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ARMA Estimation and Forecasting · Tutorial 380 of 952

Prediction Intervals for ARMA Forecasts

Construct and interpret ARMA prediction intervals while keeping the conditioning information and innovation-distribution assumptions explicit.

Intermediate 12 min read

What You'll Learn

  • State the assumptions that make forecast errors conditional on observed history equal weighted future innovations.
  • Construct Gaussian prediction intervals from forecast-error variances.
  • Explain why forecast uncertainty is nondecreasing with horizon and when it approaches a finite limit.
  • Use simulation or bootstrap quantiles when innovations are nonnormal.
  • Distinguish uncertainty about future observations from uncertainty about estimated model parameters.

Why Prediction Intervals Need an Information Set

An ARMA forecast gives a point prediction for a future observation, but it does not describe how far that observation may fall from the prediction. A prediction interval adds that uncertainty. Its calculation depends on the distribution of the forecast error, and that distribution must be defined relative to the information available when the forecast is made.

For ARMA models, a useful representation expresses observations as weighted sums of innovations. The forecast-error variance is then built from the weights on future innovations. However, this calculation is valid for prediction from observed history only when the innovations used in the representation are recoverable from that history. Causality alone does not guarantee this; invertibility matters.

Definition: Let \(\mathcal{F}_T^X\) denote the information in the observed process through time \(T\). The \(h\)-step forecast is \(\hat{X}_{T+h\mid T}=E(X_{T+h}\mid\mathcal{F}_T^X)\). The forecast error is \(e_{T+h\mid T}=X_{T+h}-\hat{X}_{T+h\mid T}\). A prediction interval is an interval for \(X_{T+h}\) designed to contain that future observation with a specified predictive coverage, conditional on the stated model and information.

Throughout, suppose the process is a stationary, purely nondeterministic ARMA process with known parameters and mean \(\mu\), and has the innovation representation

$$ X_t-\mu=\sum_{j=0}^{\infty}\psi_j\varepsilon_{t-j},\qquad \psi_0=1. $$

Assume the process is causal and invertible, so that with the infinite observed past the innovations are recoverable from observed history. Also assume the innovations are independent over time, have mean zero and variance \(\sigma^2\), and are independent of the past. For Gaussian prediction intervals, add the assumption \(\varepsilon_t\sim N(0,\sigma^2)\). Known parameters are assumed here; intervals that account for parameter estimation are a separate issue.

From Forecast-Error Variance to a Gaussian Interval

Under these conditions, the conditional forecast error consists of the weighted innovations arriving after time \(T\):

$$ e_{T+h\mid T}=\sum_{j=0}^{h-1}\psi_j\varepsilon_{T+h-j}. $$

These innovations are independent of \(\mathcal{F}_T^X\). Therefore the forecast error has conditional mean zero and variance

$$ V_h=\operatorname{Var}(e_{T+h\mid T}\mid\mathcal{F}_T^X) =\sigma^2\sum_{j=0}^{h-1}\psi_j^2. $$

If the innovations are Gaussian, the forecast error is a weighted sum of independent Gaussian variables, so its conditional distribution is \(N(0,V_h)\). Writing \(z_{1-\alpha/2}\) for the \(1-\alpha/2\) standard normal quantile, the central \(100(1-\alpha)\%\) prediction interval is

$$ \left[ \hat{X}_{T+h\mid T}-z_{1-\alpha/2}\sqrt{V_h},\ \hat{X}_{T+h\mid T}+z_{1-\alpha/2}\sqrt{V_h} \right]. $$

For a 95% interval, \(z_{0.975}\) is approximately 1.96. Under the stated assumptions, this interval has 95% conditional predictive coverage: repeated future observations generated under the model and the same information at time \(T\) fall inside it with probability 0.95. It is not an interval for the unknown mean or for an ARMA parameter.

Justification of the variance formula: The forecast error is a sum of future innovations with coefficients \(\psi_0,\ldots,\psi_{h-1}\). Independence makes every covariance between distinct terms zero, and each innovation has variance \(\sigma^2\). Thus its variance is \(\sum_{j=0}^{h-1}\psi_j^2\sigma^2\). Under Gaussian innovations, this variance and the zero mean determine the conditional normal distribution.

Worked Example: A Two-Step Moving-Average Forecast

Suppose \(X_t-\mu=\varepsilon_t+0.5\varepsilon_{t-1}\), where \(\mu=10\), the innovations are independent \(N(0,4)\), and the model is invertible. Assume the filtered value of the current innovation is \(\varepsilon_T=2\). Find 95% prediction intervals one and two steps ahead.

For one step, the forecast is \(\hat{X}_{T+1\mid T}=\mu+0.5\varepsilon_T=10+0.5(2)=11\). Only the new innovation contributes to its error, so \(V_1=4(1^2)=4\) and \(SE=\sqrt{4}=2\). The interval is \(11\pm1.96(2)\), or approximately \([7.08,\ 14.92]\).

For two steps, the future innovations have conditional mean zero, so \(\hat{X}_{T+2\mid T}=10\). The error weights are 1 and 0.5, giving \(V_2=4(1^2+0.5^2)=4(1.25)=5\), and \(SE=\sqrt{5}\approx2.236\). The interval is \(10\pm1.96(2.236)\), or approximately \([5.62,\ 14.38]\).

Both intervals are conditional on the observed history and on the assumptions given. The two-step interval is wider because the forecast error includes an additional independent future innovation.

