac{\sigma^2}{4a}B_{\text{pos}}(\tau)^2(1-e^{-2at})\right)\) and noting \(B_{\text{pos}} = -B_{\text{neg}}\):
(Note \(B_{\text{neg}}^2 = B_{\text{pos}}^2\), so the convexity term is unchanged.)
Duration identity: From \(P = e^{A + B_{\text{neg}} r}\):
Since \(B_{\text{neg}} < 0\), this correctly gives \(\partial P/\partial r < 0\) (bond prices fall when rates rise).
Yield formula: \(y(t,T) = -\frac{\ln P}{\tau} = -\frac{A + B_{\text{neg}} r}{\tau} = -\frac{A}{\tau} - \frac{B_{\text{neg}}}{\tau}r = -\frac{A}{\tau} + \frac{B_{\text{pos}}}{\tau}r\), which is identical to the positive convention result.
All pricing results are numerically identical because \(e^{A - B_{\text{pos}}r} = e^{A + B_{\text{neg}}r}\) by definition of \(B_{\text{neg}} = -B_{\text{pos}}\).
Exercise 7. Using the relationship \(B'(\tau) = 1 - aB(\tau)\), show that \(\int_0^\tau B(s)\,ds = \frac{\tau - B(\tau)}{a}\). Then use this result to simplify the \(A\)-function quadrature \(\int_t^T \theta(u)B(u,T)\,du\) when \(\theta\) is constant.
Solution to Exercise 7
First part: proving \(\int_0^\tau B(s)\,ds = \frac{\tau - B(\tau)}{a}\).
From \(B'(\tau) = 1 - aB(\tau)\), we can write \(aB(\tau) = 1 - B'(\tau)\), so:
since \(B(0) = 0\). Dividing by \(a\):
Second part: simplifying the quadrature when \(\theta\) is constant.
When \(\theta\) is constant (Vasicek model), the \(A\)-function quadrature involves:
Substituting \(s = T - u\) (so \(du = -ds\)), with limits \(s=0\) to \(s=T-t=\tau\):
Therefore the complete \(A\)-function in the Vasicek case becomes:
Using the closed-form integral of \(B^2\):
This clean expression avoids any numerical integration and is the standard Vasicek bond price formula.