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I am trying to calculate the auto-covariance of time series $Z_t$. Given a weakly stationary process $Y_t$; $t \in \mathbb{Z}$: $$Z_t=a+Y_t$$ Now, my goal is to show that $Z_t$ is also a stationary process. I know that $\mathbb{E}[Y_t]$ should be constant, and it is, $\mathbb{E}[Y_t] = a + m$. Then, I need to prove that the auto-covariance $\newcommand{\cov}{\rm Cov} \cov(Z_t, Z_{t+h})$ depends only on $h$.

I am trying to prove it by calculating the covariance as: $$\cov(Z_t, Z_{t+h}) =\mathbb{E}[Z_tZ_{t+h}] - (m+a)^2$$ but I am not getting anywhere. I am not sure what should I show.

I hope I was clear. If you have any insight please let me know.

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  • $\begingroup$ Do you have any distributional assumption for $Y_t$? You mentioned the expectation, but what about variance of $Y_t$ ? $\endgroup$ – Sheikh May 3 '15 at 18:22
  • $\begingroup$ I know about the stationary status of $Y_t$, hence I know that its expectation and variance are constant, they do not depend on $t$. $\endgroup$ – user11291 May 4 '15 at 18:00
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I assume your goal is to show that $Z_t$ is also a weakly stationary process.

First, you showed ${Z_t}$ has a constant mean:

$$ \mu(Z) = E(Z_t) = E(a + Y_t) = a + E(Y_t) = a + \mu(Y), $$

where $\mu(Y)$ is constant given that $Y_t$ is weakly stationary.

Then, you show that $\gamma_k(Z)$ the covariance the process of ${Z_t}$ only depends on the lag $h$, and not on time $t$:

$$ \gamma_h(Z) = cov(Z_t, Z_{t+h}) = cov(a + Y_t, a + Y_{t+h}) = cov(a, a) + cov(a, Y_{t+h}) + cov(Y_t, a) + cov(Y_t, Y_{t+h}) = cov(Y_t, Y_{t+h}) = \gamma_h(Y), $$

where the last equality is true given that $Y_t$ is weakly stationary.

I believe the answer is complete, unless somebody can point out something missed out.

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