Prediction-Interval Derivation For Simple Smoothing

Why Prediction-Interval Width Depends on the Horizon

A simple exponential smoothing forecast carries the latest estimated level forward unchanged. That does not mean the uncertainty is unchanged: future observations can contain new random errors, and each of those errors can also affect the smoothed level used for later forecasts. To construct a prediction interval, we need to account for both effects.

The derivation below uses the innovation form of simple exponential smoothing. It gives a forecast-error variance for any horizon $h$, conditional on the information available at time $T$. The variance formula follows from independence and constant variance; an exact normal prediction interval requires the additional assumption of normally distributed errors.

Model and assumptions: Let $y_t$ be the observation at time $t$, and let $l_t$ be the smoothed level after observing $y_t$. Simple exponential smoothing with smoothing parameter $0\leq\alpha\leq1$ can be written as $$ y_t=l_{t-1}+\varepsilon_t,\qquad l_t=l_{t-1}+\alpha\varepsilon_t, $$ where the innovations $\varepsilon_t$ have mean zero, common variance $\sigma^2$, and are independent across time. For an exact normal prediction interval, also assume $\varepsilon_t\sim N(0,\sigma^2)$. The forecast is conditional on the observed history through time $T$; the fitted level, $\alpha$, and $\sigma^2$ are treated as known.

The first equation says an observation is the previous level plus a new innovation. The second says that the level is updated by a fraction $\alpha$ of that innovation. The $h$-step forecast made at time $T$ is $\hat y_{T+h\mid T}=l_T$ for every $h\geq1$. The point forecast is constant across horizons, but its prediction-error variance is not.

Deriving the Forecast-Error Variance

Write the forecast error as the future observation minus its forecast. To derive it, first express the future level in terms of the level at time $T$ and the innovations that arrive after $T$. Repeated application of the level update gives

l_{T+h-1} =l_T+\alpha\sum_{j=1}^{h-1}\varepsilon_{T+j}.

When $h=1$, the sum is empty and the future level is just $l_T$. The observation at time $T+h$ adds one more innovation, $\varepsilon_{T+h}$. Therefore the $h$-step forecast error is

e_{T+h\mid T} =y_{T+h}-\hat y_{T+h\mid T} =\alpha\sum_{j=1}^{h-1}\varepsilon_{T+j}+\varepsilon_{T+h}.

This expression shows why the horizon matters: there are $h-1$ innovations that update the level before the target observation, as well as the target observation’s own innovation. All these innovations are independent and have variance $\sigma^2$. The variance of a sum of independent random variables is the sum of their variances; a constant multiplier $c$ multiplies variance by $c^2$. Hence:

Result: Under the stated independent, constant-variance innovation assumptions, the conditional forecast-error variance at horizon $h$ is

\operatorname{Var}(e_{T+h\mid T}\mid\mathcal F_T) =\sigma^2\left[1+(h-1)\alpha^2\right],

where $\mathcal F_T$ denotes the information available through time $T$. If the innovations are also normal, then $$ e_{T+h\mid T}\mid\mathcal F_T \sim N\left(0,\ \sigma^2[1+(h-1)\alpha^2]\right). $$

The $1$ in the variance formula is the contribution of the target observation’s innovation. Each of the preceding $h-1$ innovations contributes $\alpha^2\sigma^2$, because it changes the level by $\alpha\varepsilon_t$, and that level change remains in the later forecast error. This argument derives the formula rather than assuming that forecast errors at different horizons have the same variance.

Constructing a Horizon-Specific Prediction Interval

Suppose the innovations are normal and the model parameters and fitted level are treated as known. Let $z_{1-\gamma/2}$ be the standard normal quantile with cumulative probability $1-\gamma/2$. A $100(1-\gamma)\%$ prediction interval for the future observation is

\hat y_{T+h\mid T} \ \pm\ z_{1-\gamma/2}\, \sigma\sqrt{1+(h-1)\alpha^2}.

