Is weak stationary of bivariate series equivalent to weak stationary of every element of it? If X1t and X2t are both weak-stationary time series, can I always get Xt is weak stationary bivariate series?
If Xt is weak stationary, can we always get X1t and X2t stationary?

 A: A vector random process is said to be a wide-sense stationary process (also called a weakly stationary process) if the mean vector has constant value (that is, $E[\mathbf X(t)]$ equals the same vector for all $t$) and the covariance matrix of $\mathbf X(t)$ and $\mathbf X(t-i)$ depends only on $i$ and does not depend on the value of $t$ at all. Now, the $(j,k)$-th term of the covariance matrix is $\operatorname{cov}(X_j(t), X_k(t-i))$ and this depends only on $i$ and not on $t$ at all. In particular, $\operatorname{cov}(X_j(t), X_j(t-i))$ is a function of $i$ alone (and the mean of $X_j(t))$ is a constant, that is, does not depend on $t$) and so $\{X_j(t)\}$ is individually a wide-sense stationary process. By a similar argument, so is $\{X_k(t)\}$  individually a wide-sense stationary process. In short, if a vector random process is wide-sense stationary, the individual processes comprising it are individually wide-sense stationary, and any two individual processes are jointly wide-sense stationary.
