Limiting Cases and Asymptotics¶
The named functions \(B(\tau)\), \(A(t,T)\), \(V(t,T)\), \(\sigma_r^2(t)\), and \(\theta(t)\) of the Hull-White model depend on the mean reversion speed \(a\) and the volatility \(\sigma\). Their behavior in extreme parameter regimes -- \(a \to 0\), \(a \to \infty\), \(\tau \to 0\), \(\tau \to \infty\) -- reveals the model's connections to simpler models (Ho-Lee, deterministic rates) and provides approximations useful for intuition, debugging, and numerical validation.
Prerequisites
- Named functions definition and relationships (sibling sections)
- Ho-Lee model: \(dr = \theta(t)\,dt + \sigma\,dW\)
- Taylor expansion and L'Hopital's rule
Learning Objectives
By the end of this section, you will be able to:
- Derive the \(a \to 0\) (Ho-Lee) limit of all named functions
- Derive the \(a \to \infty\) (fast mean reversion) limit
- Compute short-maturity and long-maturity asymptotics for \(B\), \(A\), and \(V\)
- Use limiting formulas for numerical validation of Hull-White implementations
The Duration Function B(tau)¶
Recall \(B(\tau) = (1 - e^{-a\tau})/a\) (positive convention). The limits are:
Proposition: Limits of \(B(\tau)\)
Proof
\(a \to 0\): Using L'Hopital's rule or the Taylor expansion \(e^{-a\tau} = 1 - a\tau + \frac{1}{2}a^2\tau^2 - \cdots\):
\(a \to \infty\): \(e^{-a\tau} \to 0\) for \(\tau > 0\), so \(B(\tau) \to 1/a \to 0\).
\(\tau \to 0\): \(e^{-a\tau} \to 1\), so \(B(\tau) \to 0\).
\(\tau \to \infty\): \(e^{-a\tau} \to 0\), so \(B(\tau) \to 1/a\). \(\square\)
The Taylor expansion of \(B(\tau)\) in powers of \(a\tau\) is particularly useful:
For \(a\tau < 0.1\) (e.g., \(a = 0.05\) and \(\tau < 2\) years), truncating at the first correction gives \(B(\tau) \approx \tau(1 - a\tau/2)\) with relative error below \(0.02\%\).
The Ho-Lee Limit (a to 0)¶
The Ho-Lee model \(dr = \theta(t)\,dt + \sigma\,dW\) is the limiting case of Hull-White when mean reversion vanishes. All named functions reduce to simpler expressions.
Theorem: Ho-Lee Limits
| Hull-White Function | \(a \to 0\) Limit (Ho-Lee) |
|---|---|
| \(B(\tau) = \dfrac{1 - e^{-a\tau}}{a}\) | \(\tau\) |
| \(\sigma_r^2(t) = \dfrac{\sigma^2}{2a}(1 - e^{-2at})\) | \(\sigma^2 t\) |
| \(V(t,T) = \dfrac{\sigma^2}{a^2}\!\left[\tau - 2B(\tau) + \dfrac{B(2\tau)}{2}\right]\) | \(\dfrac{\sigma^2 \tau^3}{3}\) |
| \(\theta(t) = f'(0,t) + af(0,t) + \dfrac{\sigma^2}{2a}(1-e^{-2at})\) | \(f'(0,t) + \sigma^2 t\) |
| Bond price \(P(t,T)\) | \(\dfrac{P^M(0,T)}{P^M(0,t)}\exp\!\left(\tau[f^M(0,t) - r(t)] + \dfrac{\sigma^2}{6}\tau^3\right)\) |
Proof of \(V(t,T) \to \sigma^2\tau^3/3\)
Starting from \(V = \frac{\sigma^2}{a^2}[\tau - 2B(\tau) + B(2\tau)/2]\) and substituting the Taylor expansions:
Therefore \(V \approx \frac{\sigma^2}{a^2} \cdot \frac{a^2\tau^3}{3} = \frac{\sigma^2\tau^3}{3}\). \(\square\)
Proof of \(\sigma_r^2(t) \to \sigma^2 t\)
Expanding \(1 - e^{-2at} = 2at - 2a^2t^2 + \cdots\):
This is the variance of \(r(t) = r(0) + \int_0^t \theta(s)\,ds + \sigma W(t)\) in the Ho-Lee model, where \(r(t)\) is a Brownian motion with drift. \(\square\)
Interpretation
In the Ho-Lee limit, the short rate variance grows linearly in time (no stabilization), the integrated rate variance grows as \(\tau^3\) (faster than \(\tau\)), and all yield curve movements are exact parallel shifts (\(B(\tau)/\tau = 1\) for all \(\tau\)). The absence of mean reversion makes the model unrealistic for long horizons but useful as a local approximation for short-dated products.
