Forecasts at More Than One Horizon
An ARIMA forecast depends on both the information available at the forecast origin and the horizon of interest. A one-step forecast may use the latest observed value and the latest estimated innovation directly. At longer horizons, some of those inputs are no longer known: forecasts must use their own earlier forecasts, while uncertainty accumulates as future innovations enter the series.
For a risk model, this distinction matters when a decision depends on several future periods. A forecast of next month's claim count is not interchangeable with a forecast three months ahead, and an interval for the expected count is not the same as an interval for the count that will actually be observed. The calculations below treat scenarios and data as hypothetical.
Impulse Responses and Forecast-Error Variance
Consider an ARIMA model written as \(\phi(B)(1-B)^dY_t=c+\theta(B)\varepsilon_t\), where \(B Y_t=Y_{t-1}\), \(\phi(B)\) and \(\theta(B)\) are the autoregressive and moving-average polynomials, and \(d\) is the differencing order. Deterministic terms, such as a constant or drift, affect the forecast level but not the response of the series to innovations.
Assume the ARMA model for the differenced series is causal, so its autoregressive roots lie outside the unit circle. Write the coefficients \(g_j\) as those in the power-series expansion \(\theta(B)/[\phi(B)(1-B)^d]=\sum_{j=0}^{\infty}g_jB^j\). For known model parameters and observed history, the forecast error at horizon \(h\) can then be written in terms of future innovations. To use this as a conditional-expectation forecast, assume the innovations are a martingale difference sequence relative to the information filtration: \(E[\varepsilon_t\mid\mathcal F_{t-1}]=0\). Also assume \(E[\varepsilon_t^2\mid\mathcal F_{t-1}]=\sigma^2\). These assumptions imply the needed conditional means and variances; uncorrelatedness alone would not imply the conditional-mean statement.
The variance formula follows because the forecast error is a weighted sum of future martingale-difference innovations. For two different future times \(s<t\), the martingale-difference property gives \(E[\varepsilon_s\varepsilon_t\mid\mathcal F_n]=0\): condition first on \(\mathcal F_{t-1}\), which contains \(\varepsilon_s\), and use \(E[\varepsilon_t\mid\mathcal F_{t-1}]=0\). Each future innovation has conditional variance \(\sigma^2\), so only the squared-weight terms remain in the variance.
Condition on the observed history and set the conditional means of future innovations to zero. Include known past innovations where the model calls for them.
Use the coefficients \(g_j\) for the original series. Differencing changes these weights: for an integrated model, they are not generally the coefficients for the differenced series alone.
Add \(\sigma^2g_j^2\) for \(j=0,\ldots,h-1\). Do not include weights beyond the requested horizon.
If future innovations are independent Gaussian variables with variance \(\sigma^2\), the forecast error is conditionally Gaussian, and a normal prediction interval follows. With only the martingale-difference and variance assumptions, the variance formula holds, but an exact normal interval does not follow.
For a stationary ARMA model (\(d=0\)), the weights are the coefficients of \(\theta(B)/\phi(B)\). For an integrated model, the factor \((1-B)^{-d}\) accumulates the effects of shocks on the original scale. For example, in an ARIMA(0,1,1) model with \(\Delta Y_t=\varepsilon_t+\theta\varepsilon_{t-1}\), the level-response weights are \(g_0=1\) and \(g_j=1+\theta\) for \(j\geq1\). Thus later horizons include multiple future shocks with those weights.
Worked Forecasts and Intervals
Worked Example: A Stationary AR(1) at Three Horizons
A hypothetical monthly severity index follows the known-parameter model \(Y_t-10=0.5(Y_{t-1}-10)+\varepsilon_t\). The innovations are independent \(N(0,4)\), and the latest value is \(Y_n=14\). Find the forecasts and 95% prediction intervals at horizons one, two, and three. Round interval endpoints to two decimal places.
