Simple Exponential Smoothing Equations

What Simple Exponential Smoothing Updates

A time series may fluctuate from period to period even when it has no persistent trend or seasonal pattern. Simple exponential smoothing (SES) represents its current underlying level with a single number. When a new observation arrives, the method revises that level by combining the observation with the level estimated before it arrived.

The method is intended for a series whose central level may change over time, but for which a trend or seasonal structure is not being modeled. It does not require that every observation equal the level: observations can be above or below it. The smoothing equation specifies how the level is updated; it does not, by itself, establish that SES is appropriate for every series.

Definition: Let $y_t$ be the observed value at time $t$, and let $L_{t-1}$ be the estimated level immediately before observing $y_t$. For a smoothing parameter $\alpha$ between 0 and 1, the simple exponential smoothing level update is

L_t=\alpha y_t+(1-\alpha)L_{t-1}.

Here $L_t$ is the updated level after observing $y_t$. Before $y_t$ is observed, the one-step forecast for it is $L_{t-1}$; after the update, the one-step forecast for the next observation is

\hat{y}_{t+1\mid t}=L_t.

The parameter $\alpha$ is the weight placed on the newest observation. The remaining weight, $1-\alpha$, is placed on the previous level. Both weights are nonnegative and sum to one when $0\leq\alpha\leq1$, so the updated level is a weighted average of the new observation and the previous level. In ordinary smoothing applications, $0<\alpha<1$; the endpoints are useful limiting cases. At $\alpha=1$, the updated level equals the newest observation. At $\alpha=0$, the level does not change.

The subscripts describe the timing. $L_{t-1}$ is available before the current observation $y_t$ is known. Once $y_t$ is recorded, the equation produces $L_t$. That updated value is then used as the forecast of $y_{t+1}$. The level is an estimate carried from one period to the next, not another observed data point.

Why the Update Has This Form

The level before observing $y_t$ is also the forecast for $y_t$. The difference $y_t-L_{t-1}$ is the forecast error, with the convention that a positive error means the observation exceeded the forecast. SES adjusts the old level by a fraction $\alpha$ of that error. This gives an equivalent form of the update.

Start from the prior level.
Before the current observation, the forecast for time $t$ is $L_{t-1}$.

Measure the forecast error.
The observed departure from that forecast is $y_t-L_{t-1}$.

Apply the smoothing fraction.
Add $\alpha$ times the error to the prior level:

L_{t-1}+\alpha(y_t-L_{t-1}) =L_{t-1}+\alpha y_t-\alpha L_{t-1} =\alpha y_t+(1-\alpha)L_{t-1} =L_t.

The final expression is the weighted-average update. Thus the two forms are algebraically identical.

This form also explains the direction of the adjustment. If $y_t>L_{t-1}$, then the error is positive and $L_t$ rises. If $y_t<L_{t-1}$, the error is negative and $L_t$ falls. For $0\leq\alpha\leq1$, the change in level is $\alpha$ times the observed error, not the entire error unless $\alpha=1$.

Using the Update and Forecast Equations

A reliable calculation follows the sequence in the definition: identify the previous level, identify the observation, apply the observation weight to $y_t$, and apply the complementary weight to $L_{t-1}$. Then interpret the result at the correct time point. The updated level is the forecast for the next period, not a forecast made before the current observation was available.

Worked Example: Updating a Daily Service-Call Level

A hypothetical service center tracks daily incoming calls. Its level estimate before Tuesday’s count is 240 calls per day. Tuesday’s count is 270 calls, and the smoothing parameter is $\alpha=0.30$. Calculate Tuesday’s updated level and the forecast for Wednesday. State the method and check the result.

Method. Apply the SES level update, then use the updated level as the next one-step forecast.

The previous level is $L_{t-1}=240$, the current observation is $y_t=270$, and the weight on the previous level is $1-\alpha=1-0.30=0.70$.

The weighted contributions are $0.30(270)=81$ calls per day from Tuesday’s observation and $0.70(240)=168$ calls per day from the previous level. Therefore,

$$ L_t=0.30(270)+0.70(240)=81+168=249. $$

The updated level is 249 calls per day. Since the observation for Tuesday has now been incorporated, the forecast for Wednesday is $\hat{y}_{t+1\mid t}=L_t=249$ calls.

Check. The forecast error for Tuesday was $270-240=30$ calls. The level should rise by $0.30(30)=9$ calls, and $240+9=249$, agreeing with the weighted-average calculation. This check also prevents the mistake of adjusting the prior level by the entire error.

Worked Example: A Negative Forecast Error in Thousands

A hypothetical portfolio report records monthly claim payments in thousands of dollars. Before the current month, the estimated level is $18.5$ thousand dollars. The observed payment is $16.0$ thousand dollars, and $\alpha=0.60$. Find the updated level and the next month’s one-step forecast. Keep the units explicit.