The center is the point forecast, and the margin of error uses the forecast-error standard deviation at the particular horizon $h$. For example, a 95% interval uses $z_{0.975}\approx1.96$. If $\sigma^2$ is estimated from data, substituting its estimate gives a practical interval, but the interval then treats that estimate as if it were known. The formula does not include extra uncertainty from estimating the level, smoothing parameter, or innovation variance.

Independence and constant variance justify the variance calculation, but they do not by themselves imply that the forecast error is normal. Without normal innovations, the formula still gives the forecast-error variance under the stated assumptions; using the normal quantile then gives an approximation, not necessarily an exact prediction interval. In particular, a variance alone does not determine the tails of a distribution.

Worked Example: A Four-Month Workload Forecast

A hypothetical insurer smooths a monthly claims-workload index. At the forecast origin, the fitted level is $120$ index units, $\alpha=0.40$, and the innovation variance is $\sigma^2=25$ index-units squared. Assume independent innovations $\varepsilon_t\sim N(0,25)$. Find a 95% prediction interval for the observation four months ahead. This is an index forecast, not a cash-flow valuation, so no interest rate or mortality assumption applies.

Method. Use the innovation-model forecast variance at horizon $h=4$, then apply the normal critical value $z_{0.975}=1.96$.

Step 1: Calculate the variance. There are $h-1=3$ innovations that can affect the level before the target month:

\operatorname{Var}(e_{T+4\mid T}) =25[1+3(0.40)^2] =25[1+3(0.16)] =25(1.48) =37.

Step 2: Calculate the standard deviation and margin.

\operatorname{SD}(e_{T+4\mid T})=\sqrt{37}\approx6.0828, \qquad 1.96\sqrt{37}\approx11.9222.

Step 3: Form the interval. The point forecast is $120$, so the endpoints are $120-11.9222=108.0778$ and $120+11.9222=131.9222$. Rounded to two decimals, the interval is $[108.08,131.92]$ index units.

Check. The interval is centered at $120$, and its half-width is approximately $11.92$, equal to $1.96$ times the calculated standard deviation. The method avoids using the one-step variance $25$ at a four-step horizon.

Worked Example: A 90% Interval at a Three-Step Horizon

A hypothetical service center forecasts its monthly incident count using simple exponential smoothing. Its current fitted level is $80$, $\alpha=0.50$, and its innovations are independent with distribution $N(0,16)$, so $\sigma=4$. Find a 90% prediction interval three months ahead. No interest rate or mortality assumption applies to this count forecast.

Method. For a 90% two-sided interval, use $z_{0.95}\approx1.645$. Here $h=3$, so there are two intervening innovations that update the level before the target month.

Step 1: Calculate the forecast-error variance.

\operatorname{Var}(e_{T+3\mid T}) =16[1+2(0.50)^2] =16(1+0.50) =24.

Step 2: Calculate the standard deviation and margin.

\operatorname{SD}(e_{T+3\mid T})=\sqrt{24}\approx4.8990, \qquad 1.645\sqrt{24}\approx8.0588.

Step 3: Calculate the endpoints. The interval is $80\pm8.0588$, or $[71.9412,88.0588]$. Rounded to two decimals, the 90% prediction interval is $[71.94,88.06]$ incidents.

Check. The half-width is $8.0588$, and dividing by $1.645$ recovers approximately $4.8990$, the forecast-error standard deviation. Using $1.96$ here would produce a 95% rather than a 90% interval.

Worked Example: Comparing Several Forecast Horizons

A hypothetical insurer tracks a monthly claim-reporting index. Its fitted level is $200$, the smoothing parameter is $\alpha=0.25$, and independent innovations have distribution $N(0,36)$, so $\sigma=6$. Find 95% prediction intervals at horizons $h=1$, $h=2$, and $h=5$. There is no interest rate or mortality assumption in this index-forecasting calculation.

Method. Use the same point forecast $200$ at each horizon, but recalculate the variance with the relevant number $h-1$ of level-updating innovations. Use $z_{0.975}=1.96$.

Horizon 1. There are no intervening innovations:

V_1=36[1+0(0.25)^2]=36,\quad \sqrt{V_1}=6,\quad 200\pm1.96(6)=200\pm11.76.

The interval is $[188.24,211.76]$.

