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Describe the way forecasts are produced from ARMA models.

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

The idea

Forecasts from ARMA models are conditional expectations given information at time T. Replace each future shock by its expected value of zero, keep known past shocks (residuals) at their estimated values, and replace future y values by their own forecasts. For an ARMA(1,1) the one-step forecast is ŷ₁ = δ + φ × yT + θ × eT. At two steps the shock at T + 1 is unknown and set to zero, so the forecast is ŷ₂ = δ + φ × ŷ₁. The MA term has dropped out. An MA(q) forecast therefore equals the mean beyond q steps, while an AR forecast keeps feeding on earlier forecasts. The one-step forecast error is the unforecastable shock, so its variance is σ². Longer horizons accumulate more shocks. For an AR(1) the h-step error variance is σ² × (1 + φ² + φ⁴ + … + φ²⁽ʰ⁻¹⁾). Do not reuse actual values where forecasts belong.

Formulas

One-step forecast from an ARMA(1,1)

ŷ₁ = δ + φ × yT + θ × eT

δ
the intercept
φ
the AR coefficient
yT
the latest observation
θ
the MA coefficient
eT
the latest residual

The shock at T + 1 is set to zero. At two steps the MA term drops out and ŷ₂ = δ + φ × ŷ₁; the trap is reusing actual values where forecasts belong.

h-step forecast error variance of an AR(1)

Var(error) = σ² × (1 + φ² + φ⁴ + … + φ^(2(h − 1)))

σ²
the shock variance
φ
the AR(1) coefficient
h
the forecast horizon

Grows with h towards σ² ÷ (1 − φ²) when the process is stationary. At one step it is just σ².

A practice question

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

When producing forecasts from an ARMA model, how are shocks that occur after the forecast origin treated?

  1. They are replaced by the largest past residualWhy not: Future shocks are unknown, and the best prediction of a mean-zero shock is zero, not an extreme past value.
  2. They are replaced by their expected value of zero - correct
  3. They are replaced by the sample standard deviation of the residualsWhy not: The standard deviation measures the size of the shock, not its expected value, which is zero.
  4. They are kept at their most recent observed valueWhy not: Holding the last residual constant would imply the shock persists, but white noise shocks are unpredictable.

The forecast is the conditional expectation given information at T. Future shocks e_(T+1), e_(T+2) and so on have expected value zero, while past shocks up to T are known (as residuals).

A harder one

An MA(1) model is y_t = 2 + e_t + 0.6 e_(t-1). The most recent residual is e_T = -0.5. The forecast of y_(T+2), two steps ahead, is:

  1. 1.70Why not: Gives the one-step forecast 2 + 0.6 x (-0.5); at two steps the lagged shock e_(T+1) is unknown and set to zero.
  2. 1.82Why not: Decays the MA effect geometrically with theta² (0.36 x -0.5), as if it were an AR(1); an MA effect disappears after q steps.
  3. 2.30Why not: Uses the right shock with the wrong sign: 2 - 0.6 x (-0.5).
  4. 2.00 - correct

y_(T+2) = 2 + e_(T+2) + 0.6 e_(T+1). Both shocks lie in the future, so the forecast is the mean 2.00. For an MA(q), forecasts beyond q steps equal the mean.

The hardest tier

An ARMA(1,1) model is y_t = 0.3 + 0.6 y_(t-1) + e_t - 0.5 e_(t-1). The latest observation is y_T = 2.0 and the latest residual is e_T = 0.4. The two-step-ahead forecast of y_(T+2) is:

  1. 0.88Why not: Keeps the MA term at the second step (using -0.5 x 0.4 again); by two steps e_(T+1) is unknown, so the MA term is zero.
  2. 1.50Why not: Uses the actual y_T in the second step instead of the first-step forecast: 0.3 + 0.6 x 2.0.
  3. 1.30Why not: Reports the one-step forecast, 0.3 + 0.6 x 2.0 - 0.5 x 0.4, instead of iterating once more.
  4. 1.08 - correct

One step: 0.3 + 0.6(2.0) - 0.5(0.4) = 1.30. Two steps: 0.3 + 0.6(1.30) + 0 = 1.08, because e_(T+1) has expected value zero.

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