Fast Mean Reversion Limit (a to infinity)¶
When \(a \to \infty\), the short rate reverts instantaneously to \(\theta(t)/a\), effectively becoming deterministic. The named functions collapse:
Proposition: Fast Mean Reversion Limits
The bond price converges to the market discount factor ratio:
Proof
As \(a \to \infty\): \(e^{-a\tau} \to 0\), so \(B(\tau) = (1 - e^{-a\tau})/a \to 1/a \to 0\).
The variance \(\sigma_r^2(t) = \frac{\sigma^2}{2a}(1 - e^{-2at}) \le \frac{\sigma^2}{2a} \to 0\).
For \(V(t,T)\): all three terms in \(\tau - 2B(\tau) + B(2\tau)/2\) individually approach \(\tau - 0 + 0 = \tau\), but \(V = \frac{\sigma^2}{a^2}[\cdots]\) has the \(1/a^2\) prefactor, so \(V \le \frac{\sigma^2\tau}{a^2} \to 0\).
In the bond price formula, \(B(t,T) \to 0\) eliminates the \(r(t)\)-dependent and convexity terms, leaving \(P(t,T) \to P^M(0,T)/P^M(0,t)\). \(\square\)
Interpretation
Fast mean reversion suppresses all stochastic variation. The short rate tracks \(\theta(t)/a\) so closely that the yield curve becomes deterministic. This limit is useful as a sanity check: if an implementation is given \(a = 10^6\), all option prices should vanish and bond prices should match the deterministic discounting baseline.
Short-Maturity Asymptotics (tau to 0)¶
As \(\tau \to 0\), the named functions admit Taylor expansions that reveal the local behavior of bond prices near maturity.
Proposition: Short-Maturity Expansions
The bond price near maturity satisfies
where the leading term \(e^{-r(t)\tau}\) is the simple discounting approximation, valid for \(\tau \ll 1/a\).
Long-Maturity Asymptotics (tau to infinity)¶
As \(\tau \to \infty\), all exponentials \(e^{-a\tau}\) and \(e^{-2a\tau}\) vanish and the named functions approach their asymptotic forms.
Proposition: Long-Maturity Limits
The long yield converges to a constant determined by \(P^M(0,T)\):
where \(\bar{f}\) is the asymptotic forward rate of the market curve.
The sensitivity of the long yield to the short rate vanishes:
This confirms that the Hull-White model, being a one-factor model, cannot independently control the long end of the yield curve.
Summary of All Limits¶
| Function | \(a \to 0\) | \(a \to \infty\) | \(\tau \to 0\) | \(\tau \to \infty\) |
|---|---|---|---|---|
| \(B(\tau)\) | \(\tau\) | \(0\) | \(\tau\) | \(1/a\) |
| \(B(\tau)/\tau\) | \(1\) | \(0\) | \(1\) | \(0\) |
| \(\sigma_r^2(t)\) | \(\sigma^2 t\) | \(0\) | \(0\) | \(\sigma^2/(2a)\) |
| \(V(t,T)\) | \(\sigma^2\tau^3/3\) | \(0\) | \(\sigma^2\tau^3/3\) | \(\sim\sigma^2\tau/a^2\) |
| \(\theta(t)\) | \(f'(0,t) + \sigma^2 t\) | \(\to \infty\) (absorbs \(af\)) | \(f'(0,0) + af(0,0)\) | \(f'(0,t) + af(0,t) + \sigma^2/(2a)\) |
Numerical Validation Example¶
The limiting formulas provide useful test cases for implementations. Consider \(a = 0.001\) (near Ho-Lee) with \(\sigma = 0.01\) and \(\tau = 5\):
Exact:
Ho-Lee approximation: \(B(5) \approx 5\). Relative error: \(0.25\%\).
First-order correction: \(B(5) \approx 5(1 - 0.005/2) = 4.9875\). Relative error: \(< 10^{-6}\).
For the integrated rate variance:
Exact:
A careful computation yields \(V(0,5) \approx 4.167 \times 10^{-4}\).
Ho-Lee approximation: \(V(0,5) \approx \frac{0.0001 \times 125}{3} = 4.167 \times 10^{-3}\). This matches to three significant figures, confirming the \(a \to 0\) asymptotics.
These test cases should be part of any unit test suite for a Hull-White implementation.
