Weak Stationarity of a Linear Process with a Two-Period Lag
Summary
The document examines whether a process formed from a constant and two terms of an independent, mean-zero, equal-variance normal sequence is weakly stationary. Its central lesson is to use covariance correctly: subtract the product of the means, or apply covariance bilinearity, rather than treating the raw product expectation as covariance. The constant contributes nothing to covariance.
Expanding the lagged terms shows that covariance is nonzero only when the underlying noise terms share an index. Consequently, the autocovariance depends on the lag but not on calendar time: at lag zero it combines the two squared coefficients times the innovation variance; at lags plus or minus two it is the product of the coefficients times that variance; at other lags it vanishes. The mean is constant, so these properties establish weak stationarity. The argument relies on independent innovations with common variance; the normality assumption is stronger than needed for this covariance calculation. The question’s initial calculation misses the covariance centering and therefore gives an incorrect autocovariance.
Key ideas
- Weak stationarity requires a time-invariant mean and covariance that depends only on lag.
- Covariance equals the joint product expectation minus the product of the means.
- Independent innovations contribute covariance only when the same innovation appears in both process values.
- For the two-lag construction, nonzero autocovariance occurs at lag zero and at lags matching the offset.
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Full text
# Show that $\text{Cov}[X_r,X_s]=\text{Cov}[X_{r+h},X_{s+h}]$ for $X_t=a+bZ_t+cZ_{t-2}.$
# Show that $\text{Cov}[X_r,X_s]=\text{Cov}[X_{r+h},X_{s+h}]$ for $X_t=a+bZ_t+cZ_{t-2}.$
> Problem: Let $\{Zt\}$ be a sequence of independent normal random variables, each with mean $0$ and variance $\sigma^2$, and let $a$, $b$, and $c$ be constants. Is $X_t=a+bZ_t+cZ_{t-2}$ a (weakly) stationary process?
I want to do this in two ways. First one is to just calculate the autocovariance function $\gamma_X(h)$ and the mean function $\mu_X(t)$ and show that both are independent of time. The other way is to show that $\gamma_X(r,s)=\gamma_X(r+h,s+h).$
Method 1: We have since $\mu_Z(t)=0$ we have
- $E[X_t]=E[a+bZ_t+cZ_{t-2}]=a+bE[Z_t]+cE[Z_{t-2}] = a+b\mu_Z(t)+c\mu_Z(t) = a.$
- And for the ACF \begin{align} \gamma_{X}(h)&=\text{Cov}[X_t,X_{t+h}]=E[X_tX_{t+h}]=E[(a+bZ_t+cZ_{t-2})(a+bZ_{t+h}+cZ_{t+h-2})]\\ &= a^2+abE[Z_{t+h}] + acE[Z_{t+h-2}]+abE[Z_t]+b^2E[Z_tZ_{t+h}] + bcE[Z_tZ_{t+h-2}]\\ &+ acE[Z_{t-2}]+ bcE[Z_{t-2}Z_{t+h}] + c^2E[Z_{t-2}Z_{t+h-2}] = a^2. \end{align}
So since both $\mu_X(t)$ and $\gamma_X(h)$ are just a constant it's independent of $t$ and thus $X_t$ is is weakly stationary.
Question: Due to the independence of the $Z_t$ is it ok to conclude that $\gamma_X(h)=a^2$ or do I need to set $h=0,1,2$ and calculate the ACF for the 3 cases?
Method 2: In this method, I don't understand really how to do it. This is my start.
\begin{align} \gamma_X(r,s)&=E[(X_r-\mu_X(t))(X_s-\mu_X(t))]=E[X_rX_s]-aE[X_r]-aE[X_s]+a^2\\ &= E[X_rX_s]-a^2-a^2+a^2 = E[X_rX_s]-a^2... \end{align}
But now I don't see how to come to $\gamma_X(r+h,s+h)$ in the RHS. Any help is appreciated.
## Answer by Cettt (score 2, accepted)
https://quant.stackexchange.com/a/63601
in method 1 you did not use the correct definition of the covariance. For two random variables $X$ and $Y$ we have that $$ Cov(X, Y) = E[XY] - E[X]E[Y]. $$
Also, we can use that the covariance is linear in both arguments (this is the bilinearity of the covariance). Therefore
\begin{align} \gamma_X(h) &= Cov(X_{t+h}, X_t) = Cov(a+ bZ_{t + h} + cZ_{t + h -2}, a+ bZ_{t} + cZ_{t -2}) \\[3mm] &=Cov(a, a) + Cov(a, bZ_t) + Cov(a, cZ_{t-2}) \\ & \quad+ Cov(bZ_{t+h}, a) + Cov(bZ_{t+h}, b Z_t) + Cov(bZ_{t+h}, cZ_{t-2}) \\ & \quad+ Cov(cZ_{t+h - 2}, a) + Cov(cZ_{t+h - 2}, b Z_t) + Cov(cZ_{t+h - 2}, cZ_{t-2}) \\[3mm] &= 0 + 0 + 0 + 0 + b^2\sigma^2 \cdot 1_{\{h = 0\}} + bc\sigma^2 \cdot 1_{\{h = -2\}} + 0 + bc\sigma^2 \cdot 1_{\{h = 2\}} + c^2\sigma^2 \cdot 1_{\{h = 0\}} \\[3mm] & = \begin{cases} b^2\sigma^2 + c^2\sigma^2, & h = 0, \\ bc\sigma^2, & h \in \{-2, 2\}, \\ 0, & \text{otherwise}, \end{cases}. \end{align}
In the last step we have used that for $s, t$ we have that $$ Cov(Z_s, Z_t) = \begin{cases} 0, & s \neq t, \\ \sigma^2, & s = t \end{cases}. $$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
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