Charta is on Google Play: practise on the go. Get the app
Charta

Define moving average (MA) processes, describe their properties and compute the series for an MA process from given parameters.

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

The idea

A moving average process is built from current and past shocks. The MA(1) is y(t) = μ + ε(t) + θ × ε(t − 1). MA(q) uses q lagged shocks. Because it is a finite sum of white noise it is always covariance stationary, whatever the θ values. For MA(1), the variance of the series comes from the current shock and the lagged shock: Variance = σ² × (1 + θ²). The lag-1 autocovariance is the covariance between a value and the one before it: Covariance = θ × σ². The lag-1 autocorrelation, ACF, is that covariance divided by the variance: ACF = θ ÷ (1 + θ²). The lag-1 autocorrelation can never exceed 0.5 in absolute value. The autocorrelation function is exactly zero beyond lag q, the signature that identifies an MA, while the partial autocorrelation decays. Invertibility, which for MA(1) means the absolute value of θ is below one, is a separate condition. Both θ and 1 ÷ θ give the same autocorrelations. To compute the series, add the mean, the current shock and each lagged shock times its θ.

Formulas

Variance of an MA(1)

γ(0) = σ² × (1 + θ²)

σ²
the variance of the white noise shock
θ
the MA(1) coefficient

Any θ, since an MA is always stationary. The lag-1 autocovariance is θ × σ² and all higher ones are zero.

Lag-1 autocorrelation of an MA(1)

ρ(1) = θ ÷ (1 + θ²)

θ
the MA(1) coefficient

The shock variance cancels. θ and 1 ÷ θ give the same value, so pick the invertible root with |θ| below one.

A practice question

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

The autocorrelation function of an MA(2) process is:

  1. a geometric decay that never reaches zeroWhy not: Describes an AR process; an MA(q) has an autocorrelation that cuts off after lag q.
  2. nonzero at lag 1 only and zero from lag 2 onwardWhy not: Applies the MA(1) cut-off; an MA(2) has two lagged shocks and so can be correlated up to lag 2.
  3. nonzero at lags 1 and 2 and exactly zero beyond lag 2 - correct
  4. zero at every nonzero lagWhy not: Describes white noise; the lagged shocks in an MA(2) create correlation between nearby observations.

An MA(q) value shares at most q shocks with its neighbours, so correlation exists only up to lag q. For q = 2 the autocorrelation vanishes from lag 3 onward, while the partial autocorrelation decays.

A harder one

An MA(1) process is y_t = 1 + e_t + 0.5 e_(t-1). The shocks are e_0 = 0.4, e_1 = -0.2, e_2 = 0.6 and e_3 = 0.1. The value of y_3 is:

  1. 1.00Why not: Applies the 0.5 weight to e_1 (two periods back) instead of e_2, a misaligned lag.
  2. 1.65Why not: Applies theta to the current shock and gives the lagged shock a weight of one, reversing their roles.
  3. 0.40Why not: Omits the mean of 1 from the process, leaving only e_3 + 0.5 e_2.
  4. 1.40 - correct

y_3 = 1 + e_3 + 0.5 e_2 = 1 + 0.1 + 0.5 x 0.6 = 1.40. Only the current and one lagged shock enter an MA(1).

The hardest tier

The lag-1 sample autocorrelation of a series is 0.40, and an analyst fits an MA(1) to it. Solving theta / (1 + theta²) = 0.40 gives two solutions. Which value of theta should be used for forecasting, and why?

  1. 0.40, because the autocorrelation equals the MA coefficientWhy not: Treats rho(1) as theta; the relationship is rho(1) = theta / (1 + theta²).
  2. 2.00, because the larger solution gives the better fitWhy not: Both solutions fit identically; the larger one is not invertible, so it cannot be expressed in terms of past observations.
  3. 0.67, from solving theta / (1 + theta) = 0.40Why not: Drops the square in the denominator, so it solves the wrong equation.
  4. 0.50, because it is the solution with absolute value below one (the invertible one) - correct

0.4 theta² - theta + 0.4 = 0 has roots 0.5 and 2.0 (reciprocals). Both give the same autocorrelations, but only theta = 0.5 has |theta| < 1 and is invertible, the convention for estimation and forecasting.

In the app: 6 practice questions on this outcome, spaced review that brings it back before you forget it.

Start free →

More in Stationary Time Series