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Exam SRM · Smoothing and Forecasting · Tutorial 508 of 926

A Worked Forecast For A Changing Level And Trend

Calculate and compare four point-forecast methods, tracking how each responds to recent observations and projects a persistent trend.

Intermediate 13 min read

What You'll Learn

  • Calculate naive and simple exponential smoothing forecasts from a short series.
  • Update Holt level and trend estimates observation by observation.
  • Calculate damped-trend forecasts across several future periods.
  • Compare how smoothing parameters and trend damping affect forecast paths.
  • Avoid common errors in multi-horizon trend calculations.

Why Compare Forecasts on the Same Series?

A business-volume series can rise overall while still changing pace from period to period. A forecast that simply repeats the latest observation ignores a trend; a trend model can project that trend forward, but may extend it too aggressively. Comparing methods on the same observations makes these differences visible and helps explain what each forecast is saying.

This tutorial works with equally spaced observations and point forecasts. The examples assume no seasonal pattern and use additive changes in the series’ units. They do not assume a probability distribution for future observations: a point forecast is a central prediction from the specified method and state, not a prediction interval.

Definition: Let \(y_t\) be the observed value at time \(t\), \(\ell_t\) the estimated level after observing \(y_t\), and \(b_t\) the estimated trend per period. The forecast made at time \(t\) for \(t+h\) is denoted \(F_{t+h|t}\). In these examples, the smoothing parameters satisfy \(0\leq\alpha,\beta\leq1\). A damped-trend parameter satisfies \(0<\phi<1\).

Four Forecasts From a Common Set of Observations

The methods below differ in what they carry forward from the past. A naive forecast carries forward only the most recent observation. Simple exponential smoothing (SES) carries forward a smoothed level. Holt’s method updates both a level and a trend, projecting the trend without damping. Damped-trend smoothing also estimates a level and trend, but reduces the trend’s contribution farther into the future.

Forecast formulas: The naive forecast is \(F_{t+h|t}=y_t\) for every \(h\geq1\). Under SES, \(\ell_t=\alpha y_t+(1-\alpha)\ell_{t-1}\), and \(F_{t+h|t}=\ell_t\). Under Holt’s method, \(\ell_t=\alpha y_t+(1-\alpha)(\ell_{t-1}+b_{t-1})\), \(b_t=\beta(\ell_t-\ell_{t-1})+(1-\beta)b_{t-1}\), and \(F_{t+h|t}=\ell_t+hb_t\). Under damped-trend smoothing, replace the one-period projected level with \(\ell_{t-1}+\phi b_{t-1}\) in the level update, and use \(b_t=\beta(\ell_t-\ell_{t-1})+(1-\beta)\phi b_{t-1}\). Its forecast is \(F_{t+h|t}=\ell_t+b_t\sum_{j=1}^{h}\phi^j\).

The updates require initial level and trend values. For a short illustration, a common simple initialization is to set the first level to the first observation and the first trend to the change between the first two observations. Other initialization choices are possible; the resulting estimates and forecasts can differ. The examples state their initial values so the calculations are reproducible.

Worked Example: Comparing All Four Methods

A hypothetical distributor records weekly order volume, in hundreds of units: \(100,110,108,120,124\). Calculate the next-week forecast using the naive method, SES with \(\alpha=0.5\), Holt’s method with \(\alpha=0.5,\beta=0.4\), and damped-trend smoothing with \(\alpha=0.5,\beta=0.4,\phi=0.8\). For both trend methods, initialize \(\ell_1=100\) and \(b_1=10\). There is no seasonal component or assumed error distribution.

Naive method. The last observed value is \(y_5=124\), so \(F_{6|5}=124\).

SES. Set \(\ell_1=y_1=100\). Applying the level update at each observation gives

$$ \begin{aligned} \ell_2&=0.5(110)+0.5(100)=105,\\ \ell_3&=0.5(108)+0.5(105)=106.5,\\ \ell_4&=0.5(120)+0.5(106.5)=113.25,\\ \ell_5&=0.5(124)+0.5(113.25)=118.625. \end{aligned} $$

SES gives the same forecast for every future horizon, so \(F_{6|5}=118.625\) hundreds of units.

Holt’s method. Start with \(\ell_1=100,b_1=10\). At time 2, the one-step prediction before seeing \(y_2\) is \(100+10=110\). The update is \(\ell_2=0.5(110)+0.5(110)=110\) and \(b_2=0.4(110-100)+0.6(10)=10\). At time 3, the prediction is \(110+10=120\), so \(\ell_3=0.5(108)+0.5(120)=114\), and \(b_3=0.4(114-110)+0.6(10)=7.6\).

