Regimes with regular durations
A Vasicek short rate whose mean level switches between two regimes that each last a nearly fixed time. The yield curve then depends on how long the current regime has lasted.
The model
The short rate reverts to a level set by a hidden regime $y_t \in \{1, 2\}$:
$$dr_t = \kappa\,(\theta_{y_t} - r_t)\,dt + \sigma\,dW_t .$$In the other examples a regime ends at a constant rate, so its duration is exponential and its age says nothing about when it will end. Here regime $i$ lasts an Erlang time: the sum of $k$ independent exponential stages, each with mean $m_i/k$. The mean duration is $m_i$ and its standard deviation is
$$s_i = \frac{m_i}{\sqrt k} .$$With $k = 1$ this is the usual Markov switch. As $k$ grows the regimes last close to $m_1$ and $m_2$, like a policy cycle or a clock.
The slow variable is the short rate, which reverts over $1/\kappa$. The fast variable is the regime, which changes every $m_1$ or $m_2$. The expansion is in the durations.
Parameters: $\kappa = 0.5$, $\sigma = 0.01$, $\theta = (0.07, 0.01)$, $m = (0.3, 0.6)$ years.
Stages make the regime Markov
The regime alone is not a Markov process: the chance that it ends soon depends on how long it has lasted. The stage is Markov.
Label the stages $1, \dots, k$ in regime 1 and $k+1, \dots, 2k$ in regime 2, arranged in a ring. Stage $j$ moves to stage $j+1$ at rate $k/m_1$ in regime 1 and $k/m_2$ in regime 2, and stage $2k$ returns to stage $1$.
All stages of a regime share its mean level $\theta_i$, so no price observed today can tell them apart. They differ only in how soon the regime will change. Observing a function of a Markov chain that is constant on groups of states is called lumping.
The stage chain has constant hazards, so everything on the engine page applies to it without change.
Reduction to a linear system
Let $P_j(T, r)$ be the price of a zero-coupon bond maturing at $T$ when the chain is in stage $j$. As in the regime-switching example,
$$P_j(T, r) = e^{a_j(T) - B(T)\,r}, \qquad B(T) = \frac{1 - e^{-\kappa T}}{\kappa},$$ $$a' = (Q + \operatorname{diag} g)\,a, \qquad a(0) = \mathbf 1, \qquad g_j(t) = -\kappa\,\theta_j\,B(t) + \tfrac12\sigma^2 B(t)^2 ,$$where $Q$ is the generator of the ring and $\theta_j$ is the level of the regime that stage $j$ belongs to.
Averaging
The regime spends a fraction $\pi_i$ of the time in state $i$:
$$\pi_1 = \frac{m_1}{m_1 + m_2}, \qquad \pi_2 = \frac{m_2}{m_1 + m_2}, \qquad \bar\theta = \pi_1\theta_1 + \pi_2\theta_2 .$$When the regimes are short the bond price tends to Vasicek's price with the level $\bar\theta$:
$$\log P_{\text{Vas}} = -B\,r - \kappa\bar\theta\,I_1 + \tfrac12\sigma^2 I_2,$$ $$I_1 = \int_0^T B = \frac{T - B}{\kappa}, \qquad I_2 = \int_0^T B^2 = \frac{1}{\kappa^2}\Big(T - 2B + \frac{1 - e^{-2\kappa T}}{2\kappa}\Big).$$This limit does not depend on $k$: the averaged model sees only the fractions of time.
The first correction
The regime enters the price through the integral
$$Z = -\kappa\int_0^T \theta_{y_s}\,B(T - s)\,ds .$$To first order in the durations the logarithm of $\mathbb E\,e^{Z}$ is the mean of $Z$ plus half its variance. Each piece gives one term.
The variance term
Over a long horizon the variance of $\int (\theta_{y_s} - \bar\theta)\,h(s)\,ds$ is
$$2\,(\theta_1 - \theta_2)^2 K_1 \int h^2 ,$$where $K_1$ is the Green–Kubo number of the regime indicator, the integral of its autocovariance. For an alternating renewal process it depends on the durations only through their means and variances:
$$K_1 = \frac{m_2^2 s_1^2 + m_1^2 s_2^2}{2\,(m_1 + m_2)^3} .$$For Erlang durations $s_i^2 = m_i^2/k$, so
$$K_1 = \frac{m_1^2\,m_2^2}{k\,(m_1 + m_2)^3} .$$This is the Markov value divided by $k$. Regular regimes average out faster, because a long stretch of time holds close to the right share of each regime.
The start term
The mean of $Z$ differs from its averaged value because the regime today is known. Most of the difference comes from the first few regimes, while $B(T - s)$ is still close to $B(T)$. So the difference is
$$-\kappa\,B(T)\,w, \qquad w = \mathbb E\int_0^\infty (\theta_{y_s} - \bar\theta)\,ds,$$where the expectation is given the current regime and how long it has lasted.
To compute $w$, follow the chain to the next start of regime 1. Let $R$ be the expected remaining time in the current regime.
Starting in regime 1, the level exceeds $\bar\theta$ by $\pi_2(\theta_1 - \theta_2)$ for the remaining time, then falls short by $\pi_1(\theta_1 - \theta_2)$ for a whole regime 2. Starting in regime 2 there is only the shortfall over the remaining time.
