Every ergodic additive functional hides a martingale. Eg:

$$ Y_{t + 1} - Y_t = \kappa (X_t, W_{t +1}). $$

where $X_t$ is stationary & ergodic Markov State, can be decomposed into

$$ Y_t = \left(Y_0 + g(X_0)\right) + t\nu + \sum_{j = 0}^{t - 1}\kappa_m(X_{j}, W_{j + 1}) - g(X_t). $$
So later we can do CLT on the time series: $1/\sqrt{N}(Y_N - N\nu) \xrightarrow{d} N(0, E[\kappa_m \kappa_m'])$.

  • Let $\kappa_2(X_t, W_{t + 1})= \kappa(X_t, W_{t + 1}) - E[\kappa(X_t, W_{t + 1})\mid X_t]$

    $\kappa_2$ is the direct shock hit into $\Delta Y_t$

  • Let $f(X_t) = E[\kappa(X_t, W_{t + 1}) \mid X_t] - E[\kappa(X_t, W_{t + 1})]$

    Let $g(X_t) = \sum_{j = 0}^\infty E[f(X_{t + j})\mid X_t]$

    Let $\kappa_1(X_t, W_{t + 1}) = g(X_{t + 1}) - (g(X_t) - f(X_t))$

    $\kappa_1$ is the indirect shock from $X_{t}$ that hit forever

Finally, obtain the martingale component by

$$ \kappa_m(X_t, W_{t + 1}) = \kappa_1 (X_t, W_{t + 1})+ \kappa_2(X_t, W_{t + 1}). $$

Here’s a simulation to see the decomposition:

  • $Y_{t+1} − Y_t = ν + d·X_t + F·W_{t+1}$
  • $X_{t+1} = a·X_t + W_{t+1}$

We extracts a martingale in four steps.

Drag the parameters, resample the shock path, and watch each step of the construction update on the same draw.

Full martingale decomposition pipeline: parameter controls, four extraction step charts on one shock path, and the resulting four-component decomposition of Y with a central limit theorem check.

Step 1 — demean: κ₂ = κ − E[κ|Xₜ]

Split each increment into what you knew at t (amber) and pure surprise (blue). κ₂ is already a martingale difference. The amber wiggle is the remaining problem.

κ = Yₜ₊₁−Yₜ E[κ|Xₜ] = ν+dXₜ κ₂ = FWₜ₊₁
Step 1 chart.

Steps 2–3 — f(Xₜ) = E[κ|Xₜ]−ν and g(Xₜ) = Σⱼ E[f(Xₜ₊ⱼ)|Xₜ]

f = predictable mean-zero part. g = its total future contribution: same shape, amplified 1/(1−a) = 10.0×. Push a up and watch g explode while f stays put.

f(Xₜ) = dXₜ g(Xₜ) = d/(1−a)·Xₜ
Steps 2 and 3 chart.

Step 3b — κ₁ = g(Xₜ₊₁) − g(Xₜ) + f(Xₜ)

The trick: adding the forecast revision converts persistent amber into iid teal. Same information, repackaged as surprise. Autocorrelation drops from ≈a to ≈0 (see cards below).

f(Xₜ) persistent κ₁ = (d/(1−a))Wₜ₊₁ iid
Step 3b chart.

Step 4 — κₘ = κ₁ + κ₂: indirect + direct channel of one shock

One iid increment carrying the full permanent effect of Wₜ₊₁: through the state (κ₁) and directly (κ₂). Identity: κ = ν + κₘ − [g(Xₜ₊₁)−g(Xₜ)].

κ₁ κ₂ κₘ = κ₁+κ₂
Step 4 chart.

Outcome — Yₜ = (Y₀+g(X₀)) + tν + Σκₘ − g(Xₜ)

Sum the identity over t. The four lines add up to Yₜ exactly at every date. Only the martingale grows in variance — it alone survives 1/√N scaling in the CLT.

Yₜ trend tν martingale Σκₘ −g(Xₜ) stationary constant Y₀+g(X₀)
Outcome chart.
κₘ coefficient d/(1−a)+F
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CLT variance E[κₘ²]
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Autocorr: f(Xₜ) vs κₘ
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Max |parts − Yₜ|
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