Show that $\text{Cov}[Z_t,Z_{t+h}]=\text{Cov}[Z_s,Z_{s+h}].$

Question

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 1

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 $j$.