Suppose I have 2 time-series processes.

If they are jointly weakly stationary then the linear combination is weakly stationary.

If the linear combination is non-stationary does it mean at least one of the two processes is non-stationary?

I can't come up with an example of where this last statement is false.


2 Answers 2


Here's a simple counterexample (for discrete time). Let $X_t$ and $Z_t$ be iid standard Normal sequences. Let $\alpha_t$ be a sequence of numbers in $(-1,1)$. Define $Y_t=\alpha_t X_t+\beta_t Z_t$. Now

  • $Y_t$ is independent for different times.
  • The variance of $Y_t$ is $\alpha_t^2+\beta_t^2$ so given any $\alpha_t$ with $|\alpha_t|<1$ we can (and do) choose $\beta_t$ to make $Y_t$ standard Normal
  • therefore $Y_t$ is weakly stationary: its distributions are all standard Normal
  • But $\mathrm{cov}[X_t,Y_t]= \alpha_t$

Now consider the linear combination $Y_t-\alpha X_t=\beta_t Z_t$. This series is not weakly stationary because $\beta_t$ changes over time. The variance at time $t$ is $\beta_t^2$, which is not constant.

The condition you'd need for weak stationarity of linear combinations is that the pair $(X_t, Y_t)$ were individually weak-stationary and that their covariance was constant over time. You could say they were "jointly weak stationary", though I don't know whether this is standard terminology.

Two final notes: first, $X_t$ and $Y_t$ in this example are strongly stationary as well as weakly stationary. Second, $X_t$ and $Y_t$ are each uncorrelated over time, but that would be easy to change.


Let $$Z_t=aX_t+bY_t$$ Given that $Z_t$ is is NOT stationary, but $X_t$ and $Y_t$ are stationary. So, we can say that $\exists \,\ s \neq t$ such that $$E(Z_t)\neq E(Z_s)$$

$$\implies aE(X_t) + bE(Y_t) \neq aE(X_s) + bE(Y_s)$$

However, $E(X_t)= E(X_s)$ and $E(Y_t)= E(Y_s)$, $\forall \,\, t,s$

So there is a contradiction.

EDIT: Apologies for childish answer above. Another attempt. Let me try using logic statements.

Statement $A$: $Z_t$ is non-stationary.

Statement $B$: At least one of $X_t$ and $Y_t$ is non-stationary.

Let's assume that what you say is correct, i.e.,

$$A \implies B $$ So, $$!B \implies !A $$

This would mean that each of $X_t$ and $Y_t$ being stationary (statement !B) would imply stationarity of linear combination of $X_t$ and $Y_t$. We know this to be not always true. So $A \implies B $ is also not true.

  • $\begingroup$ If the processes are zero mean then this is not enough. I think $X_t$ and $Y_t$ can be individually stationary, but not jointly. $\endgroup$
    – shani
    Oct 18, 2019 at 8:19
  • 2
    $\begingroup$ $E(Z_t)\neq E(Z_s)$ is one special case of nonstationarity. The condition is sufficient for nonstationarity but not necessary. Hence, your proof does not address the general case. $\endgroup$ Oct 18, 2019 at 9:06
  • $\begingroup$ @RichardHardy: Thanks for pointing this out. This is obviously a stupid mistake. $\endgroup$
    – Dayne
    Oct 18, 2019 at 11:05
  • $\begingroup$ @shani: Correct. I made another attempt. $\endgroup$
    – Dayne
    Oct 18, 2019 at 12:42
  • $\begingroup$ weakly stationary (WS ) refers only to the means and variance of the series being constant and independent of $t$.. We know the mean of the resulting series in a linear combination of the individual means so that should still be constant and independent of $t$. The variance is a linear combination of the variances but includes a covariance. Therefore, the linear combination of the two series is not necessarily WS because one doesn't know if the covariance of the two series is time invariant. If it is not, then the resulting variance is not constant and WS does not hold. $\endgroup$
    – mlofton
    Apr 9, 2022 at 12:20

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