Method. This is a stationary AR(1), so the response weights are \(g_j=0.5^j\). The forecast is \(10+0.5^h(14-10)\). The variance is \(4\sum_{j=0}^{h-1}(0.5^j)^2\). Independent Gaussian innovations make each forecast error conditionally Gaussian, so use \(\hat Y_{n+h|n}\pm1.96\sqrt{\operatorname{Var}(e_{n+h|n}\mid\mathcal F_n)}\).
At horizon 1, the forecast is \(10+0.5(4)=12\). The variance is \(4(1)=4\), and the standard error is \(\sqrt4=2\). The interval is \(12\pm1.96(2)=12\pm3.92\), or \([8.08,15.92]\).
At horizon 2, the forecast is \(10+0.5^2(4)=11\). The variance is \(4(1+0.5^2)=4(1.25)=5\), and the standard error is \(\sqrt5=2.2361\). The margin is \(1.96(2.2361)=4.38\) to two decimals, giving \([6.62,15.38]\).
At horizon 3, the forecast is \(10+0.5^3(4)=10.5\). The variance is \(4(1+0.25+0.0625)=4(1.3125)=5.25\), and the standard error is \(\sqrt{5.25}=2.2913\). The margin is \(1.96(2.2913)=4.49\), giving \([6.01,14.99]\).
Check. Each variance includes one additional squared response weight at the next horizon: \(4\), then \(5\), then \(5.25\). The forecasts move toward the stationary mean of 10, while the prediction intervals widen.
Worked Example: An Integrated MA Model with a Known Past Innovation
A hypothetical monthly index follows the known-parameter ARIMA(0,1,1) model \(\Delta Y_t=\varepsilon_t+0.5\varepsilon_{t-1}\), with independent \(\varepsilon_t\sim N(0,4)\). At time \(n\), \(Y_n=100\) and the latest innovation is known to be \(\varepsilon_n=2\). Find the forecasts and 95% prediction intervals for horizons one through three. Round interval endpoints to two decimals.
Method. At horizon one, \(E[\Delta Y_{n+1}\mid\mathcal F_n]=0.5\varepsilon_n=1\), so the forecast is 101. At later horizons, the expected future increments are zero, so the level forecast remains 101. The level-response weights are \(g_0=1\) and \(g_j=1.5\) for \(j\geq1\). Apply the forecast-error variance formula and, because the innovations are independent Gaussian variables, use the normal 1.96 multiplier.
At horizon 1, the variance is \(4(1^2)=4\), so the standard error is 2. The interval is \(101\pm1.96(2)=101\pm3.92\), or \([97.08,104.92]\).
At horizon 2, the variance is \(4(1^2+1.5^2)=4(3.25)=13\). The standard error is \(\sqrt{13}=3.6056\); the margin is \(1.96(3.6056)=7.07\). The interval is \([93.93,108.07]\).
At horizon 3, the variance is \(4(1^2+1.5^2+1.5^2)=4(5.5)=22\). The standard error is \(\sqrt{22}=4.6904\); the margin is \(1.96(4.6904)=9.19\). The interval is \([91.81,110.19]\).
Check. The expected level forecast remains 101, but its forecast-error variance grows from 4 to 13 to 22. The known past innovation affects the forecast level; it is not an additional future innovation in the variance sum.
Prediction Intervals Versus Uncertainty in the Fitted Mean
The variance above describes future-observation uncertainty when model parameters are treated as known. In practice, parameters are estimated, so the fitted conditional mean is itself uncertain. An interval for that mean accounts for uncertainty in the estimated parameters; it does not, by itself, include the future innovation that determines how the next observation departs from its conditional mean.
A first-order approximation for uncertainty in an estimated forecast mean can be obtained by the delta method. If the forecast is \(m(\hat\beta)\), where \(\hat\beta\) is the estimated parameter vector, its approximate variance is \(\nabla m(\hat\beta)^\mathsf{T}\widehat{\operatorname{Cov}}(\hat\beta)\nabla m(\hat\beta)\). This is different from the forecast-error variance generated by future innovations. A full predictive interval may include both sources. Adding their variances is justified when the components are uncorrelated; independence is a sufficient condition, not something to assume automatically.
