Why Initialization Matters
Simple exponential smoothing updates an estimate of a series level as each new observation arrives. The update rule is straightforward, but it needs a starting value before it can be applied. That starting value is the initial level. It affects early fitted values, so a calculation is incomplete unless the initialization convention and its time index are clear.
This method is intended for a series with a level but no systematic trend or seasonal pattern. It is not a model for every time series. The smoothing parameter determines how quickly the level responds to new observations; initialization determines where the sequence of level estimates begins.
The index attached to the initial level depends on the convention. One common convention sets \(l_1=y_1\) and starts the updates at \(t=2\). Another specifies a level before an observation and updates from that point. An early-sample average convention typically sets the level at the end of a short initialization window, then begins updating with the next observation. These conventions can all be used, but their indices must not be mixed.
Two Initialization Choices
A first-observation start is convenient when there is no separate initialization window: set \(l_1=y_1\), then apply the recurrence for \(t=2,3,\ldots\). The fitted value for the second observation is \(l_1\). The first observation itself has no fitted value from an earlier smoothed level under this convention.
An early-sample average can be useful when the first few observations are intended to provide a more stable starting estimate. If the first \(m\) observations are used for initialization, set \(l_m=(y_1+\cdots+y_m)/m\), then start updating with \(y_{m+1}\). The observations used in that average are initialization data; they are not then processed again in the update recursion. This convention does not prescribe a universally best value of \(m\): the window must be stated, and the result can depend on the data chosen.
Worked Example: Starting With the First Observation
A hypothetical monthly service-count series has observations \(y_1=84\), \(y_2=92\), \(y_3=88\), and \(y_4=100\). Use \(\alpha=0.5\) and the convention \(l_1=y_1\). Find the fitted values for months 2 through 4 and the level after month 4.
Method. Initialize with \(l_1=84\). For each later month, the fitted value is the level from the preceding month, and the update is \(l_t=0.5y_t+0.5l_{t-1}\).
For month 2, the fitted value is \(l_1=84\). After observing \(y_2=92\), \(l_2=0.5(92)+0.5(84)=46+42=88\).
For month 3, the fitted value is \(l_2=88\). After observing \(y_3=88\), \(l_3=0.5(88)+0.5(88)=44+44=88\).
For month 4, the fitted value is \(l_3=88\). After observing \(y_4=100\), \(l_4=0.5(100)+0.5(88)=50+44=94\).
Check and trap avoided. Each fitted value uses the level before that month's observation, not the level after updating with it. The fitted values for months 2 through 4 are 84, 88, and 88, and the final level is 94. With \(\alpha=0.5\), each new level is the midpoint of the current observation and the previous level, as these calculations confirm.
Worked Example: Initializing With an Early-Sample Average
A hypothetical portfolio's weekly claim counts during an initialization window are 50, 56, and 54. The next three observations are 60, 52, and 58. Use the average of the first three observations as the initial level and set \(\alpha=0.25\). Find the fitted values for the next three observations and the level after the last one.
Method. The initialization window ends at time 3, so first calculate \(l_3\). Then update using observations 4, 5, and 6 in order. The fitted value for each of those observations is the level immediately before its update.
The initial level is \(l_3=(50+56+54)/3=160/3=53\frac{1}{3}\). Thus the fitted value for observation 4 is \(53\frac{1}{3}\). Updating with \(y_4=60\), \(l_4=0.25(60)+0.75(53\frac{1}{3})=15+40=55\).
The fitted value for observation 5 is \(l_4=55\). Updating with \(y_5=52\), \(l_5=0.25(52)+0.75(55)=13+41.25=54.25\).
The fitted value for observation 6 is \(l_5=54.25\). Updating with \(y_6=58\), \(l_6=0.25(58)+0.75(54.25)=14.5+40.6875=55.1875\).
Check and trap avoided. The fitted values are \(53\frac{1}{3}\), 55, and 54.25; the final level is 55.1875. The first three observations contribute to the initial average, and the recursion begins with observation 4. Applying the update again to observation 1 would use the wrong time index for this convention.
Why Different Starting Levels Become Closer
Suppose two calculations use the same observations and the same \(\alpha\), but have different levels at the same starting time. At each update, both levels include the same weighted observation. Their difference is therefore multiplied only by the weight on the previous level, \(1-\alpha\). Repeating this argument shows exactly how the initial difference carries forward.
Let \(l_t\) and \(l'_t\) use the same observation \(y_t\) and smoothing parameter \(\alpha\): \(l_t=\alpha y_t+(1-\alpha)l_{t-1}\) and \(l'_t=\alpha y_t+(1-\alpha)l'_{t-1}\).
