Portfolio choice with fast regimes

Merton's problem, the Markowitz frontier and a regime-blind investor when bull and bear markets alternate over weeks to months: each correction is one Green–Kubo number.

The model

A stock and a bank account. The interest rate $r$, the stock's expected return $\mu$ and its volatility $\sigma$ depend on a market regime $y_t$:

The regime is a Markov chain with generator $Q$ and stationary law $\pi$, independent of $W$. An investor holds the fraction $w_t$ of wealth $X_t$ in the stock:

The Sharpe ratio in regime $i$ is $\lambda_i = (\mu_i - r_i)/\sigma_i$. There are two time scales: the investment horizon $T$, measured in years, and the life of a regime, measured in weeks or months.

The example has two regimes. A bull market with a Sharpe ratio of one lasts four months on average. A bear market with falling prices and high volatility lasts about six weeks.

The regime changes at rate $q_1$ out of the bull market and $q_2$ out of the bear market, so the bull market holds a share $\pi_1 = q_2/(q_1 + q_2)$ of the time.

Parameters: $r = 3\%$; bull $\mu_1 = 15\%$, $\sigma_1 = 12\%$ ($\lambda_1 = 1$); bear $\mu_2 = -5\%$, $\sigma_2 = 30\%$ ($\lambda_2 = -0.267$); $q_1 = 3$, $q_2 = 9$ per year, so $\pi = (0.75,\ 0.25)$; relative risk aversion $\gamma = 3$; horizon $T = 5$ years for utility and $T = 1$ year for the Markowitz problem. A speed factor $m$ multiplies both rates.

The quantities

Three investors are compared.

• The observed-regime investor sees $y_t$ and maximizes expected CRRA utility $U(x) = x^{1-\gamma}/(1-\gamma)$ of terminal wealth.
• The Markowitz investor also sees $y_t$ and minimizes the variance of $X_T$ for a given mean.
• The regime-blind investor has CRRA utility, cannot see $y_t$ and holds a constant fraction $w$.

Utility is reported as a certainty-equivalent rate: the constant growth rate that a riskless investment would need to give the same expected utility,

The Markowitz investor's answer is the efficient frontier: the smallest attainable $\operatorname{Var} X_T$ for each value of $\mathbb E X_T$.

Averaging

When the regime switches infinitely fast, each investor sees an averaged market, but not the same one.

The investor who sees the regime can still act on it, holding Merton's fraction for each regime. This investor's growth rate in regime $i$ is

and in the limit the investor earns the average $\bar\psi = \pi\cdot\psi$. The investor's market is Merton's with the square-averaged Sharpe ratio $\langle\lambda^2\rangle = \pi\cdot\lambda^2$, as in the fast-factor limit of Fouque, Sircar and Zariphopoulou (2017).

The regime-blind investor sees a single stock with the averaged return $\bar\mu$ and the averaged variance $\bar s = \pi\cdot\sigma^2$. The investor's Merton fraction is

The gap between the two, $\langle\lambda^2\rangle \ge (\bar\mu - r)^2/\bar s$, is a Cauchy–Schwarz inequality. In the example it is the difference between a certainty-equivalent rate of 15.7% and one of 5.3%. The corrections below are what finite regime lifetimes add to each.

Reduction to a linear system

Let $V_i(t, x)$ be the observed-regime investor's value in regime $i$ with wealth $x$ at time $t$. The Hamilton–Jacobi–Bellman system couples the regimes through $Q$:

Try $V_i = \frac{x^{1-\gamma}}{1-\gamma}\,\Psi_i(T - t)$. Then $x\,\partial_x V_i$ and $x^2\,\partial_{xx}V_i$ are multiples of $x^{1-\gamma}\Psi_i$, and the bracket becomes

attained at Merton's fraction $w_i = \lambda_i/(\gamma\sigma_i)$. Dividing by $x^{1-\gamma}/(1-\gamma)$ leaves a linear system in time to go $\tau$:

This is the system that the engine solves to all orders, with the constant forcing $g = (1-\gamma)\psi$. It is also the Feynman–Kac form of

The regime-blind investor leads to the same system. Given the regime path, $\log X_T$ is Gaussian, and its exponential moment gives

Here $h_i(w)$ is the certainty-equivalent rate of the fixed fraction $w$ in regime $i$; the observed investor's $\psi_i$ is its maximum over $w$. The two problems differ only in the vector that plays the part of the forcing.

