Why Filter a Stationary Process?
A time series is often transformed by combining its current value with values at nearby dates. A moving average, for example, can smooth short-run fluctuations; a difference can emphasize changes. Such transformations are linear filters. Their coefficients determine how much each time point contributes to the output.
A central question is whether filtering a weakly stationary process produces another weakly stationary process. The answer is yes under a useful regularity condition: the filter coefficients must be absolutely summable. This condition also makes the output’s mean and autocovariance well defined. We will derive the result and then calculate the covariance patterns of finite moving averages of white noise.
Definition of a Linear Filter
Let \(\{X_t\}\) be a real-valued weakly stationary process with finite second moments. Write its mean as \(\mu_X=E[X_t]\) and its autocovariance at lag \(h\) as \(\gamma_X(h)=\operatorname{Cov}(X_t,X_{t+h})\). Weak stationarity means that the mean is constant over time and the autocovariance depends only on the lag, not on the date \(t\).
A finite filter has only finitely many nonzero coefficients and is therefore absolutely summable. One-sided filters, which use the present and past values only, are included by setting \(\psi_j=0\) for \(j<0\). Allowing a two-sided filter makes the stationarity result no harder to state.
Absolute summability controls the total weight placed on the input series. In particular, it guarantees that the filtered sum converges in mean square: the centered input values all have the same finite standard deviation, and the sum of the absolute coefficient weights is finite. It also ensures that the covariance formula below is an absolutely convergent series.
How Filtering Changes the Mean and Autocovariance
The output mean is constant because it is a fixed weighted sum of the constant input mean. For the covariance, the constant \(c\) has no effect, and each pair of input terms contributes according to the lag between their dates. That lag is \(h+j-k\) when the output values are \(Y_t\) and \(Y_{t+h}\).
Write \(Y_t-E[Y_t]=\sum_j\psi_j(X_{t-j}-\mu_X)\). The constant part of the filter disappears from the covariance.
For finite partial sums, bilinearity gives a sum over coefficient pairs \(j,k\), with input covariance \(\operatorname{Cov}(X_{t-j},X_{t+h-k})\).
The input covariance is \(\gamma_X(h+j-k)\). Since \(|\gamma_X(r)|\leq\gamma_X(0)\) for every lag \(r\), the absolute sum is bounded by \(\gamma_X(0)(\sum_j|\psi_j|)^2\). Thus the covariance series converges, and the limit depends on \(h\), not on \(t\).
The inequality in Step 3 follows from the Cauchy–Schwarz inequality applied to two centered input values, each with variance \(\gamma_X(0)\). The bound permits the covariance of the finite partial filters to converge to the displayed double sum. Since the mean is constant and covariance depends only on the lag, the filtered process is weakly stationary.
This result concerns second-order properties. It does not say that filtering makes a process Gaussian or independent. For example, if the input is white noise, its values are uncorrelated at different dates, but they need not be independent. A filter can create nonzero autocovariances even when the input has none.
Worked Example: Filtering a Stationary Input
Worked Example: Filtering a Stationary Input
Suppose a weakly stationary input has mean \(\mu_X=4\), variance \(\gamma_X(0)=9\), and autocovariances \(\gamma_X(1)=3\) and \(\gamma_X(2)=1\). Consider \(Y_t=X_t+0.5X_{t-1}\). The filter is finite, so its coefficients are absolutely summable. Its output is weakly stationary by the theorem.
The mean is
The variance is the covariance at lag zero. Expanding the two terms gives
Here \(\gamma_X(-1)=\gamma_X(1)=3\), because autocovariances of a real stationary process are symmetric in the lag. For lag one, the pairs of terms give
Thus the output autocorrelation at lag one is \(\rho_Y(1)=\gamma_Y(1)/\gamma_Y(0)=8.75/14.25\approx0.6140\). Filtering has changed the variance and lag-one autocorrelation, but neither depends on the date \(t\).
