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Outline what part mean reversion plays in forecasts over long horizons.

FRM® Part I · Quantitative Analysis · Stationary Time Series · FRM1.QA.10.m

The idea

In a stationary model, forecasts pull toward the unconditional mean as the horizon grows. For an AR(1) the h-step forecast is ŷh = μ + φʰ × (yT − μ), so the gap to the mean shrinks geometrically. The more persistent the series, the closer φ is to one and the slower the convergence. For large h the point forecast is essentially μ, which adds no information beyond the long-run average. At the same time the forecast error variance rises toward the unconditional variance of the series. Unconditional variance = σ² ÷ (1 − φ²). It stops growing, unlike a random walk, whose error variance grows without limit. Practical consequences: long-horizon forecasts from a stationary model are close to the mean with intervals as wide as the data's own spread. A frequent error is to scale the one-step standard deviation by the square root of the horizon.

Formulas

h-step forecast of a stationary AR(1)

ŷh = μ + φ^(h) × (yT − μ)

μ
the unconditional mean, δ ÷ (1 − φ)
φ
the AR(1) coefficient
h
the horizon
yT
the latest observation

Stationary AR(1). The gap closes geometrically; the number of periods to cut it to a fraction f is ln(f) ÷ ln(φ).

A practice question

From the question bank, with the reason each wrong option is wrong.

As the forecast horizon increases, the point forecast from a covariance-stationary ARMA model converges to:

  1. the last observed valueWhy not: Persistence in the last value fades; the forecast does not stay at the last value unless the series has a unit root.
  2. the unconditional mean of the series - correct
  3. zeroWhy not: Converges to the unconditional mean, which is zero only if the intercept is zero or the series is mean-adjusted.
  4. the unconditional variance of the seriesWhy not: The variance is a measure of spread, not a level; the point forecast converges to the mean.

For a stationary model, the effect of the current state decays at a geometric rate, so long-horizon forecasts approach the unconditional mean delta / (1 - sum of the phi).

A harder one

A series follows an AR(1) with phi = 0.8. After how many periods is the expected gap to the long-run mean (the forecast deviation) reduced to 10 percent of its present size?

  1. 5.0 periodsWhy not: Uses 1 / (1 - phi), the persistence horizon of the series, rather than solving 0.8^h = 0.10.
  2. 1.4 periodsWhy not: Solves ln(0.10) / ln(0.20), using 1 - phi (0.2) instead of phi as the base of the decay.
  3. 10.3 periods - correct
  4. 2.3 periodsWhy not: Computes only -ln(0.10) = 2.30 without dividing by ln(0.8).

The gap after h periods is phi^h = 0.8^h. Setting 0.8^h = 0.10 gives h = ln(0.10) / ln(0.8) = 10.32.

The hardest tier

A stationary AR(1) with phi = 0.9, mean zero and shock variance 1 is used to forecast five steps ahead. The 95% forecast interval half-width (1.96 times the forecast error standard deviation) is closest to:

  1. 4.50Why not: Uses the unconditional standard deviation, which overstates the five-step error because the forecast error variance has not yet reached its limit.
  2. 4.38Why not: Scales the one-step standard deviation by the square root of 5, a random-walk rule that ignores the decaying weights phi^j.
  3. 1.96Why not: Uses the one-step forecast error standard deviation of 1, ignoring accumulated shocks.
  4. 3.63 - correct

Error variance = (1 - 0.9^10) / (1 - 0.81) = 3.428, standard deviation 1.851, half-width 3.63. The unconditional limit would be 5.263.

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