The Markowitz problem is linear-quadratic, as Zhou and Yin (2003) show with regimes. With a constant rate $r$ and target $d$, the smallest $\mathbb E(X_T - d)^2$ from regime $i$ is $P_i(T)\,(x - d\,e^{-rT})^2$, where

Write $\beta_i = e^{-2rT}P_i(T) = \mathbb E_i\exp\big(-\int_0^T\lambda^2_{y_s}\,ds\big)$. The frontier is

It has the shape of the no-switching frontier of Zhou and Li (2000), with $e^{-\lambda^2 T}$ replaced by $\beta$. It is the constant-rate case of the frontier of Zhou and Yin, as restated in Yin and Zhou (2004), eq. (32): with a deterministic rate their function $H$ is the same in every regime, their $\theta$ vanishes, and $P(0,i_0)H(0,i_0)^2 = \beta_{i_0}$. The certificate also derives it from the Lagrangian of the constrained problem and checks it by Monte Carlo.

The first-order rule

All three quantities have the form $\mathbb E\exp\big(\int_0^T g_{y_s}\,ds\big)$ for a constant vector $g$. With the regime drawn from $\pi$, the integral has mean $T\bar g$. Over a horizon much longer than a regime its variance is close to

where $Q^{\#}$ is the group inverse of the generator from the solvability page and $\tilde f = f - \pi\cdot f$. For a nonreversible chain $\mathcal K(f,h)$ need not equal $\mathcal K(h,f)$. Its covariance-rate Gram form is the symmetric part

The integral is nearly Gaussian, so the exponential moment adds half the variance to the mean:

This is the rule of the general page with every operator a multiplication. Multiplications commute, so only the symmetric part of $\mathcal K$ enters, and the direction in which a chain cycles through its regimes cannot matter at first order.

A known starting regime $i$ adds the memory term of the first corrector:

For a single relaxation mode $-Q^{\#}\tilde g$ is $\tilde g$ times the relaxation time: the starting regime's excess lasts that long on average.

Observed regime. Put $g = (1-\gamma)\psi$ and divide by $(1-\gamma)T$:

The regime makes the average growth rate over the horizon random. For $\gamma > 1$ that is a risk the investor dislikes, and it lowers the certainty equivalent; for $\gamma < 1$ it raises it. For log utility ($\gamma = 1$) the certainty-equivalent rate is $\mathbb E\int\psi\,ds/T$ exactly, and the effect vanishes.

Regime-blind. Put $g = (1-\gamma)h(w)$:

Since $\tilde h(w) = w\,\tilde\mu - \tfrac12\gamma w^2\tilde s$, with $s = \sigma^2$, the correction is a polynomial in $w$ with three coefficients from the symmetric Green–Kubo Gram matrix:

Persistent drift regimes act like extra return variance $2\mathcal K_{\mu\mu}$ on top of $\bar s$. When the bear market is the volatile one, $\mathcal K^{\rm sym}_{\mu s} < 0$ and the cubic term adds to the penalty.

Markowitz. Put $g = -\lambda^2$:

To first order the frontier sees the chain through two numbers only: $\langle\lambda^2\rangle$ and $\mathcal K(\lambda^2, \lambda^2)$. Fluctuating Sharpe ratios raise $\beta$ and so flatten the frontier: each unit of excess mean costs more variance.

The weights

The observed-regime investor's fractions are Merton's, regime by regime, at every order:

The value factor $\Psi$ does not depend on wealth, and the chain is independent of $W$, so the stock cannot hedge changes in $\Psi$. There is no hedging demand, and the regime speed moves the value but not the policy. The certificate confirms this by maximizing over regime-dependent fractions numerically.

The regime-blind investor's best fraction does move. Setting the derivative of $\mathrm{CE}(w)$ to zero gives a cubic,

and linearizing around $w_0$ gives the first-order shift

For $\gamma > 1$ and a bear market that is both worse and more volatile, all three terms shrink the position.

The results written out

With two regimes the group inverse is $Q^{\#}\tilde f = -\tilde f/(q_1 + q_2)$, and every Green–Kubo number is a product of regime differences:

Substituting into the three rules gives the results in the model's own parameters.

