Prediction Intervals For Forecasts

Why a Forecast Needs an Interval

A point forecast gives one value for a future observation, but the observation will generally differ from that value. For planning, pricing, and risk monitoring, a single number can conceal how uncertain the forecast is. A prediction interval makes that uncertainty visible by giving a range for a future observation, with a stated coverage level under specified assumptions.

The distinction between a future observation and its mean forecast matters. A forecast may estimate the conditional mean accurately while individual future observations still vary substantially around that mean. Consequently, an interval for the mean forecast is usually narrower than a prediction interval for a future observation. The interval's meaning depends on what is being covered, the forecast horizon, and the assumptions used to quantify forecast error.

Definition: Let \(Y_{T+h}\) be the observation at time \(T+h\), and let \(\hat{Y}_{T+h|T}\) be its forecast made using information available through time \(T\). A prediction interval for \(Y_{T+h}\) is an interval intended to contain that future observation with a stated probability, conditional on the available information and the assumed forecasting model. Its width reflects uncertainty about the future observation, not merely uncertainty about the estimated mean forecast.

Normal Prediction Intervals

A common construction assumes that the forecast error at horizon \(h\), \(e_{T+h}=Y_{T+h}-\hat{Y}_{T+h|T}\), is conditionally normal with mean zero and standard deviation \(\sigma_h\). If these quantities are known, a two-sided prediction interval with nominal coverage \(1-\alpha\) is

\hat{Y}_{T+h|T}\ \pm\ z_{1-\alpha/2}\sigma_h

Here \(z_{1-\alpha/2}\) is the corresponding standard normal quantile. For example, \(z_{0.975}=1.96\) gives a 95% interval, while \(z_{0.95}\approx1.645\) gives a 90% interval. The \(\sigma_h\) in this formula is the standard deviation of the future-observation forecast error at that horizon. It is not automatically the standard error of an estimated average, nor is it necessarily the residual standard deviation from a one-step fit.

The model may provide \(\sigma_h\) directly or require it to be estimated. If it is estimated, the normal interval is generally an approximation unless the model gives a suitable exact result. The formula also depends on a correct error distribution and a correctly calculated forecast-error variance. If errors are skewed, heavy-tailed, dependent in a way the calculation ignores, or affected by changing variance, a symmetric normal interval may give poor coverage.

Justification: Under the stated conditional normal model, \(e_{T+h}/\sigma_h\) has a standard normal distribution. Therefore,

P\left(-z_{1-\alpha/2}\leq \frac{Y_{T+h}-\hat{Y}_{T+h|T}}{\sigma_h} \leq z_{1-\alpha/2}\ \middle|\ \mathcal{F}_T\right)=1-\alpha,

where \(\mathcal{F}_T\) denotes the information available through time \(T\). Multiplying all parts of the inequality by the positive number \(\sigma_h\), then adding \(\hat{Y}_{T+h|T}\), gives

P\left(\hat{Y}_{T+h|T}-z_{1-\alpha/2}\sigma_h \leq Y_{T+h}\leq \hat{Y}_{T+h|T}+z_{1-\alpha/2}\sigma_h \ \middle|\ \mathcal{F}_T\right)=1-\alpha.

This derivation shows why the interval is centered on the point forecast under a zero-mean normal forecast-error assumption. It also shows that its margin is the appropriate normal quantile multiplied by the forecast-error standard deviation at the requested horizon.

Prediction Interval or Mean-Forecast Interval?

Suppose the model's conditional mean at time \(T+h\) is \(m_{T+h|T}\). An interval for that mean concerns uncertainty in estimating \(m_{T+h|T}\); a prediction interval concerns the realized \(Y_{T+h}\). Even if the mean were known exactly, the future observation could differ from it because of new random variation. For that reason, a future-observation prediction interval must include that variation.

In a setting where the estimated mean forecast has variance \(V_{\text{mean}}\), the new observation has conditional variance \(V_{\text{new}}\) around the true mean, and these two sources are independent, the total prediction-error variance is

V_{\text{prediction}}=V_{\text{new}}+V_{\text{mean}}.

