Demonstration 1 of 4
How the past fades: geometric weights
How much does a month from a year ago count in an exponentially smoothed forecast?
Unrolling the recursion gives the newest month weight alpha, the one before alpha x (1 - alpha), and so on, a geometric decline. The moving average instead weights twelve months equally and then forgets month thirteen entirely.
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alpha is the smoothing parameter between 0 and 1; the weight of an observation k months old is alpha times (1 - alpha) to the power k. The bars are those weights; the red step is a twelve-month moving average.
Predict first. Brown's example keeps 0.8 of the weight at each step back. Does the last year then carry more or less than 90 percent of the forecast?
Choose an example
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Constructed data: the chapter's own weighting example (each step back keeps 0.8 of the weight, alpha 0.2), with alpha varied.
Calculated values
- Weight on the newest month
- 0.200
- Weight one month back
- 0.160
- Weight two months back
- 0.128
- Share of weight in the last 12 months
- 0.9313
With alpha 0.2 each step back keeps 1 - 0.2 = 0.8 of the weight before it: 0.2 x 0.8^2 = 0.128 for the month two back. The last twelve months carry 1 - 0.8^12 = 0.9313 of the total; the moving average gives each of them 1/12 = 0.083 and then drops month thirteen to nothing.
Worked steps
- Newest month: alpha = 0.2.
- One back: 0.2 x 0.8 = 0.160.
- Two back: 0.2 x 0.8^2 = 0.128.
- Last twelve months together: 1 - 0.8^12 = 0.9313.
Use the idea
When someone asks how far back a smoothed forecast looks, answer with the weights: at alpha 0.5 the last four months already carry 1 - 0.5^4 = 0.9375 of the forecast.
Where the conclusion applies
Weights shown for an unbounded history; a finite history also gives the initial level the leftover weight. The weights describe the method, not whether old data are still relevant to your series.
Check your understanding: With alpha 0.3, what weight does the observation three months back receive?
Chapter 4 source: section "Chapter 4: The Smoother".
Demonstration 2 of 4
How fast should a forecast forget?
When demand jumps to a new level, how many months does exponential smoothing need to follow it?
Each month the level moves alpha of the way toward the new observation. A large alpha closes the jump in a month or two but copies each month's noise; a small alpha ignores noise and also ignores the jump for a long time.
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The updated level is alpha times this month's observation plus (1 - alpha) times the previous level. The constructed series sits near 20 for 30 months and near 35 after, with noise of standard deviation 2.
Predict first. With alpha 0.1, will the level reach within 2 units of 35 before month 40?
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Constructed data: the chapter notebook's synthetic level shift (cell 3) and its three alpha values.
Calculated values
- Observation this month
- 35.205
- Previous level
- 21.397
- Updated level
- 28.30
- First month within 2 of 35
- 33
- Level wobble, months 41 to 60 (sd)
- 0.75
Month 31: level = 0.5 x 35.205 + 0.5 x 21.397 = 28.30. With alpha 0.5 the level first comes within 2 units of the new 35 in month 33, and once settled it wobbles with standard deviation 0.75. A middle setting reacts within a few months without following every observation.
Worked steps
- Observation: 35.205; previous level: 21.397.
- New level: 0.5 x 35.205 + 0.5 x 21.397 = 28.30.
Use the idea
When you choose a smoothing parameter, ask how fast the real level changes compared with the month-to-month noise, as the chapter's new employee must, and check both the lag after a shift and the wobble after it.
Where the conclusion applies
Constructed data with seed 20260922; the updated levels are computed after seeing each month, as in the notebook, so they are filters, not forecasts made beforehand. One shift and one noise level; other series reward other settings.
Check your understanding: The level is 30.00, the new observation is 36.00 and alpha is 0.3. What is the updated level?
Chapter 4 source: section "The Methods".
Demonstration 3 of 4
A trend need not continue forever
How much does damping change a trend forecast, and at which horizons?
Both models are fitted to the same 50 months. The undamped forecast adds one full trend step per month, a straight line. The damped forecast multiplies each further step by phi, so the line bends; with phi estimated near 1 the bend is gentle and shows mostly at long horizons.
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The forecast h months ahead starts from the final level and adds the final trend per month times a multiplier: h for Holt, phi + phi^2 + ... + phi^h for the damped trend, with damping coefficient phi between 0 and 1.
Predict first. At 30 months ahead, will the damped forecast sit more or less than 2 units below the undamped one?
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Constructed data: the chapter notebook's synthetic trend (cell 5) with its Holt and damped Holt fits.
Calculated values
- Final level
- 43.9765
- Final trend per month
- 0.4370
- Trend multiplier
- 27.784
- Damped Holt forecast at h = 30
- 56.12
- Other model at the same horizon
- 59.28
Damped Holt at 30 months ahead: 43.9765 + 27.784 x 0.4370 = 56.12, where the multiplier adds phi + phi^2 + ... + phi^30 with phi 0.995 instead of 30. The undamped model gives 59.28 at the same horizon, a gap of 3.16.
Worked steps
- Final level 43.9765, final trend 0.4370 per month.
- Trend multiplier at h = 30: 27.784 (damped).
- Forecast: 43.9765 + 27.784 x 0.4370 = 56.12.
Use the idea
For long-horizon plans built on a trend, compare the undamped and damped forecasts and ask whether the extra growth in the straight line is something you would bet on.
Where the conclusion applies
Parameters are estimated by the fitting routine on constructed data (seed 20260922); here phi reaches 0.995, the top of its allowed range, so damping is mild. The chapter's point is skepticism about long trends, not that damping always wins.
Check your understanding: A damped model has final level 100, trend 2 per month and phi 0.9. What is its forecast two months ahead?
Chapter 4 source: section "The Methods".
Demonstration 4 of 4
Seasonality earns its place on the holdout
Which member of the smoothing family forecasts two unseen years of seasonal demand best?
Each model is fitted on the first 120 months and forecasts the last 24 in one go. The figure shows the chosen model's forecast against the held-out months; the MAE scores the gap.
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MAE is the mean absolute error over the 24 held-out months, in units. Seasonal naive repeats the last observed year; SES tracks a level only; Holt-Winters adds trend and additive seasonality; Theta is the competition-winning benchmark.
Predict first. Will simple exponential smoothing beat the seasonal naive reference on this seasonal series?
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Constructed data: the companion's seeded seasonal series as fitted in the chapter notebook (cell 7), last 24 months held out.
Calculated values
- Seasonal naive MAE
- 2.88
- SES MAE
- 7.16
- Holt-Winters MAE
- 1.64
- Theta MAE
- 1.82
- Ratio to seasonal naive
- 0.57
Holt-Winters holdout MAE 1.64 against seasonal naive 2.88: 1.64 / 2.88 = 0.57, so it beats the seasonal naive reference. Seasonality earns its place only because it wins on months the models never saw.
Worked steps
- Holdout: the last 24 months, never used for fitting or choosing.
- Holt-Winters MAE over those months: 1.64.
- Ratio to seasonal naive: 1.64 / 2.88 = 0.57.
Use the idea
Score any candidate against a seasonal naive reference on months it never saw, and report the result even when the more complex method fails to win.
Where the conclusion applies
One constructed series (seed 20260922) with additive seasonality and a gentle trend, which suits Holt-Winters. A single holdout is one draw; the chapter's competitions scored thousands of series.
Check your understanding: A model's holdout MAE is 2.10 and the seasonal naive MAE is 2.80. What is the ratio, and does the model beat the reference?
Chapter 4 source: section "The Methods".