Fast regimes in any pricing equation

One first-order rule for every model whose parameters are switched by a fast hidden chain.

The examples switch a volatility, a mean level, a hazard rate or a jump intensity, and they all follow one rule. To first order, a fast chain adds a second-order term to the pricing operator, weighted by a Green–Kubo matrix of the switched parameters.

The rule has a symmetric part and an antisymmetric part. It is checked below on option smiles, on the joint survival of many names, and on a model where the order of the regimes matters.

The setting

A price $u(t, x)$ depends on time to maturity $t$ and on a slow state $x$, such as a log stock price or a short rate. With the parameters held fixed, it solves a pricing equation

$$\partial_t u = \mathcal L\,u,$$

where $\mathcal L$ is the generator of the slow process, including any discounting.

Now let some of the parameters depend on a hidden state $y_t$. The hidden state is a Markov chain with generator $Q$ and stationary law $\pi$, and it switches much faster than $x$ moves. Write the operator in regime $y$ as its average plus fluctuations:

$$\mathcal L_y = \bar{\mathcal L} + \sum_k \phi_k(y)\,A_k .$$

Here $\phi_k(y)$ is the deviation of the $k$-th switched parameter from its $\pi$-average, and $A_k$ is the operator that the parameter multiplies. Most models in finance fit this form:

modelswitched parameteroperator it multiplies
Black–Scholes, $x = \log S$variance $\sigma^2$$\tfrac12(\partial_{xx} - \partial_x)$
Black–Scholesinterest rate $r$$\partial_x - 1$
Vasicek, $x = r$mean level $\theta$, with $\kappa$ fixed$\kappa\,\partial_x$
Vasicekreversion speed $\kappa$, with $\theta$ fixed$(\theta - x)\,\partial_x$
Vasicek$\kappa$ and $\theta$ together: the products $\kappa\theta$ and $\kappa$$\partial_x$ and $-x\,\partial_x$
Vasicekvariance $\sigma^2$$\tfrac12\,\partial_{xx}$
reduced-form credithazard rate $h$$-1$
jump diffusionjump intensity $\nu$$\mathcal J - 1$, with $\mathcal J f(x) = \mathbb E f(x + Z)$
Heston, $x = (\log S, v)$long-run variance $\theta$, with $\kappa$ fixed$\kappa\,\partial_v$

A parameter that multiplies another is not a coefficient on its own. When $\kappa$ and $\theta$ switch together, the drift is $\kappa_y\theta_y - \kappa_y x$, so the coefficients are $\kappa\theta$ and $\kappa$ with the fixed operators $\partial_x$ and $-x\,\partial_x$, and the averaged drift is $\mathbb E_\pi[\kappa\theta] - \mathbb E_\pi[\kappa]\,x$.

Averaging $\kappa$ and $\theta$ separately misses $\operatorname{Cov}_\pi(\kappa, \theta)$ at leading order. With a symmetric two-state chain, $\kappa = (1, 3)$ and $\theta = (0.01, 0.09)$, the effective level is $\mathbb E_\pi[\kappa\theta]/\mathbb E_\pi[\kappa] = 0.07$, not $\mathbb E_\pi[\theta] = 0.05$. The same product $\kappa\theta$ sits in the Heston variance drift.

The regime-dependent prices $u_y$ are coupled by the chain:

$$\partial_t u_y = \mathcal L_y\,u_y + \sum_{y'} Q_{yy'}\,u_{y'} .$$

We want the price seen by someone who cannot observe the regime, so the regime is drawn from $\pi$ and the payoff does not depend on it.

Deriving the rule

Scale the chain as $Q = Q_0/\varepsilon$ and expand $u = u_0 + \varepsilon u_1 + \varepsilon^2 u_2 + \cdots$. Each order is handled with the two steps of the solvability page: impose the condition that makes the equation solvable, then solve it.

Order $\varepsilon^{-1}$. Only the chain appears:

$$Q_0\,u_0 = 0 .$$

So $u_0$ is constant across regimes. The leading price does not depend on the hidden state.

Order $1$. The equation for the first corrector is

$$Q_0\,u_1 = \partial_t u_0 - \bar{\mathcal L}\,u_0 - \sum_k \phi_k\,A_k\,u_0 .$$

Averaging under $\pi$ kills the left side and the fluctuations $\phi_k$. So a solution exists only if

$$\partial_t u_0 = \bar{\mathcal L}\,u_0 ,$$

which is the averaged model. The equation is then solved with the group inverse:

$$u_1 = \bar u_1 - \sum_k \big(Q_0^{\#}\phi_k\big)\,A_k\,u_0 .$$

The second term remembers the regime. It is large in a regime whose parameters sit above average and persist.