The independence condition matters: in general, the variance of a difference includes a covariance term. The model's correct forecast-error variance should be used rather than adding components mechanically. This variance decomposition explains why an interval for a future value is wider than one that accounts only for uncertainty about the mean.

Interval	What it is intended to cover	Variation included
Interval for a mean forecast	The conditional mean at the future time	Uncertainty in estimating that mean
Prediction interval	A future observation	New observation variation and, when applicable, uncertainty in the estimated mean

Worked Example: A 95% Interval for a Future Index Value

A hypothetical monitoring model forecasts a weekly index value three weeks ahead as 52. The model specifies that the three-week forecast error is conditionally normal with mean zero and standard deviation 9.1924 index points. Treat the forecast and standard deviation as known for this calculation. Calculate a 95% prediction interval and state what it covers.

Method. Use the normal prediction-interval formula with \(z_{0.975}=1.96\). The interval's margin is

1.96(9.1924)=18.017104\approx18.017.

Thus the unrounded endpoints are \(52-18.017104=33.982896\) and \(52+18.017104=70.017104\). To two decimal places, the 95% prediction interval is \([33.98,70.02]\) index points.

Interpretation. Under the stated conditional normal model, the interval-producing procedure has 95% coverage for the index value observed three weeks ahead. It does not say there is a 95% probability that the fixed forecast mean lies in this particular interval.

Common mistake avoided. Rounding the margin incorrectly to 18.016 would give inaccurate intermediate endpoints. The multiplication is 18.017104, which rounds to 18.017; the requested endpoints rounded to two decimal places are 33.98 and 70.02.

Worked Example: Using the Standard Deviation for the Requested Horizon

A hypothetical service receives a forecast of 240 requests for a date four days ahead. Its forecasting model gives a conditionally normal forecast error with mean zero and standard deviation 12 requests at that four-day horizon. Find a 90% prediction interval. A one-day error standard deviation is not supplied, so do not replace the given horizon-specific value.

Method. For a 90% two-sided interval, use \(z_{0.95}\approx1.645\). The margin is

$$ 1.645(12)=19.74. $$

The lower endpoint is \(240-19.74=220.26\), and the upper endpoint is \(240+19.74=259.74\). The 90% prediction interval is \([220.26,259.74]\) requests.

Check. The interval is centered at 240 because the assumed forecast error has mean zero. Its total width is \(259.74-220.26=39.48\), which equals twice the margin, \(2(19.74)=39.48\).

Interpretation and trap. The interval concerns the request count for the specified future date, not the average request level over many dates. A common mistake is to multiply the supplied standard deviation by four because the forecast is four days ahead. The value 12 is already the standard deviation for that horizon; multiplying it again would count the horizon twice.

Accounting for Uncertainty in the Mean Forecast

Sometimes the forecast mean itself has estimation uncertainty that is not included in the stated standard deviation of new observations. If the two error components are independent, their variances add. In the next example, the problem explicitly gives both components and their independence, so the combined standard deviation can be calculated before applying a normal prediction interval.

Worked Example: Combining Two Independent Sources of Uncertainty

A hypothetical monthly enrollment forecast is 500 people. Conditional on the true mean for the target month, new-observation variation has standard deviation 20 people. Estimation uncertainty in the mean forecast has standard deviation 6 people. Assume the two components are independent and normal, and that their errors have mean zero. Calculate a 90% prediction interval for enrollment.

Method. The two variances are \(20^2=400\) and \(6^2=36\). By independence, the total prediction-error variance is \(400+36=436\), so the prediction-error standard deviation is

\sqrt{436}=20.8806\text{ people (approximately)}.

For 90% coverage, use \(z_{0.95}=1.645\). The margin is

1.645\sqrt{436}\approx1.645(20.8806)=34.3486.

The endpoints are \(500-34.3486=465.6514\) and \(500+34.3486=534.3486\). Rounded to two decimal places, the interval is \([465.65,534.35]\) people.

