Historical-Mean Forecasts

Forecasting Around a Stable Level

Some time series fluctuate from period to period without a persistent upward or downward movement. If the underlying level is reasonably stable, a natural forecast is the average of the observations seen so far. This is the historical-mean forecast: it uses the accumulated history to estimate a constant level and forecasts that level into the future.

The method is simple, but its interpretation matters. It does not predict that future observations will equal the historical average. Rather, it predicts the average level around which future observations fluctuate. As new observations accumulate, the forecast can move; each new value is incorporated into the updated average.

Definition: Let \(Y_1,\ldots,Y_n\) be the observations available at the forecast origin \(n\), and let \(y_t\) denote the observed value at time \(t\). The historical mean is \(\bar y_n=\frac{1}{n}\sum_{t=1}^{n}y_t\). The historical-mean forecast for a future observation \(h\) periods ahead is \(\hat y_{n+h\mid n}=\bar y_n\), for \(h\geq 1\). It is a constant forecast: at a given forecast origin, the point forecast is the same at every future horizon.

Model and Assumptions

A standard model for this method is a stable level plus random variation:

Y_t=\mu+\varepsilon_t,\qquad t=1,2,\ldots

Here, \(\mu\) is the constant level and \(\varepsilon_t\) is the random deviation from that level. For the historical mean to be a suitable point forecast, assume the deviations have mean zero, finite constant variance \(\sigma^2\), and are independent across time. Thus \(E[Y_t]=\mu\), and future deviations have expected value zero even after the observations so far are known. A common, stronger version of the model assumes \(Y_t\) are independent and identically distributed as \(N(\mu,\sigma^2)\); normality is not necessary for calculating the historical-mean point forecast.

Under these assumptions, the expected future observation is the stable level \(\mu\). Since \(\mu\) is usually unknown, it is estimated by \(\bar y_n\), giving \(\hat y_{n+h\mid n}=\bar y_n\). The assumptions are substantive: a trend, changing level, seasonal pattern, or dependence that makes recent observations informative about future deviations can make a constant historical mean unsuitable. Having a long history does not correct a mismatch between the model and the pattern in the series.

Why the Mean Is the Natural Constant Forecast

There are two complementary justifications. First, under squared-error loss, the conditional expectation is the forecast that minimizes expected squared error. Under the independent stable-level model, the conditional expectation of any future observation is \(\mu\), so the best forecast if \(\mu\) were known would be \(\mu\). Replacing the unknown level by its estimate gives the historical-mean forecast.

Second, among all constant values fitted to the observed series, the sample mean minimizes the sum of squared deviations. This explains why the method estimates the level with an average rather than, for example, choosing the last observation.

Justification: For a constant candidate level \(c\), define \(Q(c)=\sum_{t=1}^{n}(y_t-c)^2\). Differentiating gives \(Q'(c)=-2\sum_{t=1}^{n}(y_t-c)\). Setting this equal to zero yields \(nc=\sum_{t=1}^{n}y_t\), so \(c=\bar y_n\). Also \(Q''(c)=2n>0\), confirming that this value minimizes the sum of squared deviations. This is a fit to the observed values; the forecasting interpretation additionally relies on the stable-level assumptions.

Updating the Forecast When an Observation Arrives

Suppose the current mean is based on \(n\) observations and a new value \(y_{n+1}\) becomes available. The updated forecast is the mean of all \(n+1\) observations. It can be calculated without summing the entire series again:

\bar y_{n+1} =\frac{n\bar y_n+y_{n+1}}{n+1} =\bar y_n+\frac{y_{n+1}-\bar y_n}{n+1}.

The second form shows the direction and size of the update. If the new observation is above the previous mean, the forecast rises; if it is below, the forecast falls. The adjustment is the new observation’s difference from the previous mean, divided by \(n+1\). Consequently, for a fixed difference, a new value has less influence when it is added to a longer history.

Set the forecast origin.
Use only observations available at that time, \(y_1,\ldots,y_n\).

Calculate the current mean.
Add the \(n\) observations and divide by \(n\).

Forecast each required horizon.
Assign \(\bar y_n\) to every future period under the constant-level method.

