From Observed Losses to a Forecast
A short insurance-loss series can move irregularly from one period to the next. Forecasting from only the latest observation may overreact to a temporary jump, while averaging every observation equally may give older data too much influence. Simple exponential smoothing provides a level-based forecast that weights recent observations more heavily and gives older observations progressively less weight.
The method is appropriate when a series is represented by a changing level, without an explicit trend or seasonal component. It is a calculation procedure as well as a forecasting method: each new observation updates the level, and forecasts are based on the most recently updated level. The examples below use hypothetical insurance data and specify the initial level and smoothing parameter so the calculations are reproducible.
The parameter \(\alpha\) determines how quickly the level responds to new observations. A larger value gives the latest observation more weight in the update; a smaller value retains more of the previous level. The value of \(\alpha\) and the initial level must be known before the sequence of levels can be calculated. In each example, observations are equally spaced and measured on a consistent scale. No probability distribution is assumed for the arithmetic demonstrations.
Calculating Levels and Forecasts
Worked Example: Updating a Level and Forecasting Several Periods
A hypothetical insurer records quarterly average claim costs, in thousands of dollars, of \(80, 100, 90, 110,\) and \(100\). Set the initial level to the first observation, \(L_1=y_1=80\), and use \(\alpha=0.40\). Calculate each subsequent level, then forecast the next quarter and the two quarters after that. Compare the next-quarter forecast with the last-value naive forecast.
Method. Apply the level update to observations \(y_2\) through \(y_5\), then use the latest level for each forecast horizon. The starting level is \(L_1=80\); it is not an additional observation to be smoothed.
After observing \(y_2=100\),
After observing \(y_3=90\),
After observing \(y_4=110\),
After observing \(y_5=100\),
Therefore, the forecasts made after period 5 are
The last-value naive forecast for period 6 is \(y_5=100\), so simple exponential smoothing forecasts \(98.368\) thousand dollars while the naive method forecasts \(100\) thousand dollars. The smoothed value lies between the latest observation and the preceding level, reflecting both rather than copying the latest value. Common mistake avoided: using \(L_4\) to forecast period 6 would ignore the observed \(y_5=100\) and the update that it requires.
The multi-step forecast is constant because simple exponential smoothing has a level but no separate trend or seasonal component. This does not mean future observations must equal the forecast. It means the method has no rule for projecting a systematic rise, fall, or recurring seasonal pattern beyond the current level.
Why the Forecast Stays at the Current Level
A direct justification follows from forecasting recursively. At time \(t\), the one-step forecast is \(L_t\). For a later horizon, the method has not yet observed the future value that would change the level. In constructing a forecast, it substitutes the forecasted value for that unobserved future observation. The level update then leaves the level unchanged.
By definition, \(\hat{y}_{t+1\mid t}=L_t\).
Because \(y_{t+1}\) is not yet observed, substitute \(\hat{y}_{t+1\mid t}\) into the update: \(\hat{L}_{t+1\mid t}=\alpha\hat{y}_{t+1\mid t}+(1-\alpha)L_t\).
Substituting \(\hat{y}_{t+1\mid t}=L_t\) gives \(\hat{L}_{t+1\mid t}=\alpha L_t+(1-\alpha)L_t=L_t\). Repeating this argument leaves the forecasted level unchanged at every later horizon.
This recursive argument explains the multi-step result: \(\hat{y}_{t+h\mid t}=L_t\) for each \(h\geq1\). It is a property of the simple level-only method, not a claim that a real loss series will be flat.
Worked Example: Starting from a Given Initial Level
A hypothetical portfolio has annual average claim costs, in thousands of dollars, of \(120, 100, 140,\) and \(130\). The analyst specifies \(L_0=110\) and \(\alpha=0.30\). Calculate the four updated levels and the forecasts at horizons 1 and 3 after the fourth observation. Compare the one-step forecast with the last-value naive forecast.
Method. Here the specified initial level precedes the first listed observation. Use it as \(L_0\); do not replace it with \(y_1\).
For \(y_1=120\),
For \(y_2=100\),
For \(y_3=140\),
For \(y_4=130\),
The forecasts after the fourth observation are therefore
The period-5 last-value naive forecast is \(y_4=130\), compared with \(121.859\) from smoothing. These amounts are in thousands of dollars. Common mistake avoided: initializing with \(L_1=y_1\) would discard the explicitly supplied \(L_0=110\) and produce a different sequence of levels.