At time 4, the prediction is \(114+7.6=121.6\). Thus \(\ell_4=0.5(120)+0.5(121.6)=120.8\), and \(b_4=0.4(120.8-114)+0.6(7.6)=2.72+4.56=7.28\). At time 5, the prediction is \(120.8+7.28=128.08\). Therefore \(\ell_5=0.5(124)+0.5(128.08)=126.04\), and \(b_5=0.4(126.04-120.8)+0.6(7.28)=2.096+4.368=6.464\). The next-week forecast is \(F_{6|5}=126.04+6.464=132.504\) hundreds of units.

Damped trend. Start from the same initial level and trend, but multiply the trend contribution by \(\phi=0.8\). At time 2, the prediction is \(100+0.8(10)=108\). Hence \(\ell_2=0.5(110)+0.5(108)=109\), and \(b_2=0.4(109-100)+0.6(0.8)(10)=3.6+4.8=8.4\).

At time 3, the prediction is \(109+0.8(8.4)=115.72\). Then \(\ell_3=0.5(108)+0.5(115.72)=111.86\), and \(b_3=0.4(111.86-109)+0.6(0.8)(8.4)=1.144+4.032=5.176\). At time 4, the prediction is \(111.86+0.8(5.176)=116.0008\). Thus \(\ell_4=0.5(120)+0.5(116.0008)=118.0004\), and \(b_4=0.4(118.0004-111.86)+0.6(0.8)(5.176)=2.45616+2.48448=4.94064\).

At time 5, the prediction is \(118.0004+0.8(4.94064)=121.952912\). Therefore \(\ell_5=0.5(124)+0.5(121.952912)=122.976456\), and \(b_5=0.4(122.976456-118.0004)+0.6(0.8)(4.94064)=1.9904224+2.3715072=4.3619296\). The one-week forecast is \(122.976456+0.8(4.3619296)=126.46599968\), or \(126.466\) hundreds of units rounded to three decimal places.

Check. The forecasts are naive \(124\), SES \(118.625\), Holt \(132.504\), and damped trend \(126.466\), all in hundreds of units. Holt projects its latest trend fully; damping gives a smaller trend contribution. SES’s smoothed level is below the most recent observation because earlier lower values still influence it.

Why Damping Changes a Multi-Period Forecast

Holt’s forecast adds the same estimated trend \(b_t\) once for each future period. In contrast, a damped forecast adds \(\phi b_t\) in the first period, \(\phi^2b_t\) in the second, and so on. This is the reason that multiplying the current trend by the horizon is not the damped-trend calculation.

1
Write the period-by-period contributions.
At horizons one through \(h\), the trend contributions are \(\phi b_t,\phi^2b_t,\ldots,\phi^h b_t\).
2
Add the contributions.
The total trend addition is \(b_t\sum_{j=1}^{h}\phi^j\).
3
Use the finite geometric sum.
For \(\phi\ne1\), multiplying the sum by \(1-\phi\) gives \(\phi-\phi^{h+1}\), so the sum is \(\phi(1-\phi^h)/(1-\phi)\).

Consequently, \(F_{t+h|t}=\ell_t+b_t\phi(1-\phi^h)/(1-\phi)\). Since \(0<\phi<1\), \(\phi^h\) tends to zero as \(h\) increases. The total trend addition therefore approaches \(b_t\phi/(1-\phi)\), whereas Holt’s linear trend addition \(hb_t\) continues to grow with the horizon. Damping limits the long-run effect of a trend; it does not remove the trend from the near-term forecast.

Worked Example: How the Forecast Gap Grows With Horizon

A hypothetical service center has a fitted final level \(\ell_t=250\) cases per week and trend \(b_t=12\) cases per week. Compare Holt and damped-trend forecasts at horizons 1, 2, and 4, using \(\phi=0.75\). Treat the fitted state and parameters as given; no probability distribution is assumed.

Holt. The forecast is \(\ell_t+hb_t\). At horizons 1, 2, and 4 the forecasts are \(250+1(12)=262\), \(250+2(12)=274\), and \(250+4(12)=298\) cases per week.

