# 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 $$$$Z_t=\sum_{i=0}^na_iX_{t-i}$$$$ for some coefficients $$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.

• Yes, it is a trivial proof. You are allowed to change the name of the "bound variable" t to another name such as s. (Just like renaming a local variable inside a computer program does not affect the value the program returns). Apr 17 at 18:37

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 $$j$$.

• Ah yes I missed that part. Will change. However the expectations you have written, that follows by the stationary property right? Nothing I need to prove. Apr 17 at 20:07
• Yes, $(t+h-j) - (t-i) = (s+h-j) - (s-i)=h+i-j$ is the common shift (for the covariance stationarity equality).
– ir7
Apr 17 at 20:19