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.


1 Answer 1


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 $$ Cov(Z_s, Z_t) = \begin{cases} 0, & s \neq t, \\ \sigma^2, & s = t \end{cases}. $$

  • $\begingroup$ @Thanks for your input. Yes indeed, I for some reason, took the mean of $X_t$ to be $0$ despite showing it to be $a$. Probably because lots of questions I did earlier only involved zero-mean processes. Do you have any idea on what to do on method 2? $\endgroup$
    – Parseval
    Apr 26, 2021 at 20:09
  • $\begingroup$ For method 2 it's the same idea: express the covariance of $X$ in terms of the covariance of $Z$ and distinguish the relevant cases for $r$ and $s$. You will see that the covariance is always 0 unless either $s=r$ or $s = r + 2$ or $s = r -2$. $\endgroup$
    – Cettt
    Apr 26, 2021 at 20:14

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