Why Seasonal Effects Need a Scale
A recurring seasonal pattern can look different as a series changes level. A hypothetical service might receive about 20 more requests each winter than in a typical season, whether its usual volume is 100 or 300. Another service might receive 20% more requests each winter: the increase would be about 20 at a level of 100 and about 60 at a level of 300. The first pattern has a roughly constant seasonal effect; the second has an effect proportional to the level.
That distinction determines how a seasonal component is represented and removed. Additive seasonality is expressed in the same units as the observations. Multiplicative seasonality is expressed as a factor or percentage of the level. The terms “seasonally forecast” and “seasonally adjusted” refer to opposite operations, so it is important to distinguish them.
In the additive model, \(s_t\) is an amount in the units of \(y_t\). A seasonal effect of \(+12\), for example, adds 12 units to the underlying level. In a normalized multiplicative model, \(c_t\) is dimensionless. A factor of 1.12 means the season is typically 12% above the underlying level; a factor of 0.88 means it is 12% below. A multiplicative model requires positive levels and is most natural for positive observations whose seasonal variation scales with their size.
For a complete seasonal cycle, the conventional normalization is that additive effects sum to zero and multiplicative factors average to one. These constraints make \(T_t\) interpretable as the underlying level rather than a level that also contains an arbitrary seasonal offset or scale.
How to Recognize the Two Patterns
Compare the same season at different levels, allowing for any underlying trend. If the seasonal observations are separated from the level by about the same number of units, additive seasonality is a sensible description. If their ratios to the level are about the same, multiplicative seasonality is more appropriate. Equivalently, for a fixed seasonal position, additive effects tend to have stable differences, while multiplicative effects tend to have stable ratios.
| Feature | Additive seasonality | Multiplicative seasonality |
|---|---|---|
| Seasonal component | An amount, \(s_j\) | A factor, \(c_j\) |
| Comparison with level | Differences are roughly stable | Ratios are roughly stable |
| Seasonal variation as level rises | About the same size in original units | Typically grows in original units |
| Adjustment to remove seasonality | Subtract the seasonal effect | Divide by the seasonal factor |
A plot can help, but it does not replace checking the scale. The high-to-low seasonal spread may stay about the same in an additive pattern, whereas it may widen as the level increases in a multiplicative pattern. Real data can contain both changing levels and noise, so the comparison is about the recurring pattern, not exact equality in every observation.
Worked Example: Constant Seasonal Differences
A hypothetical quarterly measure has underlying levels of 80, 100, and 120 over three years. In each year, the observed quarterly values are as follows: year 1, 70, 75, 90, 85; year 2, 90, 95, 110, 105; year 3, 110, 115, 130, 125. Decide whether the pattern is additive or multiplicative, state the seasonal effects, and calculate the four seasonally forecast values at an underlying level of 140.
Method. Subtract each year's underlying level from the observation for the corresponding quarter. A roughly repeated set of differences supports an additive pattern.
For year 1, subtract 80 from each value: \(70-80=-10\), \(75-80=-5\), \(90-80=10\), and \(85-80=5\). The effects are \((-10,-5,10,5)\).
For year 2, subtract 100: \(90-100=-10\), \(95-100=-5\), \(110-100=10\), and \(105-100=5\). The effects are again \((-10,-5,10,5)\).
For year 3, subtract 120: \(110-120=-10\), \(115-120=-5\), \(130-120=10\), and \(125-120=5\). The effects are again \((-10,-5,10,5)\). They sum to \(-10-5+10+5=0\), as required for a normalized additive cycle.
The differences are constant across years, so the pattern is additive. At level 140, add each effect: \(140-10=130\), \(140-5=135\), \(140+10=150\), and \(140+5=145\). The seasonally forecast values are 130, 135, 150, and 145 units.
Check. Their deviations from 140 are \((-10,-5,10,5)\), and the deviations sum to zero. The trap avoided is treating these forecast values as adjusted values: they include the seasonal effects.
Seasonal Forecasting and Seasonal Adjustment
A seasonally forecast value includes the seasonal component. To obtain it from a level, add the additive effect or multiply by the multiplicative factor. A seasonally adjusted value has the seasonal component removed. To calculate it from an observation, subtract the additive effect or divide by the multiplicative factor.
The adjustment operations follow directly from solving each seasonal model for its underlying level. Ignoring the error term for this calculation, the additive model says \(y_t=T_t+s_j\); subtracting \(s_j\) from both sides gives \(T_t=y_t-s_j\). The multiplicative model says \(y_t=T_t c_j\); dividing both sides by \(c_j\) gives \(T_t=y_t/c_j\), provided \(c_j\ne 0\). A normalized seasonal factor is positive, so that division is well-defined.
This distinction also prevents a common direction error. Multiplication by a seasonal factor applies seasonality; division by that factor removes it. Likewise, adding an additive effect applies seasonality, while subtracting it removes it.
Worked Example: Proportional Seasonal Variation
A hypothetical quarterly volume has underlying levels of 200, 300, and 400 units over three years. The four quarterly factors are 0.8, 0.9, 1.1, and 1.2. Calculate the seasonal observations at each level and seasonally adjust an observed value of 440 from the quarter with factor 1.1.
Method. Multiply the underlying level by each seasonal factor to obtain seasonally forecast values. To adjust an observation, divide by its factor.
At level 200, the four calculations are \(200(0.8)=160\), \(200(0.9)=180\), \(200(1.1)=220\), and \(200(1.2)=240\).
At level 300, they are \(300(0.8)=240\), \(300(0.9)=270\), \(300(1.1)=330\), and \(300(1.2)=360\).