After that point every path is the same, which adds a constant. The constant is fixed by requiring $w$ to average to zero over the stationary distribution, in which the stage is uniform within each regime. The result is
$$w_1 = (\theta_1 - \theta_2)\,\pi_2\big(R - \pi_1 m_2 - \pi_1 c\big), \qquad w_2 = -(\theta_1 - \theta_2)\,\pi_1\big(R + \pi_2 c - \pi_1 m_2\big),$$ $$c = \frac{(m_1 - m_2)(k + 1)}{2k} .$$For $k = 1$ the remaining time is always $m_i$, whatever the age, and the start terms reduce to the Markov values
$$w_1 = (\theta_1 - \theta_2)\,\frac{\pi_2\,m_1 m_2}{m_1 + m_2}, \qquad w_2 = -(\theta_1 - \theta_2)\,\frac{\pi_1\,m_1 m_2}{m_1 + m_2} .$$The remaining time given the age
The stage is hidden but the age $\tau$ of the regime is observed. Given the age, the number of completed stages is Poisson with mean $k\tau/m_i$, conditioned on fewer than $k$ stages having been completed. So
$$R(\tau) = \frac{m_i}{k}\;\frac{\sum_{n=0}^{k-1} (k - n)\,\mu^n/n!}{\sum_{n=0}^{k-1} \mu^n/n!}, \qquad \mu = \frac{k\,\tau}{m_i} .$$At $\tau = 0$ this is $m_i$. As the age grows it falls toward $m_i/k$, the mean of one stage.
The bond price written out
In regime $i$, at age $\tau$, with short rate $r$ and maturity $T$:
$$\log P_i(T, r, \tau) = -B\,r - \kappa\bar\theta\,I_1 + \tfrac12\sigma^2 I_2 + \kappa^2(\theta_1 - \theta_2)^2 K_1\,I_2 - \kappa\,B\,w_i\big(R(\tau)\big) + O(m^2),$$ $$\begin{aligned} B &= \frac{1 - e^{-\kappa T}}{\kappa}, \\ I_1 &= \frac{T - B}{\kappa}, \\ I_2 &= \frac{1}{\kappa^2}\Big(T - 2B + \frac{1 - e^{-2\kappa T}}{2\kappa}\Big), \\ \bar\theta &= \pi_1\theta_1 + \pi_2\theta_2, \qquad \pi_i = \frac{m_i}{m_1 + m_2}, \\ K_1 &= \frac{m_1^2\,m_2^2}{k\,(m_1 + m_2)^3}, \\ w_1 &= (\theta_1 - \theta_2)\,\pi_2\big(R - \pi_1 m_2 - \pi_1 c\big), \\ w_2 &= -(\theta_1 - \theta_2)\,\pi_1\big(R + \pi_2 c - \pi_1 m_2\big), \\ c &= \frac{(m_1 - m_2)(k + 1)}{2k}, \\ R &= \frac{m_i}{k}\;\frac{\sum_{n=0}^{k-1} (k - n)\,\mu^n/n!}{\sum_{n=0}^{k-1} \mu^n/n!}, \qquad \mu = \frac{k\,\tau}{m_i} . \end{aligned}$$A worked example
Take the parameters above with $r = 3\%$ and $T = 1$, starting in the high regime. The averaged model gives a one-year yield of 299.9 basis points. The yields $-\log P/T$ in basis points:
| stages $k$ | age $\tau$ | remaining $R$ | first order | numerical solution |
|---|---|---|---|---|
| 1 | any | 0.300 | 331.3 | 326.0 |
| 4 | 0 | 0.300 | 325.4 | 325.4 |
| 4 | 0.15 | 0.182 | 306.8 | 308.3 |
| 4 | 0.3 | 0.135 | 299.5 | 301.4 |
| 4 | 0.6 | 0.105 | 294.8 | 296.8 |
| 16 | 0 | 0.300 | 324.0 | 323.2 |
| 16 | 0.15 | 0.151 | 300.6 | 301.5 |
| 16 | 0.3 | 0.064 | 286.8 | 288.1 |
| 16 | 0.6 | 0.033 | 281.9 | 283.2 |
With Markov regimes the one-year yield in the high regime is the same at every age. With 16 stages it falls by 40 basis points between the start of the regime and twice its mean duration. By then the high regime is almost certainly about to end, and the yield sits below the averaged value.
At one year the variance term is small (0.09 basis points for $k = 1$) and the start term carries the effect. At long maturities the variance term grows like $T$ and the start term fades like $1/T$.
What the yield curve identifies
To first order the variance term sees $K_1$ alone, so regular durations cannot be told apart from Markov durations with shorter means. The start term separates them: under Markov switching the curve does not move with the age of the regime, and under regular switching it does. The age of a regime is observable, for example the time since the last policy change, so it can be used as a regressor.
The same stage construction covers any duration distribution built from exponential stages, which can approximate any distribution of durations. At first order only the means and variances of the durations enter the variance term.
Certificate: verify_erlang.py. It checks $K_1$ and $w$ against the group inverse of the stage chain to machine precision, and checks that the first-order yield error falls like the square of the durations.