Check. The combined standard deviation, about 20.88, exceeds 20 because the mean-estimation variance adds to the new-observation variance. The calculated margin is consequently greater than the margin \(1.645(20)=32.9\) that would omit mean-estimation uncertainty.

Common mistake avoided. Adding standard deviations to obtain \(20+6=26\) is not the correct combination under the stated independence assumption. Variances add, giving \(\sqrt{20^2+6^2}\), not the sum of the standard deviations.

Reading Coverage Correctly

A nominal 95% prediction interval is a statement about the long-run performance of the interval procedure under its assumptions. If the same forecasting setup were applied repeatedly, with the model and conditions remaining appropriate, approximately 95% of the resulting intervals would contain their corresponding future observations. For a particular interval and a future observation not yet seen, it is reasonable to describe the interval as having 95% model-based predictive coverage. It is not a guarantee that this particular observation will fall inside.

An observation outside a 95% interval does not, by itself, prove the forecasting method is invalid; such outcomes are expected occasionally even under a correct model. Repeatedly seeing too many observations outside intervals, however, is evidence that the assumed uncertainty may be too small or the model may not represent the process well. Conversely, excessively wide intervals may offer little practical guidance even if they cover reliably.

Coverage is conditional on what was known at forecast time and on the model assumptions. A normal interval may be unsuitable for a strongly skewed outcome, and a symmetric interval for a count can have fractional or negative endpoints. Such results are a warning to consider an interval construction that respects the outcome distribution, not permission to interpret impossible values literally. Estimated parameters, changing patterns, and dependence not reflected in \(\sigma_h\) can also affect actual coverage.

Common Exam Traps

Using uncertainty for the mean alone. A future observation includes new variation around the mean. Check that the standard deviation is for the future-observation forecast error.
Using the wrong horizon. Forecast-error variability can depend on \(h\). Use the standard deviation for the requested horizon and do not scale it again unless the model's formula explicitly requires that calculation.
Confusing coverage with certainty. A 95% interval does not guarantee that the next observation is inside, nor does it mean that 95% of observations fall within these particular fixed endpoints.
Ignoring assumptions. The symmetric normal formula requires an appropriate normal forecast-error distribution and a valid forecast-error standard deviation. It is not automatically correct just because a point forecast is available.
Rounding too early. Keep adequate precision in the standard deviation and margin, then round the final endpoints as requested.

Key takeaway: A prediction interval targets a future observation. Under a conditionally normal, zero-mean forecast-error model, calculate it as the point forecast plus or minus a normal quantile times the forecast-error standard deviation for that horizon. Interpret its coverage as conditional and model-dependent.

Check Your Understanding

For each problem, identify the quantity the interval covers and use only the assumptions stated.

A hypothetical weekly demand forecast is 72 units, with a normal forecast-error standard deviation of 5.5 units at the target week. Find a 90% prediction interval using \(z_{0.95}=1.645\). (Answer: \([62.95,81.05]\) units.)
A model forecasts 310 inspections for a particular future day. The forecast-error distribution is normal with mean zero and standard deviation 18 inspections for that day. Calculate a 95% prediction interval using \(z_{0.975}=1.96\). (Answer: \([274.72,345.28]\) inspections.)
A future monthly index value has a point forecast of 46 and a conditionally normal forecast-error standard deviation of 4. Find an 80% prediction interval using \(z_{0.90}=1.282\). (Answer: \([40.87,51.13]\).)
A hypothetical forecast has mean 120. New-observation variation has standard deviation 10, and independent uncertainty in the estimated mean has standard deviation 3. Assume both components are normal with mean zero. Find a 95% prediction interval using \(z_{0.975}=1.96\), rounding endpoints to two decimal places. (Answer: \([99.54,140.46]\).)
A procedure is described as producing nominal 90% prediction intervals for next-month observations. In a large number of repeated forecasts made under the model assumptions, how should the 90% coverage be interpreted? (Answer: Approximately 90% of the intervals should contain their corresponding future observations in the long run; it does not guarantee coverage for any one observation.)

What You'll Learn