At level 400, they are \(400(0.8)=320\), \(400(0.9)=360\), \(400(1.1)=440\), and \(400(1.2)=480\). The factors have average \((0.8+0.9+1.1+1.2)/4=4/4=1\), satisfying the multiplicative normalization. For each quarter, the ratio of seasonal observation to level is the same across years, while the differences grow with the level. This supports multiplicative seasonality.
The observed value 440 belongs to a quarter with factor 1.1. Its seasonally adjusted value is \(440/1.1=400\) units. As a check, applying the factor back to the adjusted value gives \(400(1.1)=440\).
Common mistake. Multiplying \(440(1.1)=484\) does not remove seasonality; it applies the factor again. The correct adjustment is division, which returns the underlying value of 400.
A Direct Comparison of Differences and Ratios
The same seasonal observations can be examined in two ways. If the series level doubles and the seasonal deviations from that level also double, but the seasonal ratios remain unchanged, that is evidence for a multiplicative pattern. If the deviations stay similar while ratios change, that points toward an additive pattern. Neither comparison alone proves a model is perfect; random variation can make observed differences or ratios imperfect.
Worked Example: Comparing the Same Seasonal Profile at Two Levels
A hypothetical quarterly claim count has underlying levels of 100 and 200 in two periods. Its quarterly observations are 90, 100, 110, and 120 at level 100, and 180, 200, 220, and 240 at level 200. Determine the more suitable seasonal representation and adjust the second period's third-quarter observation.
Method. Calculate both the deviations from the level and the ratios to the level for each quarter.
At level 100, the differences are \(90-100=-10\), \(100-100=0\), \(110-100=10\), and \(120-100=20\). The ratios are \(90/100=0.9\), \(100/100=1.0\), \(110/100=1.1\), and \(120/100=1.2\).
At level 200, the differences are \(180-200=-20\), \(200-200=0\), \(220-200=20\), and \(240-200=40\). The ratios are \(180/200=0.9\), \(200/200=1.0\), \(220/200=1.1\), and \(240/200=1.2\).
The differences double, but the ratios repeat. The factors average \((0.9+1.0+1.1+1.2)/4=4.2/4=1.05\), so they are not normalized to mean one. Their mean is 1.05 because these stated levels do not correspond to the average level across the four quarters. For a normalized multiplicative representation, divide each factor by 1.05, giving \(6/7,\ 20/21,\ 22/21,\ 8/7\); these have mean one. The corresponding underlying level is multiplied by 1.05, so the fitted seasonal values are unchanged.
The second period's third-quarter observation is 220. Using the original ratio \(1.1\), its underlying level is \(220/1.1=200\). Equivalently, using the normalized factor \(22/21\) and normalized underlying level \(210\), the adjusted value is \(220/(22/21)=210\). This illustrates that the definition of the underlying level depends on normalization; the seasonal pattern is represented by consistent ratios, with factors conventionally centered to mean one.
Check. The normalized factors sum to \(6/7+20/21+22/21+8/7=18/21+20/21+22/21+24/21=84/21=4\). Their mean is therefore 1. The common mistake avoided is comparing differences alone and concluding the pattern is additive without checking how those differences change with the level.
Interpretation and Common Traps
The choice between additive and multiplicative seasonality is a modeling decision about scale. If a count, volume, or amount grows and its seasonal swings grow in proportion, a multiplicative representation may be appropriate. If seasonal swings remain about the same size in the original units, an additive representation may be more interpretable. A multiplicative model can be awkward or unsuitable when observations or levels can be zero or negative. An additive model does not have that restriction.
- Do not call a seasonalized value adjusted. Adding an effect or multiplying by a factor puts seasonality into a level; subtracting or dividing removes it.
- Do not compare seasonal spreads without considering the level. A larger spread at a higher level can be consistent with multiplicative seasonality.
- Check normalization. Additive effects over a cycle should sum to zero; multiplicative factors should average to one. If factors are not centered, the underlying level is scaled by their average.
- Keep the units straight. Additive effects have the units of the series; multiplicative factors are unitless.
These representations explain the seasonal component without specifying how a level or trend is updated over time. Once the scale of the seasonal effect is understood, a forecasting method can incorporate that component; the interpretation and adjustment rules remain the same.
Check Your Understanding
For each question, identify the relevant seasonal operation or pattern and show the key arithmetic before checking the answer.
- A hypothetical monthly index has an underlying level of 75 and additive effects \(-6,-2,2,6\). Calculate the four seasonally forecast values and verify the seasonal effects are centered. (Answer: 69, 73, 77, 81; the effects sum to 0.)
- A hypothetical quarterly measure is 160 units before seasonality, with the quarter's factor equal to 1.1. Calculate its seasonally forecast value. If the observed value is 220 in that same quarter, calculate its seasonally adjusted value. (Answer: forecast value \(160(1.1)=176\); adjusted value \(220/1.1=200\).)
- At underlying levels 100 and 200, the same season has observations 92 and 184. Compare the differences and ratios, and state which seasonal representation these observations support. (Answer: differences are \(-8\) and \(-16\), while both ratios are 0.92; this supports a multiplicative pattern.)
- A three-season additive pattern has observed seasonal effects \(-4,-6\), \(-1,1\), and \(5,5\) across two cycles, respectively. Find the average effect for each season and the seasonally forecast values at level 80. (Answer: average effects are \(-5,0,5\), which sum to 0; forecast values are 75, 80, 85.)
- A hypothetical observation is 288 in a season with multiplicative factor 1.2. A student calls \(288(1.2)=345.6\) the seasonally adjusted value. State the correct operation, compute the adjusted value, and verify by reapplying seasonality. (Answer: divide, \(288/1.2=240\); checking gives \(240(1.2)=288\).)