Moving averages

Averages over a sliding window of consecutive observations that smooth out short-term noise; trailing windows use only the past, centred windows estimate the trend.

\text{MA}_t = \frac{1}{k}\sum_{j=0}^{k-1} y_{t-j}

Trailing moving average of order k

$y_t$: observation at time $t$
$k$: window length (order); the first $k - 1$ positions have no value

\hat T_t = \frac{1}{m}\sum_{j=-(m-1)/2}^{(m-1)/2} y_{t+j} \;\; (m \text{ odd}), \qquad \hat T_t = \frac{1}{2m}y_{t-m/2} + \frac{1}{m}\sum_{j=-m/2+1}^{m/2-1} y_{t+j} + \frac{1}{2m}y_{t+m/2} \;\; (m \text{ even})

Centred moving average; for even m the 2×m-MA

$m$: window length; set it to the seasonal period to remove seasonality
$\hat T_t$: estimated trend-cycle at time $t$ ; missing at both ends of the series

When to use

Trailing MA: monitoring the recent level (e.g. 7-day average of daily sales) or as a naive forecast — it uses only data available at time $t$ .
Centred MA: estimating the trend in a historical series, and as the first step of classical decomposition. With $m$ equal to the seasonal period (12 for monthly, 4 for quarterly) the seasonal pattern averages out; even periods need the 2×m-MA so the window is centred.

How to interpret

Larger windows give smoother curves but react more slowly to genuine changes; a trailing MA lags the series by about $(k-1)/2$ periods.
The values at the ends are missing by construction: no data exist beyond the end of the series to centre the window.

Common mistakes

Using a centred MA in a forecasting or monitoring setting: it looks into the future.
A window that is not a multiple of the season, which leaves seasonal ripples in the "trend".
Reading the smoothed peak as the actual peak: smoothing flattens peaks and troughs.

Example

Series 12, 15, 14, 18, 17, 21, 20, 24.

Trailing 3-MA at $t = 3$ : $(12 + 15 + 14)/3 = 13.67$ ; at $t = 8$ : 21.67.
Centred 3-MA at $t = 2$ : 13.67 (same window, placed in the middle).
2×4-MA at $t = 3$ : $(12/2 + 15 + 14 + 18 + 17/2)/4 = 15.375$ ; it exists only for $t = 3, \ldots, 6$ .