Horizon 2. There is one intervening innovation:

V_2=36[1+(0.25)^2]=36(1.0625)=38.25,\quad \sqrt{V_2}\approx6.1847,\quad 1.96\sqrt{V_2}\approx12.1219.

The interval is $[187.88,212.12]$, rounded to two decimals.

Horizon 5. There are four intervening innovations:

V_5=36[1+4(0.25)^2]=36(1.25)=45,\quad \sqrt{V_5}\approx6.7082,\quad 1.96\sqrt{V_5}\approx13.1481.

The interval is $[186.85,213.15]$, rounded to two decimals. The three intervals share the same center, $200$, while their widths increase with the horizon. This is consistent with the variance formula: for $\alpha>0$, each additional step adds $\alpha^2\sigma^2$ to the forecast-error variance.

Check. The variances are $36$, $38.25$, and $45$, respectively; they are nondecreasing. This calculation avoids the mistake of assuming that a constant point forecast implies a constant prediction interval.

Common Exam Trap: Geometric Weights for Future Innovations

A tempting but incorrect calculation is to treat future innovations as though their effects on the forecast error decay geometrically. For example, with $\alpha=0.40$, $\sigma^2=25$, and $h=4$, an incorrect approach might use the error variance

25\left[1+(0.40)^2\{1+(0.60)^2+(0.60)^4\}\right] =25(1.238336) =30.9584.

That calculation assigns successively smaller coefficients to innovations that arrive during the forecast period. But an innovation that updates the level by $\alpha\varepsilon_t$ remains part of every later level. The correct four-step forecast error contains three level contributions with coefficient $\alpha$, and the target observation innovation with coefficient $1$:

e_{T+4\mid T} =0.40\varepsilon_{T+1} +0.40\varepsilon_{T+2} +0.40\varepsilon_{T+3} +\varepsilon_{T+4}.

Thus the correct variance is $25[1+3(0.40)^2]=37$, not $30.9584$. Geometric weights occur when expressing a smoothed level in terms of past observations; they are not the coefficients of successive future innovations in this forecast error.

Using the Result Carefully

The derivation is conditional on a fitted level at time $T$. It quantifies variation from future innovations, not every possible source of uncertainty in a fitted forecasting procedure. If the level or smoothing parameter is estimated imprecisely, or if the innovation variance is estimated, the displayed interval does not automatically incorporate that additional uncertainty.

The assumptions also matter for interpretation. Serial dependence among innovations changes the variance of their sum because covariance terms may no longer be zero. Changing innovation variance means the terms in the sum do not all contribute the same $\sigma^2$. And if innovations are not normal, a normal interval is an approximation unless a separate argument justifies its coverage.

Key takeaway: Under independent, mean-zero, constant-variance innovations, the $h$-step forecast-error variance for simple exponential smoothing is $\sigma^2[1+(h-1)\alpha^2]$. For an exact normal prediction interval, additionally assume normal innovations and use the forecast-error standard deviation at that horizon.

Check Your Understanding

For each problem, use the innovation-model assumptions stated in the question. Round interval endpoints to two decimals where needed.

A hypothetical monthly operations index uses $\alpha=0.20$, with independent innovations of variance $9$. What is the forecast-error variance at horizon $h=4$? (Answer: $10.08$)
A hypothetical call-volume series uses $\alpha=0.60$ and independent innovations with variance $16$. Find the forecast-error variance at horizon $h=3$. (Answer: $27.52$)
A hypothetical inventory-demand index has fitted level $40$, $\alpha=0.50$, and independent normal innovations $N(0,4)$. Find a 95% prediction interval at $h=2$, using $z_{0.975}=1.96$. (Answer: $[35.62,44.38]$)
A hypothetical insurer’s reporting index has $\alpha=0.25$ and independent innovations with variance $9$. Find the forecast-error variance at $h=5$. (Answer: $11.25$)
A hypothetical monthly service count has fitted level $70$, $\alpha=0.40$, and independent normal innovations $N(0,25)$. Find a 90% prediction interval at $h=3$, using $z_{0.95}=1.645$. (Answer: $[60.55,79.45]$)

What You'll Learn