Why a Moving Average of White Noise Is Weakly Stationary
Summary
The problem asks whether a finite linear filter of weak white noise has a time-invariant mean and autocovariance. The attempted proof correctly aims to use the fact that the input process has a constant mean and covariance structure, but it incorrectly expands the product of two sums as a sum of matching-index terms, omitting cross-products.
The accepted answer corrects this expansion: multiplying the filtered values requires summing over every pair of input indices. Linearity of expectation then reduces the claim to showing that each input mean and each relevant pairwise product expectation is unchanged when both time indices are shifted by the same amount. Weak stationarity supplies those properties. The result is that the finite moving average is weakly stationary; the argument does not require the input observations to be independent or Gaussian.
Key ideas
- A product of two filtered sums contains cross-products across all pairs of indices.
- Linearity of expectation lets the covariance proof be checked term by term.
- Weak stationarity preserves input means and pairwise product expectations under a common time shift.
- A finite linear filter of a weakly stationary process has a time-invariant mean and autocovariance.
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Full text
# Show that $\text{Cov}[Z_t,Z_{t+h}]=\text{Cov}[Z_s,Z_{s+h}].$
# Show that $\text{Cov}[Z_t,Z_{t+h}]=\text{Cov}[Z_s,Z_{s+h}].$
> Problem: If $X\sim\text{WN}(\mu,\sigma^2).$ Let then $Z$ be the process defined by \begin{equation} Z_t=\sum_{i=0}^na_iX_{t-i} \end{equation} for some coefficients $a_1,...,a_n\in\mathbb{R}$ with $a_0=1.$ Show that $\text{Cov}[Z_t,Z_{t+h}]=\text{Cov}[Z_s,Z_{s+h}]$ and that $\mathbb{E}[Z_t]=\mathbb{E}[Z_{t+h}].$
Attempt:
\begin{align} \text{Cov}[Z_t,Z_{t+h}]&=\mathbb{E}[Z_tZ_{t+h}]-\mathbb{E}[Z_t]\mathbb{E}[Z_{t+h}]\\ &=\mathbb{E}\left[\sum_{i=0}^{n}a_i^2X_{t-i}X_{t+h-i}\right]-\mathbb{E}\left[\sum_{i=0}^{n}a_iX_{t-i}\right]\mathbb{E}\left[\sum_{i=0}^{n}a_iX_{t+h-i}\right]\\ &=\mathbb{E}\left[\sum_{i=0}^{n}a_i^2X_{s-i}X_{s+h-i}\right]-\mathbb{E}\left[\sum_{i=0}^{n}a_iX_{s-j}\right]\mathbb{E}\left[\sum_{i=0}^{n}a_iX_{s+h-j}\right]\\ &=\mathbb{E}[Z_sZ_{s+h}]-\mathbb{E}[Z_s]\mathbb{E}[Z_{s+h}] = \text{Cov}[Z_s,Z_{s+h}]. \end{align} We can just replace $t$ with $s$ in $X_t$ for a weakly stationary time series, the mean and variance do not vary with time. Same reasoning can easily be done to show the expectations.
Question: Is this correct? It feels too easy that we can just replace things inside of expectations like that.
## Answer by ir7 (score 2, accepted)
https://quant.stackexchange.com/a/63444
Note that:
$$ Z_tZ_{t+h} = \left(\sum_{i=0}^{n}a_iX_{t-i}\right) \left(\sum_{j=0}^{n}a_jX_{t+h-j}\right) \not =\sum_{i=0}^{n}a_i^2X_{t-i}X_{t+h-i} $$
Yes, as the expectation operator is linear, all we need is for:
$$ E[X_{t-i}]=E[X_{s-i}]$$
and
$$ E[X_{t-i}X_{t+h-j}] = E[X_{s-i}X_{s+h-j}] $$
to hold for all $h$, $i$, and $j$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.