13.5 Regime switching: the Hamilton filter

Setup#

$s_t\in\{0,1\}$ is a hidden Markov chain with transition matrix $P_{ij} = \Pr(s_t = j\mid s_{t-1} = i)$, rows summing to 1. Given $s_t = j$, $y_t\sim N(\mu_j,\sigma_j^2)$ independently of the past (Hamilton 1989 used an autoregression; the i.i.d.-within-regime case keeps the algebra clean). Let $Y_t = (y_1,\dots,y_t)$, $f_j(y) = \phi((y-\mu_j)/\sigma_j)/\sigma_j$, and define the vectors

• predicted probabilities $\xi_{t\mid t-1}(j) = \Pr(s_t = j\mid Y_{t-1})$;
• filtered probabilities $\xi_{t\mid t}(j) = \Pr(s_t = j\mid Y_t)$;
• smoothed probabilities $\xi_{t\mid n}(j) = \Pr(s_t = j\mid Y_n)$.

Proof

(13.23) is Bayes' rule: $\Pr(s_t=j\mid Y_t) \propto \Pr(s_t=j\mid Y_{t-1})\,p(y_t\mid s_t=j, Y_{t-1})$, and $p(y_t\mid s_t=j,Y_{t-1}) = f_j(y_t)$ by conditional independence. The normaliser is $p(y_t\mid Y_{t-1})$. For (13.24), $\Pr(s_{t+1}=j\mid Y_t) = \sum_i\Pr(s_{t+1}=j\mid s_t=i, Y_t)\Pr(s_t=i\mid Y_t) = \sum_i P_{ij}\xi_{t\mid t}(i)$, using the Markov property ($s_{t+1}$ depends on $Y_t$ only through $s_t$). (13.25) is the prediction-error decomposition $p(Y_n) = \prod_t p(y_t\mid Y_{t-1})$ with the normaliser of (13.23).∎

The usual initial distribution is the ergodic one, $\pi^\top P = \pi^\top$. For two states,

and the expected duration of a stay in state $j$ is $1/(1-p_{jj})$ (the stay length is geometric with success probability $1-p_{jj}$).

Proof

$\Pr(s_t=i\mid Y_n) = \sum_j\Pr(s_t=i\mid s_{t+1}=j, Y_n)\Pr(s_{t+1}=j\mid Y_n)$. The key step: given $s_{t+1}$, the future observations $y_{t+1},\dots,y_n$ carry no further information about $s_t$, because they depend on $s_t$ only through $s_{t+1}$. So $\Pr(s_t=i\mid s_{t+1}=j, Y_n) = \Pr(s_t=i\mid s_{t+1}=j, Y_t) = \Pr(s_{t+1}=j\mid s_t=i)\Pr(s_t=i\mid Y_t)/\Pr(s_{t+1}=j\mid Y_t) = P_{ij}\,\xi_{t\mid t}(i)/\xi_{t+1\mid t}(j)$ by Bayes. Substitute. Kim (1994) gives the general version.∎

Filtered versus smoothed: a look-ahead trap#

$\xi_{t\mid t}$ is a function of $Y_t$ only: you could have computed it at the close of day $t$. $\xi_{t\mid n}$ uses all data to $n$. Smoothed probabilities are sharper and look better on a chart, so a backtest that trades on $\xi_{t\mid n}$ shows an excellent regime-timing strategy that could never have been run. The test test_filtered_probs_use_only_past_data rewrites $y_{t+1},\dots,y_n$ and asserts that $\xi_{s\mid s}$ for $s\le t$ is bit-for-bit unchanged while $\xi_{s\mid n}$ changes.

There is a second, subtler leak. Even $\xi_{t\mid t}$ is computed with parameters estimated on the full sample. If the stressed regime's volatility was estimated using March 2020, then the filter "knew" in 2019 how bad a stressed regime looks. A clean backtest re-estimates $(\mu,\sigma,P)$ on data up to $t$ (expanding or rolling windows) and only then filters.

Worked example: one filter step#

Daily Nifty-like returns in percent: calm regime $\mu_0 = 0.05$, $\sigma_0 = 0.7$; stressed regime $\mu_1 = -0.10$, $\sigma_1 = 2.0$; $p_{00} = 0.98$, $p_{11} = 0.95$ (illustrative). Expected durations: 50 days calm, 20 days stressed. Ergodic probabilities from (13.26): $\pi_0 = 0.05/0.07 = 0.714$, $\pi_1 = 0.286$.

Yesterday's filtered probabilities were $(0.9, 0.1)$. Predicted for today: $\xi_{t\mid t-1} = P^\top(0.9,0.1)^\top = (0.9\times0.98 + 0.1\times0.05,\ 0.9\times0.02 + 0.1\times0.95) = (0.887, 0.113)$.

Today the index falls 2.5%. The densities are $f_0(-2.5) = \phi(-3.643)/0.7 = 7.48\times10^{-4}$ and $f_1(-2.5) = \phi(-1.2)/2.0 = 0.0971$. Joint terms: $0.887\times7.48\times10^{-4} = 6.64\times10^{-4}$ and $0.113\times0.0971 = 0.01097$. Their sum, $p(y_t\mid Y_{t-1}) = 0.01164$, is today's likelihood contribution. Filtered: $\Pr(\text{stressed}\mid Y_t) = 0.01097/0.01164 = 0.943$. One bad day moves the probability from 11% to 94%, because a 2.5% fall is a 3.6σ event in the calm regime and a 1.2σ event in the stressed one. Had the index risen 0.3% instead, the filtered stress probability would fall to 0.045.