With two regimes the three Green–Kubo numbers of the blind investor satisfy $(\mathcal K^{\rm sym}_{\mu s})^2 = \mathcal K_{\mu\mu}\mathcal K_{ss}$, and the correction is a perfect square: it is $(1-\gamma)\,\mathcal K(h, h)$ with $h_1 - h_2$ written out. With more regimes the three numbers are free, subject to the symmetric matrix they form being positive semidefinite.

Why the symmetric cross coefficient matters

For any real $a,b$, the long-run variance of the additive functional $a\mu+b s$ is $2\mathcal K(a\mu+b s,a\mu+b s)$. Therefore

The antisymmetric part of the directed Green–Kubo matrix cancels from every scalar contraction $\mathcal K(h,h)$. Consequently reversing a regime cycle can transpose $\mathcal K$ but cannot change the blind investor's first-order certainty equivalent.

This cancellation does not say that direction is invisible to every problem: it survives when distinct, noncommuting slow operators multiply the two feature directions. Under detailed balance the directed form is already symmetric, the Markov-chain version of Onsager's reciprocal relation.

The certificate makes the distinction visible with a clockwise-biased three-state chain. It finds

yet the symmetric expansion reproduces $\mathcal K(h,h)=0.01688688$ to machine precision. Substituting only $\mathcal K(\mu,s)$ gives $0.00278355$, an $83.5\%$ understatement. For $Q_m=mQ$, the exact matrix-exponential coefficient $m\{\mathrm{CE}_m-\bar h\}$ converges to the symmetric prediction $-0.03377375$; at $m=512$ it is $-0.03376714$, while the one-sided formula predicts $-0.00556710$.

Exact Gram rank and maturity aggregation

The positive-semidefinite statement has an exact identification converse. Let $Q$ be the generator of any finite irreducible chain, let $F\in\mathbb R^{n\times p}$ collect regime features, and center its columns as $H=(I-\mathbf 1\pi)F$.

Define the symmetric Green–Kubo Gram matrix by $G_{ab}=\mathcal K^{\rm sym}(H_a,H_b)$. Then

This does not require reversibility. Indeed, for any coefficient vector $c$, put $h=Hc$ and $\phi=-Q^\#h$, so that $-Q\phi=h$. Stationarity gives the Dirichlet identity

The right side vanishes only when $\phi$ is constant on the strongly connected state graph. The group-inverse normalization makes that constant zero, and hence $h=0$. Thus the Gram matrix loses precisely the feature combinations that are already constant across regimes; the antisymmetric directed sector changes neither this nullspace nor its rank.

There is an important scope distinction. If the centered loading matrix depends on maturity and a nonnegative measure $\nu$ is used to aggregate it, then, subject to integrability,

Consequently the integrated rank can exceed every fixed-maturity rank. For the two-state chain $Q=\left(\begin{smallmatrix}-1&1\\1&-1\end{smallmatrix}\right)$, contrast $z=(1,-1)^\top$, and loadings $H(\tau)=z(1,\tau)$, one has $\mathcal K(z,z)=1/2$. Every $G(\tau)$ has rank one, while uniform integration over $0\le\tau\le1$ gives

The certificate verifies the Dirichlet identity on the nonreversible three-state chain to machine precision and reproduces this integrated matrix by Gauss–Legendre quadrature. The first theorem is a fixed-generator, fixed-feature statement; it must not be used as a rank bound after combining different maturity-loading directions.

Quantitative coercivity after maturity integration

The same argument gives more than rank. On the centered real Hilbert space $L^2_0(\pi)$, let

For a measurable centered loading family $H(\tau)$ with finite second moment, define the ordinary loading Gram matrix and the integrated Green–Kubo Gram matrix by

where the adjoints use the $L^2(\pi)$ inner product in state space. Then

Indeed, the pointwise inequality $a\|H(\tau)c\|_\pi^2\le \langle H(\tau)c,\mathscr R H(\tau)c\rangle_\pi \le b\|H(\tau)c\|_\pi^2$ integrates for every coefficient vector $c$. Consequently, with eigenvalues in increasing order,

and, on the common nonzero support,

This is a quantitative persistent-excitation theorem: variation in the maturity loadings determines $M$, while the fixed Markov driver changes its scale and conditioning only through $[a,b]$. It is the Green–Kubo analogue of the Gram/observability viewpoint introduced in Kalman's original 1960 control paper. It does not say that any fixed-maturity Gram matrix has the integrated rank.

