Why Exponential Smoothing Uses Geometric Weights

Why the Recursion Creates Geometric Weights

Simple exponential smoothing updates a level by combining the latest observation with the previous level. That update is compact, but it can hide how much each earlier observation still contributes. Expanding the recursion makes those contributions explicit: the most recent observation has the largest weight, and each step further back multiplies the weight by the same factor.

The expansion also shows why the initial level matters, especially early in a series. Its contribution declines over time, but it is part of the level unless the smoothing parameter makes its weight zero. The results below are algebraic: they do not require a particular probability distribution for the observations.

Definition: Let $y_t$ be the observation at time $t$, let $\ell_t$ be the smoothed level after observing $y_t$, and let $\ell_0$ be the specified initial level. Simple exponential smoothing uses a fixed smoothing parameter $0<\alpha<1$ and the recursion $$ \ell_t=\alpha y_t+(1-\alpha)\ell_{t-1}, \qquad t=1,2,\ldots. $$ The one-step-ahead forecast made after time $t$ is $\ell_t$. Write $q=1-\alpha$; then $0<q<1$.

Each update gives the new observation weight $\alpha$ and retains a fraction $q$ of the previous level. Since that previous level already combines earlier observations, repeatedly substituting the recursion gives older observations successively smaller powers of $q$.

Expanding the Recursion

Substitute the update for $\ell_{t-1}$ into the formula for $\ell_t$, then continue substituting until the initial level appears. The result is

\ell_t =\alpha y_t+\alpha qy_{t-1}+\alpha q^2y_{t-2} +\cdots+\alpha q^{t-1}y_1+q^t\ell_0 =\alpha\sum_{j=0}^{t-1}q^j y_{t-j}+q^t\ell_0.

Thus $y_t$, the latest observation, has weight $\alpha$; $y_{t-1}$ has weight $\alpha q$; and $y_{t-j}$ has weight $\alpha q^j$. The initial level has weight $q^t$. These statements use $j=0,\ldots,t-1$, so the oldest observed value in this finite series is $y_1$.

Theorem: For fixed $0<\alpha<1$, the level after $t$ observations is $$ \ell_t=\alpha\sum_{j=0}^{t-1}(1-\alpha)^j y_{t-j} +(1-\alpha)^t\ell_0. $$ The weights on the observed values sum to $1-(1-\alpha)^t$, and the weight on the initial level is $(1-\alpha)^t$. Therefore, all weights together sum to one.

The weight-sum result follows from the finite geometric series:

\sum_{j=0}^{t-1}\alpha q^j =\alpha\frac{1-q^t}{1-q} =1-q^t, \qquad (1-q^t)+q^t=1,

where $1-q=\alpha$. This is also a useful check on an expansion: a missing term or an incorrect exponent will often make the weights fail to sum to one.

Why the Formula Holds

The formula can be justified by induction, which verifies the pattern for every positive integer $t$, not only for the first few updates.

Base case.
At $t=1$, the recursion gives $\ell_1=\alpha y_1+q\ell_0$. This is exactly the expansion with one observed value and initial-level weight $q^1$.

Assume the expansion at time $t$.
Suppose $\ell_t=\alpha\sum_{j=0}^{t-1}q^j y_{t-j}+q^t\ell_0$.

Apply one more update.
Using $\ell_{t+1}=\alpha y_{t+1}+q\ell_t$, substitute the assumed expansion: $$ \ell_{t+1} =\alpha y_{t+1}+\alpha\sum_{j=0}^{t-1}q^{j+1}y_{t-j}+q^{t+1}\ell_0. $$ The first term is the weight-$\alpha$ contribution from the newest observation. In the sum, set $k=j+1$; its observations become $y_{t+1-k}$, with weights $\alpha q^k$ for $k=1,\ldots,t$. Together the terms give $\alpha\sum_{k=0}^{t}\!q^k y_{t+1-k}+q^{t+1}\ell_0$, the required expansion at time $t+1$.

The induction establishes the expression for every $t\geq1$. The geometric pattern is a direct consequence of retaining the same fraction $q$ of the previous level at every update.

Worked Examples

Worked Example: Expand and Check a Three-Observation Level

A hypothetical monitoring series has initial level $\ell_0=100$, smoothing parameter $\alpha=0.25$, and observations $y_1=108$, $y_2=96$, and $y_3=112$. Find $\ell_3$ by recursion and by the geometric-weight expansion. The values are fixed inputs for this calculation; no probability distribution is assumed.

Method: recursion. Here $q=1-\alpha=0.75$. Update one time at a time:

\ell_1=0.25(108)+0.75(100)=27+75=102,

\ell_2=0.25(96)+0.75(102)=24+76.5=100.5,

\ell_3=0.25(112)+0.75(100.5)=28+75.375=103.375.

Method: expanded weights. At time 3, $y_3,y_2,y_1,\ell_0$ have respective weights $0.25$, $0.25(0.75)=0.1875$, $0.25(0.75)^2=0.140625$, and $(0.75)^3=0.421875$. Their weighted contributions are

0.25(112)=28,\quad 0.1875(96)=18,\quad 0.140625(108)=15.1875,\quad 0.421875(100)=42.1875.

Adding them gives $28+18+15.1875+42.1875=103.375$, agreeing with the recursive calculation. The weights also check: $0.25+0.1875+0.140625+0.421875=1$. The common mistake avoided here is to omit the initial-level contribution.

Worked Example: Track the Weight of Each Observation

A hypothetical series starts at $\ell_0=50$, uses $\alpha=0.20$, and records $y_1=60$, $y_2=55$, $y_3=70$, and $y_4=65$. Find the final level and identify the share of that level attributable to the initial value. Treat the observations as given values, with no distributional assumption.