Update when data arrive.
Use \(\bar y_{n+1}=\bar y_n+(y_{n+1}-\bar y_n)/(n+1)\), then use the updated mean for forecasts from the new origin.

Worked Example: Forecasting a Stable Monthly Level

A hypothetical benefit plan records monthly reimbursements, in thousands of dollars. The first five observations are \(42,46,44,48,40\). Assume these observations and future reimbursements follow independent \(N(\mu,\sigma^2)\) distributions with unknown stable level \(\mu\) and unknown variance \(\sigma^2>0\). Calculate the historical-mean forecasts for the next three months. Then update the forecasts when the next observed reimbursement is \(50\).

Method. Calculate the sample mean at the current forecast origin. Under the stated stable-level model, this estimates the common expected reimbursement, and the point forecast is constant across future horizons.

Initial mean. The sum of the five observed values is \(42+46+44+48+40=220\). Thus \(\bar y_5=220/5=44\) thousand dollars.

Three forecasts. The one-, two-, and three-month-ahead point forecasts are each \(44\) thousand dollars. A forecast of the same level at each horizon does not assert that the actual reimbursements will all be \(44\).

Update. The new observation is \(50\), so the updated sum is \(220+50=270\), and the updated mean is \(270/6=45\) thousand dollars. Using the recursive formula gives the same result: \(44+(50-44)/6=44+1=45\).

Check and trap avoided. The updated mean lies between the previous mean \(44\) and new value \(50\), as it must. Using \(50\) alone as the new forecast would discard the first five observations; that would be a last-value forecast, not a historical-mean forecast.

How New Data Change the Mean

The update equation makes clear why the historical-mean forecast changes as the series grows. An observation above the current mean pulls the updated mean upward, but its effect is divided by the new sample size. The historical mean is not a fixed estimate that ignores new evidence; it changes systematically as each new observation is included.

The weights also change over time. At time \(n\), each of the first \(n\) observations has weight \(1/n\). After a new observation arrives, each old observation has weight \(1/(n+1)\), and the new observation also has weight \(1/(n+1)\). Thus the update incorporates the new value while reducing the weight assigned to every earlier value.

Worked Example: Updating After Two New Values

A hypothetical service center records daily numbers of completed assessments. The first four counts are \(18,22,20,24\). Assume an independent \(N(\mu,\sigma^2)\) stable-level model, with unknown \(\mu\) and \(\sigma^2>0\). The next two counts are \(15\) and \(21\). Find the forecast after each new count arrives, and explain the direction of each update.

Method. Start with the historical mean, then apply the recursive update using the new sample size at each step.

Initial forecast. The first four counts sum to \(18+22+20+24=84\). Therefore, \(\bar y_4=84/4=21\) assessments per day.

After the count of 15. The new count is below the existing mean. The updated mean is \(21+(15-21)/5=21-6/5=19.8\). Checking directly, the five observations sum to \(84+15=99\), and \(99/5=19.8\).

After the count of 21. The new count is above the updated mean, so the mean rises. It becomes \(19.8+(21-19.8)/6=19.8+1.2/6=20\). Directly, the six observations sum to \(99+21=120\), and \(120/6=20\).

Check and trap avoided. The result after the second update is \(20\), between \(19.8\) and \(21\). Dividing the second update by \(5\) would be an error: the second new value is the sixth observation, so the update denominator is \(6\), not the previous sample size.

Point Forecasts and Forecast Uncertainty

A point forecast gives one central value, not a guarantee about the next observation. Even when the stable-level model is appropriate, future random variation remains. Estimating the level from a finite history introduces additional uncertainty because the sample mean may differ from \(\mu\).

If the errors are independent with variance \(\sigma^2\), then the variance of the future-observation forecast error is

\operatorname{Var}(Y_{n+h}-\bar Y_n) =\operatorname{Var}(\varepsilon_{n+h})+\operatorname{Var}(\bar Y_n) =\sigma^2+\frac{\sigma^2}{n} =\sigma^2\left(1+\frac{1}{n}\right).

