ARMA processes, causality and invertibility
Fix polynomial signs and mean treatment before using root locations to identify causal and invertible stationary representations.
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The question for today
What you’ll understand
- Write a centered ARMA model using the stated lag-polynomial signs.
- Distinguish stationary existence, causal dependence and invertible innovation recovery.
- Calculate filter coefficients and covariance without confusing an intercept with the mean.
An autoregression expresses a current value through earlier values, but the equation alone does not guarantee that its stationary solution depends only on past shocks. Causality specifies that extra direction of dependence. Invertibility asks a different question: can shocks be recovered from present and past observations? Root conditions answer both questions once polynomial signs, common factors and the mean are fixed. These conventions are particularly important when software reports an intercept under one parameterization and a long-run mean under another.
Center before assigning polynomial coefficients
Let B shift a series backward by one time step. We write phi(B)(X_t−mu)=theta(B)epsilon_t, where phi(z)=1 minus the sum of autoregressive coefficients times powers of z and theta(z)=1 plus the sum of moving-average coefficients times powers of z. Noise has mean zero, positive finite variance and zero autocovariance at nonzero lags. Independence or Gaussianity is an additional assumption, not part of weak white noise by definition.
The constant mu is the stationary mean. An uncentered AR equation may instead include an intercept c=phi(1)mu when that quantity is nonzero. Thus in an AR(1), c=(1−phi)mu. A coefficient near one can make a modest intercept correspond to a large mean. Estimating the mean, conditioning on an initial value and assuming a stationary distribution are distinct choices that affect the likelihood and forecasts built from the model.
Causality and invertibility constrain different roots
For coprime polynomials, a causal stationary ARMA solution with absolutely summable impulse coefficients exists exactly when every zero of phi lies outside the closed unit disk. Expanding theta divided by phi then gives a present-and-past noise representation. Invertibility of that representation requires every zero of theta outside the closed unit disk, allowing the inverse expansion to recover current noise from present and past centered observations. Roots on the unit circle require separate analysis and are excluded from these stable conditions.
A stationary solution need not be causal. An AR(1) coefficient with absolute value greater than one can have a stationary solution expressed using future noise when no unit-circle root occurs. That is not the forward innovation model usually intended for forecasting. Likewise, a finite moving average is stationary for any finite coefficients, even when it is not invertible. Calling every noninvertible MA process nonstationary would confuse recoverability of shocks with existence of second moments and time-invariant covariance.
Check roots before interpreting a fitted recursion
For an AR(1), the polynomial root is one over its coefficient, so root modulus greater than one is equivalent to coefficient modulus less than one. For an MA(1) under our plus-sign convention, its root is minus one over theta, making invertibility equivalent to absolute theta less than one. The signs differ, but both conditions concern polynomial roots rather than a generic rule that coefficients must be positive.
Equivalent second-order descriptions can arise from noninvertible moving averages with adjusted noise variance. Choosing an invertible representation identifies a canonical innovation direction, but numerical root constraints do not prove that the fitted order or Gaussian distribution is correct. Residual dependence, changing parameters and omitted deterministic terms still need attention. A fitted equation should report its mean convention and initialization along with the coefficients so another analysis can reproduce its predictions.
Work it through
Expand a causal invertible ARMA(1,1)
Take phi=0.5, theta=0.25, mean zero and noise variance four. Find the first filter coefficients, stationary variance and lag-one covariance.
The AR root is two and the MA root is minus four, both outside the unit circle. Expanding (1+0.25z)/(1−0.5z) gives psi-zero one and psi-j=0.75·0.5^(j−1) for j≥1.
The sum of squared coefficients is 1+0.75²/(1−0.5²)=1.75, so stationary variance is four times 1.75, or seven.
The adjacent coefficient-product sum is 0.75 plus 0.75²·0.5/(1−0.5²)=1.125. Multiplying by four gives gamma(1)=4.5.
The model is causal and invertible with variance seven and lag-one correlation 4.5/7. Neither covariance equals the raw shock variance by default.
The living notebook
Explore the model
Interactive lab · optimization
Inspect the homogeneous recurrence separately from its shocks
Pause and reset, then set both controls to η = 0.5 and x₀ = 2. Predict values 2,1,0.5 under multiplier 0.5, one iteration per Step. For a fresh comparison, pause and reset, then set η = 1.5 and x₀ = 2; predict 2,−1,0.5 under multiplier minus 0.5. Both decay in magnitude. The instrument is gradient descent on a scalar quadratic and has no innovations; it illustrates the homogeneous AR recurrence, not a simulated ARMA sample or a stationarity test.
- Position x0
- 2
- Objective x²/2
- 2
- Iteration behavior
- Stable step
Gradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forStable homogeneous decay is a root calculation; a stochastic stationary model also needs its noise law and initialization.
Pause & reason
Check your understanding
Your turn
Try it yourself
An AR(1) is written X_t=2+0.8X_{t−1}+epsilon_t with noise variance 0.36. Find its stationary mean and variance. Separately classify the invertibility of an MA(1) with theta=2 under the plus-sign convention.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Solve the mean equation and use the centered AR variance formula. For the MA polynomial, solve 1+2z=0.
Compare with the solution
The mean is 2/(1−0.8)=10. Centering leaves a stable coefficient 0.8, so variance is 0.36/(1−0.64)=1. The intercept two is not the stationary mean. The MA polynomial with theta two has root minus one half, so that representation is not invertible. It still defines a stationary finite moving average when applied to weak white noise; the failure concerns recovery of the specified shocks from present and past observations.
Keep the idea
Take it with you
- State lag-polynomial signs, common-factor assumptions and mean treatment before checking roots.
- Causality is a past-shock representation; invertibility recovers shocks from observed history.
- Stationary existence and causal forward modeling are related but distinct questions.
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