Order $\varepsilon$. The solvability condition at the next order fixes the average $\bar u_1$. Averaging the equation under $\pi$ gives

$$\partial_t \bar u_1 = \bar{\mathcal L}\,\bar u_1 + \mathbb E_\pi\Big[\sum_j \phi_j\,A_j\,u_1\Big].$$

Substituting $u_1$ from the previous step turns the last term into

$$-\sum_{j,k} \pi\cdot\big(\phi_j\;Q_0^{\#}\phi_k\big)\,A_j A_k\,u_0 .$$

Putting $\varepsilon$ back into the chain, the coefficient is the Green–Kubo matrix

$$K_{jk} = -\,\pi\cdot\big(\phi_j\;Q^{\#}\phi_k\big) = \int_0^\infty \operatorname{Cov}_\pi\big(\phi_j(y_0),\,\phi_k(y_\tau)\big)\,d\tau .$$

The rule

$$\bar u \;\approx\; e^{t\bar{\mathcal L}} f \;+\; \int_0^t e^{(t-s)\bar{\mathcal L}}\,\Big(\sum_{j,k} K_{jk}\,A_j A_k\Big)\,e^{s\bar{\mathcal L}} f\,ds .$$

Here $f$ is the payoff. The averaged operator gains the correction

$$\bar{\mathcal L} \;\longrightarrow\; \bar{\mathcal L} + \sum_{j,k} K_{jk}\,A_j A_k ,$$

which should be applied as a first-order perturbation, as in the formula above. Exponentiating the corrected operator is not always safe, because it can contain higher derivatives with the wrong sign for a diffusion.

Covariance and order

Split the matrix into its symmetric and antisymmetric parts, $K = K^{\mathrm{sym}} + K^{\mathrm{anti}}$. The correction splits the same way:

$$\sum_{j,k} K_{jk}\,A_j A_k \;=\; \sum_{j,k} K^{\mathrm{sym}}_{jk}\,A_j A_k \;+\; \sum_{j\lt k} K^{\mathrm{anti}}_{jk}\,\big[A_j, A_k\big].$$

With a payoff that does not depend on the regime, the antisymmetric part only acts through the commutator $[A_j, A_k] = A_j A_k - A_k A_j$. So the direction of the cycle matters only when the operators do not commute: when it matters whether one parameter acts before the other.

A payoff or a running source that depends on the regime opens a second route. The race studied on the kinetics site is the example: each channel's own rate is the source of its win probability, and the asymmetry of $K$ is visible from the order of arrivals even though every operator there is a multiplication. The symmetry of $K$ under detailed balance, and its use as a witness of a hidden irreversible driver, are developed there (Cotton, working draft).

Constant-coefficient operators commute. In Black–Scholes, and whenever the level and variance of Vasicek are switched, only the covariance enters. That is why the yield curve sees a symmetric $2\times2$ matrix.

The smile from switching volatility

Switch the variance in Black–Scholes. With $x = \log S$ there is one operator,

$$A = \tfrac12\big(\partial_{xx} - \partial_x\big).$$

It is the derivative with respect to total variance $V = \sigma^2 t$: applied to a Black–Scholes price, it gives $\partial_V$. So the correction $K\,A^2$ is $K\,\partial_V^2$, and the rule gives a closed form:

$$C \;\approx\; C_{\mathrm{BS}}(\bar V) + t\,K_{\sigma^2\sigma^2}\,\frac{\partial^2 C_{\mathrm{BS}}}{\partial V^2}(\bar V),$$ $$\frac{\partial^2 C_{\mathrm{BS}}}{\partial V^2} = \frac{S\,\varphi(d_1)\,(d_1 d_2 - 1)}{4\,V^{3/2}} .$$

This is a volga term, the convexity of the price in variance. It is positive in the wings and negative near the money, so no single volatility can absorb it: it is a smile.

It is the first-order form of the mixing formula of Hull and White, with the variance of integrated variance replaced by its leading term $2t\,K$. The two differ by a term of order $\varepsilon^2$, bounded uniformly in maturity; the cumulants page gives the bound, the exact cumulants of the return, and the CIR case.

This smile is known. Sircar and Papanicolaou (1999) derive it for a fast ergodic volatility driver whose integrated fluctuations obey a central limit theorem, as a finite-state chain does. Stationarity and ergodicity alone are not enough: the covariance tail sets the correction scale. Fouque, Lorig and Sircar (2016) give the correlated fast-factor expansion to second order.

The check below confirms the rule reproduces it. A Heston example shows the antisymmetric part moving the smile.

The check uses three regimes with volatilities of 10%, 30% and 20%. The chain is twice the irreversible three-state generator

$$Q_A = \begin{pmatrix} -3 & 2 & 1 \\ 1 & -2 & 1 \\ 0.5 & 1.5 & -2 \end{pmatrix},$$

so that regimes last about two to three months.

$S = 100$, maturity 1 year, $r = 2\%$, averaged volatility 23.80%.

strikenumerical solutionruleaveraged
7024.14%24.17%23.80%
8023.83%23.81%23.80%
9023.64%23.62%23.80%
10023.57%23.55%23.80%
11023.59%23.58%23.80%
12023.69%23.67%23.80%
13023.83%23.81%23.80%

The table shows implied volatilities. The rule reproduces the smile to within a few hundredths of a volatility point. With the chain four times faster, the largest price error falls from $1.2\times10^{-3}$ to $7\times10^{-5}$, sixteen times smaller, as a second-order remainder should.

