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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?

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    $\begingroup$ 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 $t$. $\endgroup$ – Dilip Sarwate Apr 30 '13 at 15:50
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    $\begingroup$ Which stationarity concept are you using? Strict stationarity, covariance stationarity etc...? $\endgroup$ – Learner Apr 30 '13 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 '13 at 15:58
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    $\begingroup$ 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 '13 at 16:57
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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.

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  • $\begingroup$ 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 '13 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 '14 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. $\endgroup$ – Dilip Sarwate Jun 9 '14 at 19:44

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