Here $q=0.80$. The weights at time 4, listed from newest observation to oldest and then the initial level, are

y_4:0.20,\quad y_3:0.20(0.80)=0.16,\quad y_2:0.20(0.80)^2=0.128,\quad y_1:0.20(0.80)^3=0.1024,\quad \ell_0:(0.80)^4=0.4096.

Multiply each value by its weight:

0.20(65)=13,\quad 0.16(70)=11.2,\quad 0.128(55)=7.04,\quad 0.1024(60)=6.144,\quad 0.4096(50)=20.48.

The sum is $13+11.2+7.04+6.144+20.48=57.864$, so $\ell_4=57.864$. The initial value contributes $20.48$ to the level, and its weight is $0.4096$. As a check, the observation weights sum to $0.5904$; adding $0.4096$ gives one. This method avoids assigning the same weight to every observation: the weights depend on how many updates separate an observation from the current time.

Worked Example: Compare Two Initial Levels

Two calculations use the same hypothetical observations $y_1=30$, $y_2=45$, and $y_3=36$, and the same $\alpha=0.40$, but use initial levels $\ell_0^{(A)}=20$ and $\ell_0^{(B)}=50$. Find both time-3 levels and verify how much of the initial difference remains. No probability distribution is needed.

The initial levels differ by $50-20=30$, and $q=0.60$. For calculation A:

\ell_1^{(A)}=0.4(30)+0.6(20)=24,\qquad \ell_2^{(A)}=0.4(45)+0.6(24)=32.4,

\ell_3^{(A)}=0.4(36)+0.6(32.4)=14.4+19.44=33.84.

For calculation B:

\ell_1^{(B)}=0.4(30)+0.6(50)=42,\qquad \ell_2^{(B)}=0.4(45)+0.6(42)=43.2,

\ell_3^{(B)}=0.4(36)+0.6(43.2)=14.4+25.92=40.32.

The final difference is $40.32-33.84=6.48$. The initial-level weight after three updates is $q^3=0.60^3=0.216$, so the predicted remaining difference is $0.216(30)=6.48$, as calculated. The observation contributions are identical in the two calculations; only the weighted initial value differs. This comparison shows why an initial level has an effect that decays rather than disappearing after the first update.

Interpreting the Weights and Avoiding Traps

For $0<\alpha<1$, the factor $q=1-\alpha$ lies between zero and one. Consequently, each step further into the past multiplies an observation’s weight by $q$. The weights decline geometrically as observations get older; they do not decline by a fixed amount. At a fixed time $t$, the observations’ total weight is less than one whenever the initial-level weight is positive. The remainder belongs to $\ell_0$.

The initial-level influence is exactly $q^t$. If two calculations have the same $\alpha$ and observations but different initial levels, their difference after $t$ updates is $q^t$ times their original difference. This follows either from the expansion, since the observation terms cancel, or by subtracting the two recursions at each update. It tends toward zero as $t$ increases, but the rate of decline depends on $q$. The limiting cases are also informative: if $\alpha=1$, the level equals the latest observation and the initial value has zero weight after the first update; if $\alpha=0$, the level remains $\ell_0$ and observations receive no weight. The geometric-decay discussion assumes the usual interior range $0<\alpha<1$.

Reversing the order. The newest value gets $\alpha$, not $\alpha q^{t-1}$. The power of $q$ increases as the observation gets older.
Dropping the initial level. At a finite time $t$, its weight is $q^t$, generally not zero. An expansion that includes only observations will usually have weights summing to less than one.
Confusing a weight with a contribution. The contribution of $y_{t-j}$ is its weight $\alpha q^j$ multiplied by its value. The weight alone is not the amount added to the level.

For example, in the first worked example, a tempting but incorrect calculation is $0.25(112+96+108)=79$. It assigns weight $0.25$ to every observed value and omits $\ell_0$. Those weights do not match the recursion: the correct observation weights are $0.25$, $0.1875$, and $0.140625$, while the initial value has weight $0.421875$. Including the corresponding contributions gives $103.375$, not $79$.

Key takeaway: After $t$ updates, the observation $j$ periods old has weight $\alpha(1-\alpha)^j$, and the initial level has weight $(1-\alpha)^t$. The weights are geometric, ordered from largest for the most recent observation to smallest for the oldest, and they sum to one when the initial-level weight is included.

Check Your Understanding

Work each problem from the stated recursion or its expanded form. No probability distribution is assumed unless one is explicitly given.

A hypothetical series has $\ell_0=40$, $\alpha=0.30$, $y_1=50$, and $y_2=30$. Find $\ell_2$ and the weight and contribution of $\ell_0$. (Answer: $\ell_2=39.1$; initial-level weight $0.49$, contribution $19.6$.)
For $\alpha=0.20$, list the weights on $y_3,y_2,y_1,\ell_0$ in $\ell_3$, in that order, and verify their sum. (Answer: $0.20, 0.16, 0.128, 0.512$; sum $1$.)
A series uses $\ell_0=20$, $\alpha=0.50$, and observations $24,18,22$, in that order. Find $\ell_3$ using the expanded weights. (Answer: $\ell_3=21$.)
Two versions of a smoothing calculation have initial levels that differ by 80, use the same observations, and have $\alpha=0.10$. What is the difference between their levels after five updates? (Answer: $80(0.90)^5=47.2392$.)
With $\alpha=0.50$, after how many updates does the initial-level weight equal $0.125$? State the weight on the observations collectively at that time. (Answer: 3 updates; the initial-level weight is $0.125$, and the observations collectively have weight $1-0.125=0.875$.)

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