Updating When New Observations Arrive
A forecast made at one time should not be treated as the forecast from a later time if new data have arrived in between. Once a new observation is available, it updates the level; forecasts from that later origin use the updated level. A fair comparison with a last-value naive forecast follows the same timing rule: at each forecast origin, use only information available by that time.
Worked Example: Rolling Forecasts and Holdout Errors
A hypothetical insurer tracks a quarterly index of claim costs, in thousands of dollars. The first four observations are \(60, 80, 70,\) and \(90\). Use \(\alpha=0.50\) and initialize with \(L_1=y_1=60\). The next two observed values, reserved for evaluation, are \(y_5=100\) and \(y_6=95\). Make a forecast for each holdout observation using only information then available. Compare simple exponential smoothing with the rolling last-value naive method using mean absolute error (MAE).
Method. First calculate the level through period 4. Forecast period 5 from \(L_4\), then incorporate the observed \(y_5\) before forecasting period 6.
The successive training levels are
At the end of period 4, the period-5 smoothed forecast is \(L_4=80\). The naive forecast is the most recent actual value, \(y_4=90\). After period 5 is observed, update the smoothed level:
The resulting period-6 smoothed forecast is \(90\). The period-6 naive forecast uses the newly observed \(y_5=100\). The two forecasts and absolute errors are:
| Period | Actual | Smoothed forecast | Smoothed absolute error | Naive forecast | Naive absolute error |
|---|---|---|---|---|---|
| 5 | 100 | 80 | |100 - 80| = 20 | 90 | |100 - 90| = 10 |
| 6 | 95 | 90 | |95 - 90| = 5 | 100 | |95 - 100| = 5 |
Thus the two-observation MAEs are
For this holdout, the naive method has the smaller MAE. This comparison does not establish that it will always forecast better; it reports how the two methods performed on these two observations. Common mistake avoided: using \(90\) as the naive forecast for period 6 would fail to incorporate \(y_5=100\), which was available by the period-6 forecast origin.
Interpreting and Checking the Calculation
The level update is a weighted average of the new observation and the previous level. With \(0<\alpha<1\), the new level lies between those two inputs. That provides a useful arithmetic check. For example, if the previous level is \(80\), the new observation is \(100\), and \(0<\alpha<1\), a computed level outside the interval from \(80\) to \(100\) signals an error.
A frequent error is to reverse the weights: using \(\alpha\) on the old level and \(1-\alpha\) on the new observation. That would make a large \(\alpha\) retain more of the past, opposite to the stated definition. Another trap is to calculate a different forecast for each future horizon by repeatedly applying the update to the same last observed value. The update is applied when a new observation arrives; future unobserved values are not fresh data. Consequently, the multi-step forecasts remain at \(L_t\).
Check Your Understanding
For each problem, calculate from the stated initialization and use only observations available at the forecast origin.
- A quarterly hypothetical claim index is \(60\) at time 1 and \(80\) at time 2. Set \(L_1=60\) and \(\alpha=0.25\). Find \(L_2\) and the three-step-ahead forecast made at time 2. (Answer: \(L_2=65\); \(\hat{y}_{5\mid2}=65\).)
- For hypothetical annual loss amounts in thousands of dollars, \(y_1=200\), \(y_2=180\), and \(y_3=220\), use \(L_0=190\) and \(\alpha=0.40\). Calculate \(L_1,L_2,L_3\), then compare the forecast for period 4 with the last-value naive forecast. (Answer: \(L_1=194\), \(L_2=188.4\), \(L_3=201.04\); smoothed forecast \(201.04\), naive forecast \(220\), in thousands of dollars.)
- A hypothetical monthly loss index has \(L_1=40\), \(y_2=50\), and \(y_3=30\), with \(\alpha=0.50\). Calculate the final level and the next-period smoothed and naive forecasts. (Answer: \(L_2=45\), \(L_3=37.5\); smoothed forecast \(37.5\), naive forecast \(30\).)
- A smoothed forecast for period 8 was \(75\), using \(L_7=75\). The actual period-8 value is \(95\), and \(\alpha=0.40\). Calculate the updated level and the period-9 smoothed forecast. If the period-9 actual is \(85\), find the period-9 absolute error. (Answer: \(L_8=0.40(95)+0.60(75)=83\); forecast \(83\); absolute error \(2\).)
- For two holdout periods, the smoothed forecasts are \(50\) and \(54\), while the actuals are \(58\) and \(50\). The rolling naive forecasts are \(55\) and \(58\). Calculate each method's MAE and identify which is lower. (Answer: smoothed MAE \(=(8+4)/2=6\); naive MAE \(=(3+8)/2=5.5\); the naive method is lower.)