How Uncertainty Changes with the Forecast Horizon

The variance formula gives a direct comparison between successive horizons:

$$ V_{h+1}-V_h=\sigma^2\psi_h^2\geq 0. $$

Thus forecast-error variance cannot decrease as the horizon grows. It increases strictly whenever the newly added weight \(\psi_h\) is nonzero; it may stay unchanged at a particular step if that weight is zero. For a stationary causal ARMA process, the weights are square-summable, so the variance approaches the finite limit \(\sigma^2\sum_{j=0}^{\infty}\psi_j^2\). This is the process variance. As the horizon grows, the forecast generally loses information from the observed past and uncertainty approaches the unconditional variation of the process.

Worked Example: A Three-Step AR(1) Interval

Let \(X_t-\mu=0.8(X_{t-1}-\mu)+\varepsilon_t\), with \(\mu=50\), \(\varepsilon_t\sim N(0,4)\), and \(X_T=55\). The AR(1) model is stationary because \(|0.8|<1\), and its innovations are Gaussian and independent. Find a 90% prediction interval at horizon three.

The innovation weights are \(\psi_j=0.8^j\). Recursive forecasting gives \(\hat{X}_{T+3\mid T}=50+0.8^3(55-50)=50+0.512(5)=52.56\). The error variance is

$$ V_3=4(1+0.8^2+0.8^4) =4(1+0.64+0.4096) =8.1984. $$

Therefore \(SE=\sqrt{8.1984}\approx2.863\). Using \(z_{0.95}\approx1.645\), the margin is \(1.645(2.863)\approx4.710\), and the 90% interval is approximately \([47.85,\ 57.27]\), in the units of \(X_t\). It describes uncertainty about the future observation \(X_{T+3}\), not uncertainty about the estimated forecast mean.

Nonnormal Innovations: Use Predictive Quantiles

When innovations are independent but nonnormal, the weighted-sum variance formula still holds, but the forecast error need not be normal. Substituting a normal critical value then does not generally give the advertised coverage. Instead, obtain quantiles of the forecast-error distribution under the fitted innovation law. One approach simulates many independent future innovation sequences and computes the corresponding weighted sums. An innovation bootstrap does this by resampling centered fitted residuals as future innovations. These methods approximate the predictive distribution; with discrete or small samples, their realized coverage need not equal the nominal level exactly.

Worked Example: Quantiles for Skewed Innovations

Consider the invertible MA(1) model \(X_t-\mu=\varepsilon_t+0.5\varepsilon_{t-1}\), with \(\mu=20\). Suppose the innovations are independent and take the value \(-1\) with probability 0.75 and \(3\) with probability 0.25. They have mean \(0.75(-1)+0.25(3)=0\) and variance \(0.75(1)+0.25(9)=3\). Find a central 95% two-step predictive interval using the exact innovation distribution.

At horizon two the forecast is 20, and the forecast error is \(\varepsilon_{T+2}+0.5\varepsilon_{T+1}\). Its possible values and probabilities are:

Future innovationsForecast errorProbability
\((-1,-1)\)\(-1.5\)\(0.75^2=0.5625\)
\((3,-1)\)\(2.5\)\(0.25(0.75)=0.1875\)
\((-1,3)\)\(0.5\)\(0.75(0.25)=0.1875\)
\((3,3)\)\(4.5\)\(0.25^2=0.0625\)

The 2.5th percentile of this discrete error distribution is \(-1.5\), and the 97.5th percentile is \(4.5\), using the quantile definition \(q_p=\inf\{x:F(x)\geq p\}\). Adding these endpoints to the forecast gives \([20-1.5,\ 20+4.5]=[18.5,\ 24.5]\). This interval contains all four possible outcomes, so its coverage here is 100%, not exactly 95%; discreteness makes exact nominal coverage unavailable with these quantile endpoints.

Common Pitfall: Causality Is Not Enough

A tempting but incorrect argument is: “The ARMA process is causal, so the past innovations are known from past observations; therefore the error variance is \(\sigma^2\sum_{j=0}^{h-1}\psi_j^2\) for an observed-history forecast.” Causality gives a representation in terms of present and past innovations, but does not by itself make past innovations recoverable from observations. Invertibility, or another appropriate condition on the information set, is needed for that step.

For example, the causal Gaussian MA(1) process \(X_t=\varepsilon_t+2\varepsilon_{t-1}\) is not invertible in this representation. Treating \(\varepsilon_{T-1}\) as known from observed history would give a one-step error variance of \(\sigma^2\). But the same observed process has an invertible representation with coefficient 0.5 and innovation variance \(4\sigma^2\), giving observed-history one-step forecast-error variance \(4\sigma^2\). The distinction is between conditioning on past innovations and conditioning on observed history. State which information set is being used before interpreting an interval.

Key takeaway: Gaussian ARMA prediction intervals are exact under a Gaussian innovation model when the forecast-error formula is valid for the information set being used. Invertibility supports that connection for prediction from observed history; nonnormal innovations call for predictive quantiles rather than an automatic normal critical value.

Check Your Understanding

Answer each question using the stated information set and model assumptions.

  1. For a Gaussian ARMA model with forecast-error variance \(V_h\), write the central 95% prediction interval for \(\hat{X}_{T+h\mid T}\).
  2. Why is \(V_{h+1}-V_h\) nonnegative, and when can it equal zero?
  3. What additional assumption, beyond causality, is important for recovering past innovations from observed history?
  4. If innovations are independent but nonnormal, which part of the variance calculation can remain valid, and what should replace a normal critical value?
  5. In the skewed-innovation example, why does the quantile interval have coverage 100% rather than exactly 95%?
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