The equality uses independence: the future error is independent of the errors used to calculate \(\bar Y_n\). The term \(\sigma^2\) represents future-observation variation, while \(\sigma^2/n\) represents uncertainty in the estimated level. This expression does not mean that the forecast itself changes with horizon. It describes uncertainty in a future observation under the stated model. If the errors are dependent or their variance changes over time, this variance calculation need not apply.

Worked Example: Quantifying Uncertainty Under a Normal Model

A hypothetical monitoring series has observed values \(98,102,100,100\). Assume independent observations from \(N(\mu,9)\): the stable level \(\mu\) is unknown, and the variance is known to be \(9\), so the standard deviation is \(3\). Find the historical-mean forecast for the next observation and a 90% prediction interval using the known variance.

Method. Estimate the level by the sample mean. For a future observation minus that estimated mean, the error variance is \(9(1+1/n)\). Under the stated normal model, this forecast error is normally distributed with mean zero.

Forecast. The observations sum to \(98+102+100+100=400\), so \(\bar y_4=400/4=100\). The point forecast is \(100\).

Forecast-error standard deviation. With \(n=4\), the variance is \(9(1+1/4)=9(5/4)=11.25\). The standard deviation is \(\sqrt{11.25}\approx3.3541\).

Prediction interval. A central 90% normal interval uses \(z_{0.95}\approx1.645\). Its margin is \(1.645\sqrt{11.25}\approx5.5175\), giving \(100\pm5.5175\), or approximately \([94.48,105.52]\). The endpoints are rounded to two decimals.

Check and trap avoided. The interval is centered on the point forecast, \(100\), and includes uncertainty both in the future observation and in estimating \(\mu\). Using only the future-observation standard deviation \(3\) would omit the estimation component \(9/4\) from the forecast-error variance.

When a Historical Mean Is Not Suitable

A constant forecast is appropriate when the series can reasonably be treated as fluctuating around an unchanged level and when the past provides a useful estimate of that level. Before using it, inspect the time order of the observations rather than relying on the average alone. A clear sustained rise or fall, a recurring seasonal pattern, or a substantial change in level contradicts the central idea of one stable mean.

Dependence also matters. The sample mean can still estimate a long-run average in some dependent series, but the independent-error forecast argument above no longer follows automatically: recent observations might carry information about the next deviation. The historical mean should therefore not be treated as universally optimal just because it is easy to calculate. In particular, it is not a way to extrapolate a trend; that requires a different forecasting approach.

Do not confuse the average with the last observation. The historical mean uses all available observations; a last-value forecast uses only the most recent one.
Use the correct denominator when updating. Adding observation \(y_{n+1}\) requires division by \(n+1\).
Keep the forecast origin clear. A mean calculated from observations through time \(n\) must not include values that were unavailable at that forecast origin.
Separate a point forecast from an actual future value. The forecast is an estimate of the expected level, not a promise that each future observation will equal it.

Key takeaway: Under a stable-level model with independent, mean-zero deviations, the historical mean estimates the level and gives the same point forecast at every future horizon. When a new observation arrives, update the mean by adding its difference from the old mean divided by the new sample size. Check the series’ pattern and dependence before relying on the method.

Check Your Understanding

For each problem, assume a stable-level model with independent observations unless the question asks you to assess that assumption.

A hypothetical daily series has values \(6,8,10,8\). Calculate the historical-mean forecast for each of the next two days. (Answer: \(8\) for each day.)
A series has \(n=4\) observations and current mean \(25\). The fifth observation is \(13\). Use the update formula to find the new mean. (Answer: \(22.6\).)
A hypothetical series has observations \(30,33,27,30,35\). Find its historical-mean forecast, then find the updated forecast after the next observation is \(25\). (Answer: Initial forecast \(31\); updated forecast \(30\).)
For a series with \(n=9\) observations, the estimated mean is \(50\). Assume independent normal observations with known variance \(16\). What is the standard deviation of the next-observation forecast error? (Answer: \(\sqrt{160/9}\approx4.2164\).)
A series has observations \(10,12,14,16\). Calculate its historical-mean forecast, then assess whether a stable-level assumption is persuasive from these values alone. (Answer: The historical-mean forecast is \(13\); the steadily increasing sequence makes a constant-level assumption questionable.)

What You'll Learn