Sum of stationary process 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? 
 A: 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.

