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Define autoregressive (AR) processes, describe their properties and compute the series for an AR process from given parameters.

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

The idea

An autoregressive process regresses a series on its own past. The AR(1) is y(t) = δ + φ × y(t − 1) + ε(t), where ε(t) is white noise with variance σ². AR(p) adds lags up to p. It is covariance stationary when the roots of the characteristic equation lie outside the unit circle, which for AR(1) means the absolute value of φ is below one. In that case the mean of the series depends on the intercept δ and on the coefficient φ: Mean = δ ÷ (1 − φ). The variance depends on the shock variance and on the coefficient φ in the same way: Variance = σ² ÷ (1 − φ²). The autocorrelation function, ACF, at lag k is a geometric decay, a power of the coefficient φ: ACF = φ^k. It alternates in sign if φ is negative. The partial autocorrelation cuts off after lag p. To compute the series, start from a given value and apply the recursion period by period, adding the intercept and each shock. The usual error is forgetting the intercept.

Formulas

Mean of a stationary AR(1)

μ = δ ÷ (1 − φ)

δ
the intercept
φ
the AR(1) coefficient, with |φ| below one

Unconditional mean when the process is stationary. The intercept δ is the mean only when φ is zero.

Variance of a stationary AR(1)

γ(0) = σ² ÷ (1 − φ²)

σ²
the variance of the white noise shock
φ
the AR(1) coefficient

Unconditional variance, valid only if |φ| is below one. Conditional on the latest value the one-step variance is just σ².

Autocorrelation of an AR(1) at lag k

ρ(k) = φ^(k)

φ
the AR(1) coefficient
k
the lag

Geometric decay, alternating in sign when φ is negative. The autocovariance at lag k is φ^(k) times the variance.

A practice question

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

Which description fits an AR(1) process y_t = delta + phi y_(t-1) + e_t?

  1. A constant plus the current shock plus a fraction of last period's shockWhy not: Describes an MA(1), which depends on lagged shocks rather than lagged values of the series.
  2. A constant plus phi times the previous value plus a new shock - correct
  3. A constant plus the previous value with a coefficient fixed at one plus a shockWhy not: Describes a random walk with drift; an AR(1) has a coefficient phi that is estimated and below one in absolute value.
  4. An equal-weighted average of the last k values of the seriesWhy not: Describes a simple moving-average smoother of past data, not an autoregression with a shock term.

An autoregression expresses y_t as a linear function of its own lagged values plus a white noise shock. For AR(1) there is one lag with coefficient phi, and stationarity needs |phi| < 1.

A harder one

A stationary AR(1) has phi = 0.7 and shock variance 1.5. The autocovariance at lag 3 is closest to:

  1. 0.34Why not: Reports the autocorrelation phi³ instead of the autocovariance; it still needs multiplying by the variance.
  2. 0.51Why not: Multiplies phi³ by the shock variance 1.5 instead of by the series variance 1.5 / (1 - 0.49).
  3. 2.06Why not: Uses phi to the first power (the lag-1 autocovariance), not phi³.
  4. 1.01 - correct

Var(y) = 1.5 / (1 - 0.49) = 2.941. gamma(3) = phi³ x Var(y) = 0.343 x 2.941 = 1.009.

The hardest tier

A stationary AR(2) process is y_t = 0.5 y_(t-1) + 0.3 y_(t-2) + e_t. Using the Yule-Walker relations rho(1) = phi1 / (1 - phi2) and rho(2) = phi1 rho(1) + phi2, the lag-2 autocorrelation is closest to:

  1. 0.66 - correct
  2. 0.36Why not: Computes phi1 x rho(1) only, omitting the phi2 term from rho(2) = phi1 rho(1) + phi2.
  3. 0.51Why not: Squares rho(1), the AR(1) shortcut rho(2) = rho(1)², which does not hold for an AR(2).
  4. 0.55Why not: Uses phi1² plus phi2, assuming rho(1) = phi1; but rho(1) = phi1 / (1 - phi2) = 0.714.

rho(1) = 0.5 / 0.7 = 0.7143. rho(2) = 0.5 x 0.7143 + 0.3 = 0.6571.

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