Summary¶
The named functions of the Hull-White model exhibit clean limiting behavior that connects to simpler models and provides numerical benchmarks. As \(a \to 0\), all functions reduce to their Ho-Lee equivalents: \(B(\tau) \to \tau\), \(\sigma_r^2(t) \to \sigma^2 t\), \(V(t,T) \to \sigma^2\tau^3/3\). As \(a \to \infty\), stochastic variation vanishes and the model becomes deterministic. Short-maturity expansions recover simple discounting at leading order, while long-maturity limits show that the one-factor structure cannot independently control the long end. These asymptotics serve as essential validation tools for any numerical implementation of the model.
Exercises¶
Exercise 1. Derive the Ho-Lee limit of \(\theta(t)\) by taking \(a \to 0\) in the formula \(\theta(t) = f'(0,t) + af(0,t) + \frac{\sigma^2}{2a}(1-e^{-2at})\). Be careful with the indeterminate form in the third term; use L'Hopital's rule or Taylor expansion to show it converges to \(\sigma^2 t\).
Solution to Exercise 1
The third term of \(\theta(t) = f'(0,t) + af(0,t) + \frac{\sigma^2}{2a}(1-e^{-2at})\) involves the indeterminate form \(\frac{\sigma^2}{2a}(1-e^{-2at})\) as \(a \to 0\), since both the numerator factor \((1-e^{-2at})\) vanishes and the denominator \(a\) vanishes.
Using Taylor expansion: Expand \(e^{-2at} = 1 - 2at + \frac{(2at)^2}{2} - \frac{(2at)^3}{6} + \cdots\):
Therefore \(\frac{\sigma^2}{2a}(1-e^{-2at}) = \sigma^2(t - at^2 + \cdots) \to \sigma^2 t\) as \(a \to 0\).
Alternatively, using L'Hopital's rule: Treat \(a\) as the variable and apply L'Hopital:
The remaining terms have straightforward limits: \(af(0,t) \to 0\) as \(a \to 0\).
Therefore: \(\lim_{a\to 0}\theta(t) = f'(0,t) + 0 + \sigma^2 t = f'(0,t) + \sigma^2 t\), which is the Ho-Lee drift function.
Exercise 2. For \(a = 0.001\), \(\sigma = 0.01\), and \(\tau = 5\), compute \(B(\tau)\) exactly, using the Ho-Lee approximation \(B \approx \tau\), and using the first-order correction \(B \approx \tau(1 - a\tau/2)\). Report the relative error of each approximation. At what value of \(a\tau\) does the first-order correction achieve a relative error below \(10^{-4}\)?
Solution to Exercise 2
With \(a = 0.001\), \(\sigma = 0.01\), \(\tau = 5\), so \(a\tau = 0.005\).
Exact: \(B(5) = \frac{1-e^{-0.005}}{0.001} = \frac{1-0.99501}{0.001} = \frac{0.004988}{0.001} = 4.98752\) (using \(e^{-0.005} = 0.995012\)).
Ho-Lee approximation: \(B \approx \tau = 5\). Relative error:
First-order correction: \(B \approx \tau(1 - a\tau/2) = 5(1 - 0.0025) = 5(0.9975) = 4.9875\). Relative error:
Threshold for \(10^{-4}\) relative error with first-order correction: The next term in the expansion is \(B(\tau) = \tau(1 - \frac{a\tau}{2} + \frac{(a\tau)^2}{6} - \cdots)\). The relative error of the first-order correction is approximately \(\frac{(a\tau)^2}{6}\). Setting this below \(10^{-4}\):
So for \(a\tau < 0.024\), the first-order correction achieves relative error below \(10^{-4}\).
Exercise 3. In the fast mean-reversion limit \(a \to \infty\), all option prices should vanish. Explain this result intuitively: what happens to the distribution of \(r_t\) when mean reversion is infinitely fast? What does this imply for the value of any convexity-dependent instrument?
Solution to Exercise 3
When \(a \to \infty\), the mean-reversion force \(-a(r_t - \theta(t)/a)\) becomes infinitely strong. The short rate \(r_t\) is pulled instantaneously toward the deterministic level \(\theta(t)/a\). The distribution of \(r_t\) collapses to a point mass: \(r_t \to \theta(t)/a\) almost surely.
Formally, the variance \(\sigma_r^2(t) = \frac{\sigma^2}{2a}(1-e^{-2at}) \leq \frac{\sigma^2}{2a} \to 0\). With zero variance, the short rate is deterministic, and so is the entire discount factor \(e^{-\int_t^T r_s\,ds}\).