For a nonreversible, nonnormal three-state chain, the certificate obtains $[a,b]=[0.24887872,0.33445461]$. Its maturity family has pointwise rank one but integrated rank two.

The loading-Gram eigenvalues are $(0.06143693,1.57727834)$ and the integrated Green–Kubo eigenvalues are $(0.02023073,0.52555920)$; both Loewner inequalities hold with positive numerical slack. The observed condition number is $25.97827$, below the certified bound $34.50073$.

D-optimal maturity design

Rank restoration leaves a design question: which maturities should receive weight? Let the maturity set $\mathcal T$ be compact, let the pointwise symmetric Green–Kubo matrices $G(\tau)\succeq0$ vary continuously, and let $\nu$ range over probability measures on $\mathcal T$. Write

Assume at least one design has $M(\nu)\succ0$, and let $p$ be the coefficient dimension. Then a $D$-optimal design maximizing $\log\det M(\nu)$ exists. A positive-definite design $\nu_*$ is optimal if and only if its sensitivity obeys

Moreover $s_*(\tau)=p$ for $\nu_*$-almost every support point, and some optimal design is supported on at most $p(p+1)/2$ maturities. The proof is short. Moving an infinitesimal amount of mass toward $\tau$ has directional derivative $s_*(\tau)-p$; concavity of $\log\det$ makes the resulting inequalities both necessary and sufficient.

Their integral under $\nu_*$ is zero, so equality holds on its support. The supported information matrices lie in one affine hyperplane of the $p(p+1)/2$-dimensional symmetric-matrix space; Carathéodory's theorem then gives the stated finite support.

This is the Green–Kubo maturity-design specialization of Kiefer's general equivalence theory. It optimizes the integrated information matrix and does not change the fixed-maturity rank theorem.

The two-state example above is exactly solvable. Since $G(\tau)=\tfrac12(1,\tau)^\top(1,\tau)$ on $0\le\tau\le1$, equal mass at the two endpoints gives

and

with equality exactly at $0$ and $1$. This is therefore globally optimal. Uniform maturity averaging has determinant $1/48$, so the endpoint design triples determinant information.

Equivalently, $\det M(\nu)=\operatorname{Var}_{\nu}(\tau)/4$, and the endpoint law attains the sharp variance bound $1/4$ on $[0,1]$. The certificate evaluates the sensitivity on 10,001 maturities and reproduces every constant to machine precision.

Worked example

At speed $m = 1$ the regime rates are $\psi = (0.1967,\ 0.0419)$ and $\bar\psi = 0.1580$. The Merton fractions are 2.78 in the bull market and −0.30 in the bear market: leveraged long, then short. The Green–Kubo number is

so with $\gamma = 3$ the certainty-equivalent rate falls by $2\mathcal K = 7.49$ basis points, against 7.46 basis points in the numerical solution. Starting in the bull market adds $\pi_2(\psi_1 - \psi_2)/(12 \times 5) = 6.45$ basis points.

The blind investor holds $w_0 = 0.701$ in the averaged model. The regime numbers are

against $\bar s = 0.0333$. The extra effective variance $2\mathcal K_{\mu\mu}$ is 3.8% of $\bar s$.

The shift is $\delta w = −0.0440$, and the root of the cubic is 0.6609, against 0.6607 from numerical maximization. The investor's certainty-equivalent rate at $w_0$ falls from 5.45% to 5.33%.

When the regimes last four times as long ($m = 1/4$: bull markets of 16 months), the best fraction is 0.575, a cut of 18% from $w_0$, and the certainty-equivalent rate at $w_0$ is 4.98%.

For the Markowitz investor, $\langle\lambda^2\rangle = 0.7678$ and $\mathcal K(\lambda^2, \lambda^2) = 0.01348$. Over one year $\beta$ rises from 0.4640 to 0.4703 (numerical 0.4700). The standard deviation per unit of excess mean on the frontier, $\sqrt{\beta/(1-\beta)}$, rises from 0.930 to 0.942.

Checks

The observed-regime certainty-equivalent rate against the numerical solution of the linear system (matrix exponential), as the regimes speed up. The higher orders come from the engine.

