14.2 Exponential smoothing
Suppose \(x_1, x_2, \dots , x_n\) is a non-seasonal series with no trend; series with trend or seasonal patterns are dealt with below. It is natural to forecast \(X_{n+1}\) by a weighted sum of past observations, \[ \hat x(n, 1) = c_0x_n + c_1x_{n-1} + c_2x_{n-2} + \dots , \] giving the most recent values the most weight. Geometric weights are used, \[ c_i = \alpha (1 - \alpha )^i , \qquad i = 0, 1, 2, \dots , \] where \(\alpha \) is a constant with \(0 < \alpha < 1\). They add to \(1\), since \(\sum _{i\ge 0}\alpha (1 - \alpha )^i = \alpha /(1 - (1 - \alpha )) = 1\), so the forecast is a genuine weighted average. Then \begin {align*} \hat x(n, 1) &= \alpha x_n + \alpha (1 - \alpha )x_{n-1} + \alpha (1 - \alpha )^2x_{n-2} + \dots \\ &= \alpha x_n + (1 - \alpha )\bigl [\alpha x_{n-1} + \alpha (1 - \alpha )x_{n-2} + \alpha (1 - \alpha )^2x_{n-3} + \dots \bigr ] . \end {align*}
The bracket is \(\hat x(n - 1, 1)\), so \[ \hat x(n, 1) = \alpha x_n + (1 - \alpha )\hat x(n - 1, 1) . \] Starting from \(\hat x(1, 1) = x_1\), this equation can be used recursively to compute the forecasts. The method is called simple exponential smoothing. It can also be written \begin {align*} \hat x(n, 1) &= \alpha x_n + (1 - \alpha )\hat x(n - 1, 1) \\ &= \alpha \bigl [x_n - \hat x(n - 1, 1)\bigr ] + \hat x(n - 1, 1) \\ &= \hat x(n - 1, 1) + \alpha e_n , \end {align*}
where \(e_n = x_n - \hat x(n - 1, 1)\) is the error in the forecast of \(x_n\): each new forecast corrects the old one by a fraction \(\alpha \) of its error. The value of \(\alpha \) depends on the properties of the series. Small values (\(0.1 \le \alpha \le 0.3\) is common) give smooth forecasts that respond slowly; large values follow the latest observations closely. In practice \(\alpha \) is chosen to minimise the sum of squared one-step errors \(\sum e_t^2\) over the observed series. Since \(\hat x(n, k) = \hat x(n, 1)\) for all \(k\), the forecasts are flat.
Proposition 14.2.1. Simple exponential smoothing with constant \(\alpha \) gives the minimum mean squared error forecasts of the ARIMA(0,1,1) process \(X_t = X_{t-1} + Z_t - (1 - \alpha )Z_{t-1}\).
Proof. With \(\beta = -(1 - \alpha )\), the model is \((1 - B)X_t = (1 + \beta B)Z_t\), invertible since \(|\beta | < 1\). As in Example 10.2.2, \[ \pi (B) = \frac {1 - B}{1 + \beta B} = (1 - B)\sum _{j\ge 0}(-\beta )^jB^j = 1 - \sum _{j\ge 1}(1 + \beta )(-\beta )^{j-1}B^j , \] so \(X_t = \sum _{j\ge 1}\alpha (1 - \alpha )^{j-1}X_{t-j} + Z_t\), using \(1 + \beta = \alpha \) and \(-\beta = 1 - \alpha \). Given the past, \(Z_{n+1}\) has mean \(0\) and is unpredictable, so the best forecast of \(X_{n+1}\) is \(\sum _{j\ge 1}\alpha (1 - \alpha )^{j-1}x_{n+1-j} = \sum _{i\ge 0}\alpha (1 - \alpha )^ix_{n-i}\), which is the exponential smoothing forecast. (That the conditional mean is the best forecast is Theorem 14.4.1 below.) □
So exponential smoothing, though introduced as a heuristic, is optimal for a particular ARIMA model. This explains both why it often works well and when it will not: for a series that is not close to ARIMA(0,1,1), a fitted Box–Jenkins model should do better.
Example 14.2.2. Weekly demand for a product was \(20, 23, 21, 25, 24, 26, 25, 28\). Apply simple exponential smoothing with \(\alpha = 0.3\) and \(\alpha = 0.6\), compare the sums of squared one-step errors, and forecast the next week’s demand.
Solution. Start with \(\hat x(1, 1) = 20\). With \(\alpha = 0.6\): \(e_2 = 23 - 20 = 3\), so \(\hat x(2, 1) = 20 + 0.6(3) = 21.8\); then \(e_3 = 21 - 21.8 = -0.8\) and \(\hat x(3, 1) = 21.32\); and so on. The one-step forecasts for weeks \(2\) to \(8\) are:
| Week | \(x_t\) | \(\alpha = 0.3\) | \(\alpha = 0.6\) |
| 2 | 23 | 20.00 | 20.00 |
| 3 | 21 | 20.90 | 21.80 |
| 4 | 25 | 20.93 | 21.32 |
| 5 | 24 | 22.15 | 23.53 |
| 6 | 26 | 22.71 | 23.81 |
| 7 | 25 | 23.69 | 25.12 |
| 8 | 28 | 24.09 | 25.05 |
| next | 25.26 | 26.82 |
The sums of squared one-step errors are \(56.87\) for \(\alpha = 0.3\) and \(36.92\) for \(\alpha = 0.6\). The series is rising, and the small \(\alpha \) lags behind it. With \(\alpha = 0.6\) the forecast for next week is \(26.82\). (The steady rise suggests that a method allowing for trend, such as Holt’s, would do better still.)
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