Finite Moving Averages of White Noise
Suppose \(\{\varepsilon_t\}\) is white noise with mean zero, variance \(\sigma_\varepsilon^2\), and zero autocovariance at every nonzero lag. This definition does not require independence. For a finite moving-average filter
the output mean is zero. At lag \(h\geq0\), only terms that use the same white-noise date in both output values contribute to their covariance. Matching the dates gives \(j=h+k\), so
For negative lags, symmetry gives \(\gamma_Y(-h)=\gamma_Y(h)\). In particular, \(\gamma_Y(0)=\sigma_\varepsilon^2\sum_{j=0}^{q}\psi_j^2\). The output autocorrelation is \(\rho_Y(h)=\gamma_Y(h)/\gamma_Y(0)\), provided the output variance is positive. A finite filter therefore produces zero autocovariance beyond lag \(q\) when its input is white noise.
Worked Example: A Moving Average of Order One
Let \(\varepsilon_t\) be mean-zero white noise with variance \(4\), and set \(Y_t=\varepsilon_t-0.5\varepsilon_{t-1}\). The coefficients are \(\psi_0=1\) and \(\psi_1=-0.5\). The filter is finite and hence absolutely summable; the output is weakly stationary.
Its variance and lag-one autocovariance are
At every lag with absolute value greater than one, no white-noise term is shared between the two output values, so \(\gamma_Y(h)=0\). The lag-one autocorrelation is \(\rho_Y(1)=-2/5=-0.4000\). Thus a negative filter coefficient produces negative lag-one covariance in this example, even though the input has zero autocovariances at nonzero lags.
Worked Example: A Moving Average of Order Two
Let the white-noise variance be \(2\) and define \(Y_t=\varepsilon_t+0.5\varepsilon_{t-1}-0.25\varepsilon_{t-2}\). The three coefficients are \(1\), \(0.5\), and \(-0.25\), so the filtered process is weakly stationary.
The variance is
At lag one, the matched coefficient products are \(\psi_1\psi_0\) and \(\psi_2\psi_1\):
At lag two, only \(\psi_2\psi_0\) contributes, giving \(\gamma_Y(2)=2(-0.25)(1)=-0.5000\). At lags with absolute value greater than two, the covariance is zero. The corresponding autocorrelations are \(\rho_Y(1)=0.7500/2.625\approx0.2857\) and \(\rho_Y(2)=-0.5000/2.625\approx-0.1905\).
Why Absolute Summability Matters
Absolute summability is a transparent sufficient condition for a well-behaved filter. It ensures that the weighted contributions do not accumulate without bound and supports both the mean-square definition of the output and the covariance calculation. In these examples, each filter has only finitely many coefficients, so the condition holds automatically. For an infinite filter, checking \(\sum_j|\psi_j|<\infty\) is an important step before using the stationarity result.
The covariance formula also shows that filtering reshapes dependence rather than simply copying it. For general stationary input, the output covariance combines many input autocovariances with coefficient products. For white noise, all nonzero-lag input autocovariances vanish, leaving only the matching coefficient products. This is why finite moving averages of white noise have a finite covariance cutoff.
Common Mistakes
- Assuming the output covariance equals the input covariance. A filter combines values at several dates, so coefficient pairs contribute to each output lag. Use the double-sum formula or, for white noise, the matching-coefficient formula.
- Claiming independence from zero autocovariance. Zero covariance at nonzero lags means the process is uncorrelated across dates; it does not by itself imply independence.
- Forgetting the coefficient signs. Products such as \(\psi_{k+h}\psi_k\) can be negative. Keeping those signs is essential for the correct output autocovariance.
- Assuming a finite covariance cutoff for every stationary input. The cutoff beyond lag \(q\) applies to a finite moving average of white noise. A stationary input with nonzero autocovariances at larger lags can give a filtered output with nonzero autocovariances there as well.
Check Your Understanding
Use the definitions and covariance formulas to answer these questions.
- What condition on an infinite sequence of filter coefficients is sufficient for the stationarity result in this tutorial?
- Why does the constant \(c\) in a linear filter affect the output mean but not its autocovariance?
- For a finite moving average of order \(q\) applied to white noise, at what lags must the output autocovariance be zero?
- For coefficients \(\psi_0=1\) and \(\psi_1=0.25\) applied to white noise with variance \(4\), calculate the output variance and lag-one autocovariance.
- Does the finite covariance cutoff for a moving average necessarily hold when its input is a general weakly stationary process? Explain.