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Describe the action of a lag operator, derive a time series process's lag polynomial, and judge stationarity using characteristic equations.

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

The idea

The lag operator L shifts a series back one period. Applied to y(t) it gives L × y(t) = y(t − 1), and Lᵏ × y(t) = y(t − k). Applied to a constant it returns the constant. With it, any ARMA model can be written compactly. An AR(1), y(t) = φ × y(t − 1) + ε(t), becomes (1 − φL) × y(t) = ε(t). The lag polynomial is then 1 − φL. An AR(2) gives the lag polynomial 1 − φ₁L − φ₂L². Coefficients enter with the opposite sign to the equation, which is where most slips occur. Stationarity is judged from the characteristic equation, which is the lag polynomial with L replaced by z and set to zero. The process is covariance stationary only if every root lies outside the unit circle, meaning its absolute value exceeds one. A root of exactly one is a unit root, and the series is not stationary.

A practice question

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

If L is the lag operator and y_t is a time series, L² y_t equals:

  1. (y_(t-1))²Why not: Reads the exponent as squaring the lagged value; in L² the exponent counts how many times the shift is applied.
  2. y_t - y_(t-2)Why not: Confuses the lag operator with a difference over two periods; L² only shifts the series back by two periods.
  3. y_(t-2) - correct
  4. y_(t+2)Why not: Shifts the series forward; the lag operator moves values back in time.

The lag operator shifts a series back one period, L y_t = y_(t-1). Applying it twice gives L² y_t = y_(t-2).

A harder one

For the process y_t = 0.5 y_(t-1) - 0.2 y_(t-2) + e_t, the lag polynomial applied to y_t is:

  1. 1 - 0.5 L - 0.2 L²Why not: Keeps the sign of the -0.2 coefficient; moving it to the left side turns it into +0.2.
  2. 1 + 0.5 L - 0.2 L²Why not: Does not flip the signs of the coefficients when moving the terms across the equal sign.
  3. 1 + 0.5 L + 0.2 L²Why not: Flips the sign of the -0.2 term but leaves the 0.5 term unchanged, so the first coefficient has the wrong sign.
  4. 1 - 0.5 L + 0.2 L² - correct

Move both lagged terms to the left: y_t - 0.5 L y_t + 0.2 L² y_t = e_t, which gives (1 - 0.5 L + 0.2 L²) y_t = e_t.

The hardest tier

An AR(2) is y_t = 1.1 y_(t-1) - 0.3 y_(t-2) + e_t, with characteristic equation 1 - 1.1 z + 0.3 z² = 0. Which conclusion is correct?

  1. Stationary: the roots are 1.67 and 2.0, both outside the unit circle - correct
  2. Not stationary: the coefficient 1.1 exceeds oneWhy not: Applies the AR(1) rule to a single coefficient; for AR(2) the conditions concern the roots of the polynomial, and phi1 > 1 can still be stationary.
  3. Not stationary: the roots are 0.5 and 0.6, inside the unit circleWhy not: Uses the inverse roots with the criterion for the roots of the polynomial; the polynomial roots are 2.0 and 1.67.
  4. Not stationary: roots greater than one in absolute value signal explosive behaviourWhy not: Reverses the criterion: stationarity requires roots of the characteristic polynomial to lie outside the unit circle.

1 - 1.1 z + 0.3 z² = 0 has z = (1.1 +/- 0.1) / 0.6, that is 2.0 and 1.667. Both exceed one in absolute value, so the process is stationary even though phi1 = 1.1.

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