Method. Use the observation weight $0.60$ and the previous-level weight $1-0.60=0.40$. The inputs are in thousands of dollars, so the result will also be in thousands of dollars.

The contribution from the current observation is $0.60(16.0)=9.6$ thousand dollars. The contribution from the previous level is $0.40(18.5)=7.4$ thousand dollars. Thus,

$$ L_t=0.60(16.0)+0.40(18.5)=9.6+7.4=17.0 $$

The updated level is $17.0$ thousand dollars, or $\$17{,}000$. The forecast for the next month is therefore $17.0$ thousand dollars.

Check. The observed value was $16.0-18.5=-2.5$ thousand dollars below the prior level. The adjustment is $0.60(-2.5)=-1.5$ thousand dollars, so $18.5-1.5=17.0$. The level moves downward, as it should, and remains between the observation and prior level.

Worked Example: Updating Through Two Observations

A hypothetical weekly workload series starts with level $L_0=100$ tasks. Use $\alpha=0.25$. The first observed workload is $y_1=108$, and the second is $y_2=96$. Calculate both updated levels, identify each one-step forecast before its observation arrives, and give the forecast for the third week.

Method. Apply the same update in chronological order. Each new update uses the level produced by the preceding observation.

Before week 1 is observed, its forecast is $L_0=100$ tasks. After observing $y_1=108$, the weight on the previous level is $1-0.25=0.75$, so

$$ L_1=0.25(108)+0.75(100)=27+75=102. $$

The week 1 forecast error is $108-100=8$, and the adjustment is $0.25(8)=2$. Thus $L_1=100+2=102$, confirming the calculation. Before week 2 is observed, its one-step forecast is $L_1=102$.

After observing $y_2=96$, use $L_1=102$, not $L_0=100$, as the previous level:

$$ L_2=0.25(96)+0.75(102)=24+76.5=100.5. $$

The week 2 forecast error is $96-102=-6$, giving an adjustment of $0.25(-6)=-1.5$. Therefore $102-1.5=100.5$, which checks the update. The forecast for week 3 is $L_2=100.5$ tasks.

Common Mistakes and Interpretation

The update is simple, but its timing and weights are easy to misread. In particular, the symbol $\alpha$ always multiplies the new observation in the stated equation. The complementary weight $1-\alpha$ multiplies the previous level. Reversing these weights changes the method.

For the claim-payment calculation above, a reversed-weight calculation would be $0.40(16.0)+0.60(18.5)=6.4+11.1=17.5$ thousand dollars. That is not the SES update with $\alpha=0.60$: it assigns only 0.40 to the new observation and 0.60 to the previous level. The correct calculation is $0.60(16.0)+0.40(18.5)=17.0$ thousand dollars.

Using the wrong prior value. When processing $y_t$, use $L_{t-1}$, the most recently updated level, not an older level such as $L_{t-2}$.
Forecasting at the wrong time. Before observing $y_t$, the one-step forecast is $L_{t-1}$. After updating with $y_t$, the forecast for $y_{t+1}$ is $L_t$.
Applying the full forecast error. SES changes the level by $\alpha(y_t-L_{t-1})$, not by the entire error unless $\alpha=1$.
Mixing units. If observations and levels are recorded in thousands of dollars, the computed level is in thousands of dollars. Convert to dollars only after the calculation if needed.
Using SES without considering the series structure. A single-level update does not explicitly account for a trend or seasonal pattern. If these structures matter, this simple specification may not be suitable.

Key takeaway: At time $t$, combine the new observation $y_t$ with the prior level $L_{t-1}$: $L_t=\alpha y_t+(1-\alpha)L_{t-1}$. The result $L_t$ is the updated estimate and the one-step forecast for the next period.

Check Your Understanding

For each problem, use the stated observation weight and keep the timing of the forecast clear.

A hypothetical repair desk has prior level $L_{t-1}=70$ jobs per day. The current observation is $85$, with $\alpha=0.20$. Find the updated level and the next one-step forecast. (Answer: $73$ jobs per day for both.)
A hypothetical monthly payment series has prior level $40$ thousand dollars and current observation $32$ thousand dollars. With $\alpha=0.75$, find the updated level. (Answer: $34$ thousand dollars.)
A hypothetical weekly intake series begins at $L_0=50$, with $\alpha=0.40$. Its first two observations are $55$ and $45$. Find $L_1$, $L_2$, and the forecast for week 3. (Answer: $L_1=52$, $L_2=49.2$, and the week 3 forecast is $49.2$.)
Before the current observation arrives, a one-step forecast is $26$. The observed value is $31$, and $\alpha=0.30$. Find the updated level and state which period it forecasts. (Answer: The updated level is $27.5$, and it forecasts the next period after the observation.)
A hypothetical daily count series starts with $L_0=12$, uses $\alpha=0.50$, and has observations $10$ and $14$. Calculate both updates in order and the forecast after the second observation. (Answer: $L_1=11$, $L_2=12.5$, and the next one-step forecast is $12.5$.)

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