Many names from one matrix

Give each of several firms a hazard rate $h_m(y)$ that depends on the common regime, and let the defaults be independent given the regime path. The survival of a set $S$ of names is

$$P_S(t) = \mathbb E\Big[\exp\Big(-\int_0^t \sum_{m\in S} h_m(y_s)\,ds\Big)\Big].$$

Every operator is $-1$, so all of them commute, and the rule gives

$$\log P_S(t) \;\approx\; -t\sum_{m\in S}\bar h_m \;+\; t\sum_{j,k\in S} K^{\mathrm{sym}}_{jk} .$$

One matrix gives the joint survival of every subset. For a pair of names the dependence is

$$\log\frac{P_{\{1,2\}}(t)}{P_{\{1\}}(t)\,P_{\{2\}}(t)} \;\approx\; 2t\,K^{\mathrm{sym}}_{12} .$$

Two consequences follow.

Joint survival sees only the symmetric part of $K$. The order of defaults sees more.

Which name defaults first is a race under a shared driver, and two names already suffice for its law to reveal the asymmetry of $K$, provided the chain has at least three states and is not reversible. Races of this kind are studied on the kinetics site.

For two names, starting from the stationary law, the probability that name 1 defaults first is

$$p_1 = \pi^{\top}\big(\operatorname{diag}(h_1 + h_2) - Q\big)^{-1}h_1 .$$

Expanding the resolvent for a fast chain, with $\bar h = \bar h_1 + \bar h_2$,

$$p_1 = \frac{\bar h_1}{\bar h} + \frac{1}{\bar h}\Big(\frac{\bar h_1}{\bar h}\sum_{j,k}K_{jk} - K_{11} - K_{21}\Big) + O(\varepsilon^2) .$$

The entry $K_{21}$ appears without $K_{12}$. Reversing the direction of the cycle transposes $K$, and moves $p_1$ by

$$\frac{K_{12} - K_{21}}{\bar h} .$$

An example: a three-state cycle with rate 2 one way round and 0.1 the other, hazards $h_1 = (0.07, 0.01, 0.01)$ and $h_2 = (0.01, 0.07, 0.01)$, and the generator multiplied by 10. Name 1 defaults first with probability 0.500150 with the cycle one way and 0.499850 the other way. Every single-name and joint survival probability is the same for both directions.

With three or more causes, however, one unweighted race sees only the row sums of the antisymmetric matrix and can miss a nonzero cyclic current completely. The joint-default note proves that varying positive cause exposures recovers the full antisymmetric matrix and gives an exact three-cause counterexample to identification from one unweighted race.

The check uses four names and the three-state chain $Q_A$ from the smile example. The pairwise dependence over five years, in units of $10^{-4}$, is:

$$2t\,K^{\mathrm{sym}} \times 10^{4} = \begin{pmatrix} 25.3 & 16.8 & -8.5 & -6.3 \\ 16.8 & 11.9 & -4.0 & -4.9 \\ -8.5 & -4.0 & 7.1 & 0.2 \\ -6.3 & -4.9 & 0.2 & 2.4 \end{pmatrix}.$$

Its two smallest eigenvalues are zero. With the chain ten times faster, the rule matches the numerical log survival of all fifteen subsets to within $3\times10^{-6}$. At forty times faster it matches within $2\times10^{-7}$.

The order of the regimes

Now switch the reversion speed and the variance of a Vasicek short rate together. The two operators are

$$A_\kappa = (\theta - x)\,\partial_x, \qquad A_{\sigma^2} = \tfrac12\,\partial_{xx},$$

and they do not commute. A short calculation gives

$$\big[A_\kappa,\,A_{\sigma^2}\big] = \partial_{xx} .$$

So the antisymmetric part of $K$ acts as extra variance. The effective variance becomes

$$\sigma^2_{\mathrm{eff}} = \bar\sigma^2 + 2\,K^{\mathrm{anti}}_{\kappa\sigma^2} + \big(\text{symmetric terms}\big).$$

The intuition is about sequence. If high volatility tends to be followed by fast reversion, shocks are forgotten quickly and the rate is effectively calmer. If high volatility is followed by slow reversion, shocks linger.

The averages and the covariances are the same in both cases. Only the direction of the cycle differs.

The check uses a chain that cycles almost only one way, $1 \to 2 \to 3 \to 1$, at rate 2 per year, with a reverse rate of 0.1:

speeds $\kappa = (0.3,\ 1.5,\ 3.0)$, variances $\sigma^2 = (0.0004,\ 0.0036,\ 0.0016)$, $\theta = 4\%$.

For this chain the order term changes the effective variance by $-19\%$, from $0.00187$ to $0.00151$. Running the cycle the other way changes it by $+19\%$. The table shows errors in a two-year bond price, on a finite-difference grid, as the chain is sped up by a factor $m$:

approximation$m = 10$$m = 20$$m = 40$$m = 80$
averaged2.8e-61.5e-67.8e-74.0e-7
symmetric part only7.7e-63.7e-61.8e-69.1e-7
full rule4.4e-71.2e-73.0e-87.7e-9

The full rule is second order. Keeping only the covariance does worse than no correction at all, because here the order term is the larger of the two.

Higher orders and special cases

Certificate: verify_general.py.