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I just learned that a time series is called weak-sense stationary if:

1 $\mathbb{E}[X_t] = \mu$ is independent of $t$.

2 $\operatorname{Cov}(X_{t+h},X_t)$ is independent of $t$ for each $h$.

I am trying to check this for the following time series:

$X_t=b+Z_t-Z_{t-1}$ with {${Z_t}$} ~ $WN(0,\sigma^2)$.

It is quite easy to see that $\mathbb{E}[X_t] = \mu$ is independent of $t$, but how can you check condition 2? When trying to calculate $\operatorname{Cov}(X_{t+h},X_t)$, I end up with the following:

$\operatorname{Cov}(Y_t,Y_{t+h})=\operatorname{Cov}(b+Z_t-Z_{t-1},b+Z_{t+h}-Z_{t-1+h})\\=\operatorname{Cov}(Z_t,b+Z_{t+h}-Z_{t-1+h})-\operatorname{Cov}(Z_{t-1},b+Z_{t+h}-Z_{t-1+h})\\ =\operatorname{Cov}(Z_{t+h},Z_t)-\operatorname{Cov}(Z_{t-1+h},Z_t)-\operatorname{Cov}(Z_{t+h},Z_{t-1})+\operatorname{Cov}(Z_{t-1+h},Z_{t-1})$

It isn't clear to me how I can conclude that $\operatorname{Cov}(X_{t+h},X_t)$ is or is not independent of $t$ for each $h$.

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  • $\begingroup$ Hints: The $Z$'s are from a white noise process, and so $cov(Z_t,Z_s)=0$ unless $t=s$. So look at each and every one of the $Cov$ thingies that you calculated. Is it possible that the subscripts are equal? Does this equality hold only for certain choices of value of $h$ or does it hold regardless of the choice of $h$? That way, you can work out what values the four terms have. $\endgroup$ – Dilip Sarwate Oct 1 '15 at 0:11
  • $\begingroup$ Thanks for you hint. I came to the following: if h=0 you have $2\sigma^2$, if h=1 or -1 you have $-\sigma^2$. For the rest it would be independent. But does this mean that condition 2 doesn't fit? Because $cov(Y_t,Y_{t+h})$ isn't independent of $t$ for each $h$. $\endgroup$ – Alfons Ingomar Oct 1 '15 at 0:40
  • $\begingroup$ You just figured out that the covariance of $X_t$ and $X_{t+h}$ is known if you specify the value of $h$. But it doesn't depend on what $t$ is, does it? $\endgroup$ – Dilip Sarwate Oct 1 '15 at 2:08
  • $\begingroup$ Ah of course, that makes sense! So you it does satisfy with condition 2, since $t$ has doesn't influence the outcome. Thanks! $\endgroup$ – Alfons Ingomar Oct 1 '15 at 9:15
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So I figured it out with some help from the comments beneath my question. Since $Z$ are from a white noise process, $cov(Z_t,Z_s) = 0$ unless $t=s$, in that case you have $\sigma^2$.

$\operatorname{Cov}(Z_{t+h},Z_t)-\operatorname{Cov}(Z_{t-1+h},Z_t)-\operatorname{Cov}(Z_{t+h},Z_{t-1})+\operatorname{Cov}(Z_{t-1+h},Z_{t-1})$

If you try to fill each possible h (and fill this in your cov function), you get the following possibilities:

  • If $h = 0$,you have $2\sigma^2$.
  • If $h = 1 or -1$ you have $-\sigma^2$
  • 0 for al other possibilities of $h$

Since this is true for each value of $t$, you can conclude that $\operatorname{Cov}(Z_{t+h},Z_t)$ is independent of $t$ for each $h$.

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