Define \(\Delta_t=l_t-l'_t\). Then \(\Delta_t=\alpha y_t+(1-\alpha)l_{t-1}-\alpha y_t-(1-\alpha)l'_{t-1}=(1-\alpha)\Delta_{t-1}\).
After \(k\) updates from a common starting index \(r\), \(\Delta_{r+k}=(1-\alpha)^k\Delta_r\). This follows by applying \(\Delta_t=(1-\alpha)\Delta_{t-1}\) once for each of the \(k\) updates.
Because \(0\leq 1-\alpha<1\), the absolute difference between the two levels cannot grow under the same updates; it is multiplied by a factor less than one at each step. For \(\alpha=1\), one update makes the two levels equal to the latest observation. When \(\alpha\) is close to zero, the factor \(1-\alpha\) is close to one, so an initial difference can persist longer. This result concerns the effect of initialization; it does not say that either starting level is necessarily more appropriate for the data.
Worked Example: Comparing Two Starting Levels
A hypothetical daily exposure index is updated with \(\alpha=0.4\). Two analysts begin at the same time with levels \(l_0=100\) and \(l'_0=110\). The next observations are \(108\), \(96\), and \(105\). Calculate both sequences of levels and check the difference after each update.
Method. Apply the same recurrence to each starting level. The initial difference is \(l_0-l'_0=100-110=-10\), and the factor carried forward per update is \(1-\alpha=0.6\).
After observing 108, \(l_1=0.4(108)+0.6(100)=43.2+60=103.2\), and \(l'_1=0.4(108)+0.6(110)=43.2+66=109.2\). The difference is \(103.2-109.2=-6\), which is \(0.6(-10)\).
After observing 96, \(l_2=0.4(96)+0.6(103.2)=38.4+61.92=100.32\), and \(l'_2=0.4(96)+0.6(109.2)=38.4+65.52=103.92\). The difference is \(100.32-103.92=-3.6\), which is \(0.6(-6)\).
After observing 105, \(l_3=0.4(105)+0.6(100.32)=42+60.192=102.192\), and \(l'_3=0.4(105)+0.6(103.92)=42+62.352=104.352\). The difference is \(102.192-104.352=-2.16\), which is \(0.6(-3.6)\).
Check and trap avoided. After three updates, the difference must also equal \(0.6^3(-10)=-2.16\), matching the direct calculation. The sign matters: the sequence started 10 units lower, so it remains lower here, though the gap shrinks. Reversing the weight and using \(\alpha\) as the multiplier on the old level would not implement the stated smoothing rule.
Interpretation and Common Initialization Errors
Initialization is most consequential for fitted values near the start of a series. If an early-sample average is used, its value depends on both the selected observations and the length of the window. A window containing a sudden unusual observation may give a starting level that is not representative of the later level. Conversely, a very short window may be sensitive to one observation. The method specifies how to calculate a starting value; it does not remove the need to judge whether the chosen window is suitable.
A frequent exam error is to calculate an early-sample average at the end of its window, then mistakenly treat that level as if it came before the window. For example, after setting \(l_3=53\frac{1}{3}\) from observations 1 through 3, an incorrect next step would be to apply the update to \(y_1=50\), producing \(0.25(50)+0.75(53\frac{1}{3})=52.5\). That calculation processes an observation that was already incorporated in the initialization average and does not follow the stated convention. The correct next update uses \(y_4\), giving \(l_4=55\).
Another error is to report a level after an observation as that observation's fitted value. The fitted value for \(y_t\) is \(l_{t-1}\); \(l_t\) is calculated only after \(y_t\) is observed. Keeping those roles distinct is essential when checking fitted values or comparing initialization choices.
Check Your Understanding
For each problem, identify the initialization convention before applying the update rule.
- A series has \(y_1=30\), \(y_2=40\), and \(\alpha=0.2\). Set \(l_1=y_1\). What is the fitted value for \(y_2\), and what is \(l_2\)? (Answer: fitted value 30; \(l_2=32\).)
- The first two observations are 20 and 28. Use their average as \(l_2\), then update with \(y_3=32\) and \(y_4=24\) using \(\alpha=0.5\). What are \(l_2,l_3,l_4\)? (Answer: \(l_2=24\), \(l_3=28\), \(l_4=26\).)
- Two level estimates start at 70 and 80 and use the same observations, with \(\alpha=0.25\). What is their level difference after two updates? (Answer: the first level minus the second is \(-5.625\).)
- The initialization observations are 100, 110, and 90. Use their average as \(l_3\), then update with observations 120 and 100 using \(\alpha=0.3\). Give the fitted values for those two observations and the level after the second update. (Answer: fitted values 100 and 106; final level \(104.2\).)
- Two calculations have an initial level difference of 8 and use \(\alpha=0.6\) with the same subsequent observations. What is the difference after four updates? (Answer: \(8(0.4)^4=0.2048\).)