The first-order error falls by about four when the speed doubles, as a second-order remainder should. At $m = 1/4$ the bull market lasts 16 months and the horizon is five years, and the first-order rule is still within 0.6 basis points.

From a known starting regime, with the memory term:

The effect for different risk aversions at $m = 1$, in basis points of certainty-equivalent rate:

For log utility the numerical effect is zero to machine precision. The certificate also runs a three-state chain with a switching interest rate (the chain $Q_A$ of the yield-curve page, rates of 5%, 3% and 1%), where the errors fall by 3.96 and 3.98 per doubling, and confirms by numerical maximization over regime-dependent fractions that Merton's weights $(2.78,\ −0.30)$ are optimal at every speed.

The regime-blind investor. Fractions, then certainty-equivalent rates at $w_0$:

The linearized shift $w_0 + \delta w$ has error falling by four per doubling. The root of the cubic is better: it matches the numerical optimum to three or four digits even when the bull market lasts 16 months. Exact Monte Carlo over 400,000 chain paths (given the path, the expectation is exactly $e^{(1-\gamma)\int h}$) gives 5.334% ± 0.003% at $w_0$ and $m = 1$, against 5.333% from the linear system.

The Markowitz frontier. $\beta$ from the stationary start and from each regime:

The error in $\log\beta$ falls by about four per doubling. Monte Carlo of the optimal feedback $w_t X_t = -\big((\mu_{y_t} - r)/\sigma_{y_t}^2\big)\big(X_t - d\,e^{-r(T-t)}\big)$, with 400,000 paths and no time discretization (given the regime path, $X_t - d\,e^{-r(T-t)}$ is a geometric Brownian motion), reproduces the frontier at $m = 1$ with $x_0 = 1$:

Learning the regime

The blind investor is a restriction, not a model of a rational investor. An investor who filters the regime from prices, with the Wonham (1964) filter, can tilt towards the inferred regime.

Over a regime that lasts $\varepsilon$, the filter learns an amount of the same order $\varepsilon$ as the corrections above, and that gain is not in the rule. Its first-order size would need a separate expansion of the filtered problem, as solved exactly by Rieder and Bäuerle (2005), Sass and Haussmann (2004) and Honda (2003).

A switching volatility is revealed instantly by quadratic variation in continuous time. So the blind investor with switching $\sigma$ describes discrete rebalancing, or the Y-independent “practical” strategy of Fouque, Sircar and Zariphopoulou, rather than an investor who cannot see volatility.

Prior art

With an observed regime at ordinary speed, these problems are solved exactly.

Zhou and Yin (2003) give the Markowitz frontier through two systems of linear differential equations. Bäuerle and Rieder (2004) solve power utility of terminal wealth with Markov-modulated rate, drift and volatility, and compare the result with a market of averaged data. Sotomayor and Cadenillas (2009) add consumption and regime-dependent utility.

The linear system above is theirs; the Merton case goes back to Merton (1969, 1971).

For fast chains, Yin, Zhang and coauthors aggregate the chain, prove convergence to a limit problem and show that controls built from the limit are nearly optimal (Yin and Zhou 2004; Zhang and Yin 2004; Yang, Yin, Wang and Zhang 2013; Yin and Zhang 2013). The abstracts we read do not give an explicit first-order correction.

Fouque, Sircar and Zariphopoulou (2017) expand Merton's problem with a fast mean-reverting diffusion factor. The leading order uses the square-averaged Sharpe ratio, as above.

Their first correction is of order $\sqrt\varepsilon$ and proportional to the correlation between the factor and the stock; the order-$\varepsilon$ term is not computed. A chain independent of the stock has no such correlation, so its first correction is the order-$\varepsilon$ Green–Kubo term.

The closed forms $\bar\psi + (1-\gamma)\mathcal K(\psi, \psi)$ for the observed investor, $\exp\big(-T\langle\lambda^2\rangle + T\mathcal K(\lambda^2, \lambda^2)\big)$ for the frontier, and the Green–Kubo penalty and weight shift for the regime-blind investor were not found in the literature reviewed. Empirical studies of regime-switching allocation, such as Tu (2010), work at estimated regime speeds and are not asymptotic in the switching rate.

Certificate: verify_portfolio.py.