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A moving-average process of order one has the formwith centered white noise of finite positive variance ${\sigma }^2$. Each disturbance enters the current observation with weight $1$ and the next observation with weight $\theta$, then leaves the filter. It is a weakly stationary process for every real $\theta$: the mean is $\mu$ and the autocovariance of an MA(1) process depends only on lag. Strong white noise additionally makes it a strictly stationary process. No $|\theta |<1$ restriction is needed for stationarity.

For a worked observation calculation, take $\mu =10$, $\theta =0.7$, and consecutive noise values ${\epsilon }_{t-1}=-1$, ${\epsilon }_t=2$, ${\epsilon }_{t+1}=0$. Then $X_t=11.3$ and $X_{t+1}=11.4$; the same disturbance $2$ appears in both. If ${\sigma }^2=1$, the variance is $1.49$, the lag-one autocovariance is $0.7$, and the lag-one autocorrelation is $70/149\simeq 0.4698$. All higher-lag autocorrelations vanish, as shown in autocorrelation.

An MA(1) is useful for brief dependence caused by overlapping disturbances or measurements. Its name refers to a weighted moving sum of noise; its weights need not be positive or sum to one, and it differs from the data smoother in seasonal extraction by a centred moving average. Under strong white noise, observations more than one interval apart depend on disjoint independent noise families and are independent. Under weak white noise, the same cutoff establishes only uncorrelated random variables.

Invertibility of a moving-average model is a separate question: can the specified driving noise be recovered by a stable linear filter of present and past observations? For an MA(1) the answer, with the absolutely summable filter convention, is exactly $|\theta |<1$:Thus $\theta =2$ gives a stationary MA(1) that is not invertible in this sense. At $\theta =\pm 1$ it is still stationary, but the inverse coefficients are not absolutely summable. The case $\theta =0$ is simply shifted white noise. See Bartlett, lecture 5: MA(1) and stable invertibility.

For a weakly stationary process with finite second moments, let $\mu =\mathbb{E}{X}_t$ and $\gamma (h)=\mathrm{Cov}(X_{t+h},X_t)$, independent of $t$. If $\gamma (0)>0$, its autocorrelation function is $\rho (h)=\gamma (h)/\gamma (0)$. Given observations $x_1,\dots ,x_N$, one usual sample autocovariance function isThe correlogram plots $\hat{\rho }(h)$ against lag $h$. The partial autocorrelation function at lag $h$ is the correlation between $X_t$ and $X_{t-h}$ after linearly projecting each onto the intervening observations. Its sample version can be calculated by solving the Yule-Walker regression equationsthen taking the last coefficient ${\hat{\varphi }}_{h,h}$. Equivalently, use the corresponding residual correlation from the sample regressions, with a consistent covariance convention.

For a minimal causal autoregressive model of order $p$, the population partial autocorrelation function is zero after lag $p$, while the autocorrelation function follows an exponentially decaying or damped oscillatory tail. For a moving-average model of order $q$, its autocovariance and hence autocorrelation are zero after lag $q$, because observations farther apart share no driving-noise terms; its partial correlations usually tail off. These order identification by autocorrelation cutoffs patterns suggest AR or MA candidate orders. Sample cutoffs are approximate and should be judged with sampling uncertainty and residual diagnostics; the familiar white-noise bands near $\pm 1.96/\sqrt{N}$ are a null approximation, not universal confidence limits for a dependent series.

For the trend problem, write a symmetric estimator asReproducing arbitrary constant and quadratic trends requires $2a+2b+c=1$ and $8a+2b=0$; the linear-trend condition holds automatically by symmetry. These constraints follow by expanding $T_{t+j}=T_t+j(b_0+2c_{0}t)+c_0{j}^2$ for a trend $T_t=a_0+b_{0}t+c_0{t}^2$. Thus $b=-4a$, $c=1+6a$. Since the errors are uncorrelated and the estimator is unbiased, its mean squared error isThe strictly convex quadratic is minimized at $a=-3/35$, giving the five-point symmetric quadratic trend smootherThis weighted moving average allows negative coefficients; a convex average could not cancel the quadratic moment.

For three points, symmetry gives weights $(a,b,a)$. Unbiasedness for every quadratic requires $2a+b=1$ and $2a=0$, so the only solution is $(0,1,0)$. Thus there is no nontrivial three-point quadratic-preserving smoother. The literal claim that there is no unbiased estimator at all is false, since ${\hat{T}}_t=X_t$ is unbiased with error variance ${\sigma }^2$. This is the three-point quadratic reproduction forces the identity filter distinction.

For seasonal extraction by a centred moving average, use the additive decomposition $X_t=T_t+S_t+{\epsilon }_t$, where $S_t$ has period $m$ and its values over one period sum to zero. An average covering a full period removes the seasonal component and approximately retains a slowly changing trend. If $m$ is odd, take the centred length-$m$ average; if $m$ is even, centre it by averaging two adjacent length-$m$ averages, giving half weights at the two endpoints. Subtract this trend estimate from the data, average residuals within each seasonal phase over different cycles, and centre the resulting phase estimates to sum to zero. For a multiplicative decomposition, use ratios and normalize the phase estimates to average one.

Successive moving averages compose by convolution of their weights. Even an initially uncorrelated noise series becomes correlated: for weights $w_j$, the filtered noise has covariance ${\sigma }^2{\sum }_j{w}_j{w}_{j+h}$. Further smoothing changes this covariance again and can turn random variation into apparently persistent patterns while obscuring genuine short-scale features. Spectrally, repeated use of a transfer function $H$ multiplies the spectrum by $|H{|}^{2r}$ after $r$ applications, so repeated smoothing is not harmless repetition of the same estimate.

In simple exponential smoothing, choose $0<\alpha <1$ and update ${\ell }_t=\alpha X_t+(1-\alpha ){\ell }_{t-1}$. Iteration gives ${\ell }_t=\alpha {\sum }_{j=0}^{t-1}(1-\alpha {)}^j{X}_{t-j}+(1-\alpha {)}^t{\ell }_0$, so older observations receive exponentially decreasing weights. Forecast ${\hat{X}}_{t+h\mid t}={\ell }_t$ for every $h\ge 1$. This is appropriate for a locally stable level with no persistent trend or seasonality.

For a local linear trend, double exponential smoothing supplies level and slope. Holt's linear trend method usesAnother standard convention is Brown's double exponential smoothing: smooth twice with the same gain, $s_t^{(1)}=\alpha X_t+(1-\alpha )s_{t-1}^{(1)}$ and $s_t^{(2)}=\alpha s_t^{(1)}+(1-\alpha )s_{t-1}^{(2)}$, then put $a_t=2s_t^{(1)}-s_t^{(2)}$, $b_t=\alpha (s_t^{(1)}-s_t^{(2)})/(1-\alpha )$ and forecast $a_t+h{b}_t$. These variants handle a nonseasonal trend; seasonal effects need separate modelling, and a long-horizon linear extrapolation assumes the current local slope persists.