Damped trend. The trend additions are:

$$ \begin{aligned} h=1:&\quad 12(0.75)=9,\\ h=2:&\quad 12(0.75+0.75^2)=12(1.3125)=15.75,\\ h=4:&\quad 12(0.75+0.5625+0.421875+0.31640625)\\ &\quad=12(2.05078125)=24.609375. \end{aligned} $$

The damped forecasts are therefore \(259\), \(265.75\), and \(274.609375\) cases per week at horizons 1, 2, and 4. Rounded to two decimal places, the four-period forecast is \(274.61\). The gap from Holt grows with horizon: the methods differ by 3 cases at horizon 1 and by \(298-274.609375=23.390625\) cases at horizon 4. The calculation checks the intended behavior: with a positive trend and \(\phi<1\), damping produces lower forecasts than Holt at each of these horizons.

Reading the Methods Against Recent Data

The choice among these forecasts is not just a calculation choice. It is a statement about how the recent observations should influence the next forecast and how much of an estimated trend should persist.

Worked Example: Updating a Damped Trend After a New Observation

A hypothetical repair network has a prior damped-trend state \(\ell_{t-1}=60\) jobs per week and \(b_{t-1}=8\) jobs per week. The new observation is \(y_t=70\). Update the state using \(\alpha=0.5,\beta=0.25,\phi=0.75\), then calculate the next forecast. These are point-forecast calculations; no distribution for future jobs is specified.

One-step prediction before observing the new value. The damped prediction is \(\ell_{t-1}+\phi b_{t-1}=60+0.75(8)=66\).

Updated level. The level update uses that damped prediction: \(\ell_t=0.5(70)+0.5(66)=35+33=68\).

Updated trend. The level change is \(68-60=8\), while the prior damped trend contribution is \(0.75(8)=6\). Thus \(b_t=0.25(8)+0.75(6)=2+4.5=6.5\) jobs per week.

Next forecast. \(F_{t+1|t}=68+0.75(6.5)=68+4.875=72.875\) jobs per week. The new observation raised the estimated level, while the updated trend is lower than the prior undamped trend of 8.

Common mistake. A tempting but incorrect four-period damped forecast would add \(4\phi b_t=4(0.75)(6.5)=19.5\) to the level, producing \(87.5\). That repeats the same damped increment four times. The correct trend addition is \(6.5(0.75+0.75^2+0.75^3+0.75^4)=6.5(2.05078125)=13.330078125\), giving \(68+13.330078125=81.330078125\). The damped contribution shrinks each period, so the repeated-increment calculation overstates the forecast.

Common Interpretation Traps

  • Calling SES a trend forecast. SES may respond to a sustained rise by raising its level, but its multi-step forecasts remain constant at that level. It does not project a separate trend component.
  • Confusing a fitted trend with a guaranteed future change. Holt and damped-trend forecasts project estimated changes under their update rules; they do not establish that the observed trend will continue in reality.
  • Using an observation where a level is required. The naive forecast uses \(y_t\). SES, Holt, and damped trend use their updated states, which need not equal \(y_t\).
  • Applying damping only after calculating an undamped forecast. Damping affects the trend contribution in the level update as well as the multi-step forecast. The damped method must use its own state updates.
  • Ignoring units and horizon. A trend measured in cases per week is added once per projected week in Holt’s method. Damped trend adds a different, progressively smaller amount at each future week.
Key takeaway: For a common series and stated initialization, calculate each method using its own update equations before comparing forecasts. Naive repeats the last observation, SES repeats a smoothed level, Holt extends the trend linearly, and damped trend accumulates a geometrically shrinking trend contribution.

Check Your Understanding

For each question, treat observations as equally spaced and use only the stated model and parameters.

  1. A weekly count series ends with observations \(41,45,43\). What is the one-step naive forecast? (Answer: 43)
  2. For SES, let \(\alpha=0.4\), \(\ell_1=20\), \(y_2=25\), and \(y_3=22\). Calculate \(\ell_2\) and the forecast after observing \(y_3\). (Answer: \(\ell_2=22\); forecast \(=22\))
  3. A Holt state has \(\ell_t=80\) and \(b_t=5\) units per period. What is the forecast four periods ahead? (Answer: 100 units)
  4. A damped-trend state has \(\ell_t=120\), \(b_t=10\), and \(\phi=0.5\). Calculate the three-period-ahead forecast. (Answer: 128.75)
  5. A damped-trend model has \(\ell_{t-1}=60\), \(b_{t-1}=8\), \(y_t=70\), \(\alpha=0.5\), \(\beta=0.25\), and \(\phi=0.75\). Update the level and trend, then give the next forecast. (Answer: \(\ell_t=68,\ b_t=6.5,\ F_{t+1|t}=72.875\))
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