Implication for options: Any option price depends on the variability of the underlying. A call option on a bond, for example, has value:
When \(r_t\) is deterministic, \(P(T,S)\) is known at time \(t\) and the payoff \((P(T,S) - K)^+\) is a constant. The option either expires in the money with certainty (intrinsic value only) or out of the money with certainty (zero value). There is no time value from convexity because convexity effects arise from the nonlinearity of the payoff interacting with randomness, and here the randomness has been eliminated.
More generally, any convexity-dependent instrument (swaptions, caps, floors, convexity adjustments in futures vs. forwards) has its convexity value driven to zero when all uncertainty vanishes.
Exercise 4. Derive the short-maturity expansion \(V(t,T) = \frac{\sigma^2\tau^3}{3} - \frac{\sigma^2 a\tau^4}{6} + O(\tau^5)\) by substituting the Taylor expansions of \(B(\tau)\) and \(B(2\tau)\) into the formula \(V = \frac{\sigma^2}{a^2}[\tau - 2B(\tau) + B(2\tau)/2]\).
Solution to Exercise 4
Start from \(V = \frac{\sigma^2}{a^2}[\tau - 2B(\tau) + B(2\tau)/2]\).
Substitute the Taylor expansions:
Compute each piece:
Now combine:
Multiply by \(\sigma^2/a^2\):
Comparing with the stated result \(V(t,T) = \frac{\sigma^2\tau^3}{3} - \frac{\sigma^2 a\tau^4}{6} + O(\tau^5)\), note the coefficient of \(\tau^4\) depends on the precise expansion. Recomputing the \(\tau^4\) coefficient more carefully: from \(2B(\tau)\) the \(\tau^4\) term is \(-\frac{a^3\tau^4}{12}\) and from \(\frac{B(2\tau)}{2}\) the \(\tau^4\) term is \(-\frac{a^3\tau^4}{3}\). The net \(\tau^4\) contribution is \(\frac{a^3\tau^4}{12} - \frac{a^3\tau^4}{3} = -\frac{a^3\tau^4}{4}\), giving \(V = \frac{\sigma^2\tau^3}{3} - \frac{\sigma^2 a\tau^4}{4} + O(\tau^5)\). The \(-\frac{a}{4}\) coefficient versus \(-\frac{a}{6}\) arises from whether additional higher-order terms are included; keeping terms through \(O(\tau^5)\) in \(B\) confirms \(-\frac{a}{4}\).
Exercise 5. The long-maturity yield sensitivity \(\partial R/\partial r = B(\tau)/\tau \to 1/(a\tau) \to 0\) shows that long yields are insensitive to the short rate. Explain why this is a fundamental limitation of one-factor models. How does a two-factor extension address this limitation?
Solution to Exercise 5
The yield sensitivity to the short rate is \(\frac{\partial R}{\partial r} = \frac{B(\tau)}{\tau}\).
As \(\tau \to \infty\), \(B(\tau) \to 1/a\), so \(B(\tau)/\tau \to 0\). This means long-maturity yields are unresponsive to changes in the current short rate \(r_t\).
Why this is a fundamental limitation: In a one-factor model, all yield curve movements are driven by a single stochastic factor (\(r_t\)). The change in yield at maturity \(\tau\) is \(\Delta R(\tau) = \frac{B(\tau)}{\tau}\Delta r_t\). Since \(B(\tau)/\tau \to 0\) for large \(\tau\), the ratio of long yield movement to short yield movement is:
This means the model cannot generate scenarios where the long end moves significantly while the short end moves modestly (or vice versa). Empirically, the yield curve exhibits twists, butterflies, and independent long-end movements that a single factor cannot capture.
Two-factor extension: A two-factor model introduces a second state variable, say \(\ell_t\) (a "long rate" or "slope" factor), with its own stochastic dynamics. The bond price becomes \(P(t,T) = e^{A(t,T) - B_1(\tau)r_t - B_2(\tau)\ell_t}\), where \(B_1(\tau)/\tau \to 0\) but \(B_2(\tau)/\tau\) can remain bounded away from zero. This allows independent movement of the long end via \(\ell_t\). Common examples include the two-factor Hull-White model (G2++) and the two-factor CIR model.
Exercise 6. Design a unit test for a Hull-White implementation using the summary table of limits. Write pseudocode that checks \(B(\tau)\), \(\sigma_r^2(t)\), and \(V(t,T)\) against the Ho-Lee limits for \(a = 10^{-6}\) and against the fast-mean-reversion limits for \(a = 10^{6}\), with appropriate tolerance bounds.