Worked Example: Separating Mean and Observation Uncertainty
A hypothetical fitted ARIMA model gives a one-step conditional-mean forecast of 50. A parameter-covariance calculation gives an estimated standard error of 1.5 for that forecast mean. The next innovation is independent of the parameter estimation error and is \(N(0,9)\). Assume the estimated forecast mean is approximately normal. Find a 95% interval for the fitted mean and an approximate 95% prediction interval for the future observation. Round endpoints to two decimals.
Mean interval. The estimated mean has standard error 1.5, so its interval is \(50\pm1.96(1.5)=50\pm2.94\), or \([47.06,52.94]\). This interval describes uncertainty in the conditional mean, not the future observation.
Prediction interval. The variance of the estimated mean is \(1.5^2=2.25\), and the future innovation variance is 9. By the stated independence assumption, the approximate predictive variance is \(2.25+9=11.25\). Its standard error is \(\sqrt{11.25}=3.3541\), and the 95% margin is \(1.96(3.3541)=6.57\). The approximate prediction interval is \([43.43,56.57]\).
Check. The prediction interval is wider because it includes the variability of a future observation around its conditional mean as well as uncertainty in estimating that mean. The calculation relies on the stated approximate normality and independence assumptions.
Common Exam Traps
- Using only the one-step variance at every horizon. In the AR(1) example, the incorrect calculation \(4\) at horizon 3 ignores the additional response weights. The correct variance is \(4(1+0.25+0.0625)=5.25\).
- Confusing standard deviation and variance. The variance is \(\sigma^2\sum g_j^2\); the standard error used in an interval is its square root. Multiplying a normal critical value by the variance gives the wrong scale.
- Using weights for the differenced series when forecasting levels. Integration accumulates shocks. For the ARIMA(0,1,1) example, level weights after the first are 1.5, not the differenced-series MA weight 0.5.
- Calling an interval exact without a distributional assumption. Martingale-difference innovations with constant conditional variance support the variance calculation, but they do not guarantee Gaussian forecast errors. Exact Gaussian intervals require an appropriate joint conditional Gaussian assumption, such as independent Gaussian innovations.
- Treating the forecast mean as the future value. The conditional mean is a point forecast. A prediction interval must also account for future innovations; a confidence interval for the fitted mean answers a different question.
Check Your Understanding
For each problem, identify the horizon and the source of uncertainty before calculating.
- A hypothetical stationary AR(1) model is \(Y_t-20=0.4(Y_{t-1}-20)+\varepsilon_t\), with known parameters, \(\varepsilon_t\sim N(0,1)\) independently, and \(Y_n=25\). Find the two-step forecast and its forecast-error variance. (Answer: forecast \(20.80\); variance \(1.16\).)
- A hypothetical ARIMA(0,1,1) model has \(\Delta Y_t=\varepsilon_t-0.25\varepsilon_{t-1}\), independent \(\varepsilon_t\sim N(0,9)\), \(Y_n=80\), and known \(\varepsilon_n=4\). Find the two-step forecast and forecast-error variance. (Answer: forecast \(79\); variance \(14.0625\).)
- A hypothetical AR(1) model has known mean 5, coefficient 0.5, innovation variance 4, and latest value \(Y_n=13\). Find the three-step forecast and 95% prediction interval, assuming independent Gaussian innovations. (Answer: forecast \(6\); variance \(5.25\); interval \([1.51,10.49]\), to two decimals.)
- A fitted model gives a forecast mean of 40 with estimated standard error 2. The future innovation is independent Gaussian with standard deviation 5. Give the approximate 95% interval for the fitted mean and the approximate 95% prediction interval, assuming normality. (Answer: mean interval \([36.08,43.92]\); prediction interval \([29.45,50.55]\), to two decimals.)
- A hypothetical ARIMA(0,2,0) model has \(\Delta^2Y_t=\varepsilon_t\), known parameters, and independent \(\varepsilon_t\sim N(0,2)\). What is the three-step forecast-error variance? (Answer: \(28\), since the level-response weights are \(1,2,3\) and \(2(1^2+2^2+3^2)=28\).)