Suppose you have two stationary process $A_{t}$ and $B_{t}$. Suppose $Z_{t} = A_{t} + B_{t}$. Show that $Z_{t}$ is stationary. I am unsure how to solve this without knowing if the processes are independent from each other.

$\gamma_{z}(h) = E[Z_{t}Z_{t+h}] = A_{t}A_{t+h} + A_{t}B_{t+h} + B_{t}A_{t+h} + B_{t}B_{t+h} = \gamma_{A}(h) + \gamma_{BA}(h) + \gamma_{AB}(h) + \gamma_{B}(h)$.

Now if the series are independent, then the cross-covariance functions are 0 and $\gamma_{z}(h) = \gamma_{A}(h) + \gamma_{B}(h)$. So my question is, do we require that $A_{t}$ and $B_{t}$ are independent from each other in order for $Z_{t}$ to be stationary?

  • 2
    $\begingroup$ You don't need independence of the processes $\{A_t\}$ and $\{B_t\}$ but you need to assume that they are jointly stationary which is a weaker condition than independence. In particular, you need to assume that the joint distribution of $A_t$ and $B_t$ is the same for all $t$ (which forces the distribution of $Z_t$ to be the same for all $t$). Else, $A_t$ and $B_t$ might have marginal distributions not dependent on $t$ but joint distribution that varies with $t$ making the distribution of $A_t+B_t$ vary with $t$. $\endgroup$ Apr 30, 2013 at 15:50
  • 1
    $\begingroup$ Which stationarity concept are you using? Strict stationarity, covariance stationarity etc...? $\endgroup$
    – Learner
    Apr 30, 2013 at 15:52
  • $\begingroup$ weakly stationarity, i.e the mean does not depend on time and the autocovariance is only a function of the lag $\endgroup$
    – phil12
    Apr 30, 2013 at 15:58
  • 1
    $\begingroup$ For a concrete example of what @DilipSarwate explained, assume that $(C_t)_t$ is i.i.d., say centered Bernoulli, and let $A_{2t}=B_{2t}=C_{3t}$, $A_{2t+1}=C_{3t+1}$, $B_{2t+1}=C_{3t+2}$, then the distributions of $Z_{2t}$ and $Z_{2t+1}$ do not coincide. $\endgroup$
    – Did
    Apr 30, 2013 at 16:57

1 Answer 1


For weak or covariance stationarity of $\{Z_t\}$, you need the cross-covariance functions $\operatorname{cov}(A_t,B_{t+h})$ and $\operatorname{cov}(A_{t+h},B_{t})$ to be functions of $h$ alone, and not dependent on the choice of $t$. Some people call this property as joint weak stationarity, meaning that $\{A_t\}$ and $\{B_t\}$ are individually weakly stationary processes and that the cross-covariance functions have the desired property. Note that the cross-covariance functions are $0$ when $\{A_t\}$ and $\{B_t\}$ are uncorrelated processes meaning that $\operatorname{cov}(A_{t_1},B_{t_2})$ are uncorrelated for all choices of $t_1$ and $t_2$, or in words, every random variable from the process $\{A_t\}$ is uncorrelated with every random variable from the process f$\{B_t\}$. Independent processes are a subclass of uncorrelated processes. If $\{A_t\}$ and $\{B_t\}$ are uncorrelated weakly stationary processes, then their sum is a weakly stationary process.

Answer to question in comment: In general, $\operatorname{cov}(A_{t+h},B_{t})$ is a function of $h$ and $t$, and so it is of course a function of $h$. What you need to determine is whether it is also a function of $t$.

Example: Let $\Theta$ denote a random variable enjoying the property $$E[\cos(\Theta)] = E[\sin(\Theta)]=E[\cos(2\Theta)] = E[\sin(2\Theta)]=E[\cos(4\Theta)] = E[\sin(4\Theta)]=0$$ and define random processes $\{A_t \colon t \in \mathbb R\}$ and $\{B_t \colon t \in \mathbb R\}$ via $A_t = \cos(t+\Theta), B_t = \cos(t+2\Theta)$. Then, $$E[A_t]=E[\cos(t+\Theta)]=E[\cos(t)\cos(\Theta)-\sin(t)\sin(2\Theta)] = 0$$ and $$\begin{align} E[A_tA_{t+h}]&=E[\cos(t+\Theta)\cos(t+h+\Theta)]\\ &=\frac{1}{2}E[\cos(2t+h+2\Theta)+\cos(h)]\\ &=\frac{1}{2}E[\cos(2t+h)\cos(2\Theta)-\sin(2t+h)\sin(2\Theta)+\cos(h)]\\ &=\frac{1}{2}\cos(h) \end{align}$$ showing that $\{A_t\}$ is weakly stationary. A similar calculation shows that $\{B_t\}$ is also weakly stationary. However, the processes are not necessarily jointly weakly stationary because $$\begin{align} \operatorname{cov}(A_{t},B_{t+h})&= E[A_{t}B_{t+h}]\\ &=E[\cos(t+\Theta)\cos(t+h+2\Theta)]\\ &=\frac{1}{2}E[\cos(2t+h+3\Theta)+\cos(h+\Theta)]\\ &=\frac{1}{2}E[\cos(2t+h)\cos(3\Theta)-\sin(2t+h)\sin(3\Theta)+\cos(h+\Theta)] \end{align}$$ is a function of both $t$ and $h$ unless we make the additional assumption that $E[\sin(3\Theta)]=E[\cos(3\Theta)]=0$.

Returning to the original question, we note that the sum of weakly stationary random processes is not necessarily weakly stationary: additional assumptions such as joint weak stationarity are needed.

  • $\begingroup$ From this question, can we prove that $cov(A_{t+h},B_{t})$ is a function of h or do we need more information? $\endgroup$
    – phil12
    Apr 30, 2013 at 19:48
  • $\begingroup$ @Sarwate:"the sum of weakly stationary random processes is not necessarily weakly stationary: additional assumptions such as joint weak stationarity are needed." Is not it true that the ANY linear combination of two I(0) series is also I(0)? I do not understand what you mean by "joint weak stationarity"! $\endgroup$
    – user156158
    Jun 9, 2014 at 18:28
  • $\begingroup$ I have no idea what you mean by I(o) and so cannot say whether or not the sum of two I(o) series is also I(o). See Did's example in a comment on the main question for a specific example of two processes that each are stationary but are not jointly stationary and so their sum is not stationary. $\endgroup$ Jun 9, 2014 at 19:44
  • $\begingroup$ @DilipSarwate: As per definition of I(0) on wiki it seems that I(0) is equivalent to weak stationarity and therefore, it seems that a linear combination of I(0) is also I(0) only under joint weak stationarity. I am commenting on this because this is related to my question. $\endgroup$
    – Dayne
    Jan 23, 2021 at 2:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.