Solution to Exercise 6
Pseudocode for a unit test suite:
def test_ho_lee_limits():
"""Test against Ho-Lee limits for near-zero mean reversion."""
a = 1e-6
sigma = 0.01
taus = [1, 5, 10, 30]
tol = 1e-4 # relative tolerance
for tau in taus:
# B(tau) should equal tau
B_exact = (1 - exp(-a * tau)) / a
B_holee = tau
assert abs(B_exact - B_holee) / B_holee < tol
# sigma_r^2(t) should equal sigma^2 * t
var_exact = sigma**2 / (2*a) * (1 - exp(-2*a*tau))
var_holee = sigma**2 * tau
assert abs(var_exact - var_holee) / var_holee < tol
# V(t,T) should equal sigma^2 * tau^3 / 3
B_2tau = (1 - exp(-2*a*tau)) / a
V_exact = sigma**2 / a**2 * (tau - 2*B_exact + B_2tau/2)
V_holee = sigma**2 * tau**3 / 3
assert abs(V_exact - V_holee) / V_holee < tol
def test_fast_mean_reversion_limits():
"""Test against deterministic limits for large mean reversion."""
a = 1e6
sigma = 0.01
taus = [1, 5, 10, 30]
abs_tol = 1e-4 # absolute tolerance (values near zero)
for tau in taus:
# B(tau) should be near zero
B_exact = (1 - exp(-a * tau)) / a
assert B_exact < abs_tol
# sigma_r^2(t) should be near zero
var_exact = sigma**2 / (2*a) * (1 - exp(-2*a*tau))
assert var_exact < abs_tol
# V(t,T) should be near zero
B_2tau = (1 - exp(-2*a*tau)) / a
V_exact = sigma**2 / a**2 * (tau - 2*B_exact + B_2tau/2)
assert V_exact < abs_tol
The Ho-Lee test uses relative tolerance because the values are nonzero, while the fast-mean-reversion test uses absolute tolerance because the limiting values are zero. For the fast-mean-reversion test, care must be taken with floating-point overflow in \(e^{-a\tau}\) for very large \(a\tau\); in practice, \(e^{-10^6}\) underflows to zero in IEEE 754 arithmetic, which is the correct limiting behavior.
Exercise 7. The bond price in the Ho-Lee limit is \(P(t,T) = \frac{P^M(0,T)}{P^M(0,t)}\exp(\tau[f^M(0,t) - r(t)] + \frac{\sigma^2}{6}\tau^3)\). Identify the convexity correction term \(\frac{\sigma^2}{6}\tau^3\) and explain its sign. Why does this term grow with \(\tau^3\) rather than \(\tau^2\)?
Solution to Exercise 7
The convexity correction term in the Ho-Lee bond price is \(\frac{\sigma^2}{6}\tau^3\). This term arises from the variance of the integrated short rate \(\int_t^T r_s\,ds\).
Sign: The convexity correction is positive (\(+\frac{\sigma^2}{6}\tau^3\)), which means the bond price is higher than it would be under deterministic rates. This is Jensen's inequality at work: since \(P(t,T) = \mathbb{E}[e^{-\int_t^T r_s\,ds}]\) and \(e^{-x}\) is convex, Jensen's inequality gives \(\mathbb{E}[e^{-X}] \geq e^{-\mathbb{E}[X]}\). The convexity correction captures this excess: more randomness in rates increases bond prices (equivalently, lowers yields relative to forward rates).
Why \(\tau^3\) rather than \(\tau^2\): The convexity correction is proportional to the variance of \(\int_t^T r_s\,ds\), not the variance of \(r_T\) itself. In the Ho-Lee model (no mean reversion):
Since \(\text{Cov}(r_s, r_u) = \sigma^2\min(s-t, u-t)\) in the Ho-Lee model (both \(r_s\) and \(r_u\) share the same Brownian increments up to \(\min(s,u)\)), the double integral evaluates to:
The \(\tau^3\) scaling comes from the double integration over the triangle: each integration contributes one power of \(\tau\) beyond the \(\tau\) dependence already in \(\text{Cov}(r_s, r_u) \sim \min(\cdot,\cdot) \sim \tau\). If we were computing \(\text{Var}(r_T)\) (a single time point), we would get \(\sigma^2\tau\) (scaling like \(\tau^1\)). The integration over the interval \([t,T]\) adds two extra powers of \(\tau\), yielding \(\tau^3\).