Consider the following question

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How is it obvious that $\mathbb{E}(X_t) = 0$? Is it because of the following? Recursively we find that $\mathbb{E}(X_t) = \phi \mathbb{E}(X_{t-1}) = \phi^2 \mathbb{E}(X_{t-2}) = \phi^{t-1} \mathbb{E}(X_{1})$. So now we see that for $\mathbb{E}(X_t) = \mathbb{E}(X_s)$ for all $s<t$ (a requirement for stationarity) either $\phi=1$ or $\mathbb{E}(X_t) = 0$ we cannot have the former since $|\phi|<1$ thus we have $\mathbb{E}(X_t) = 0$.

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    $\begingroup$ Use stationarity and linearity of expectation to compute $0=E[Z_t] = E[X_t-\phi X_{t-1}]$ and solve for $E[X_t].$ (To guarantee a solution you need $\phi\ne 1$ and $|\phi|\gt 1$ is inconsistent with stationarity: that's where the restriction on $\phi$ comes from.) The appeal to recursion falls apart because there is no starting point: $t$ is an arbitrary integer, not just a natural number. $\endgroup$ – whuber Aug 30 '19 at 13:18
  • $\begingroup$ notice the very important hint from @whuber when it comes to $| \phi | > 1$ that would be inconsistent with stationarity too.. very often this is not well reported in textbooks.. but it is important. $\endgroup$ – Fr1 Aug 30 '19 at 13:47

You explanation seems me admissible. However a more convincing one come from a generalization like this:

$X_t = \phi_0 + \phi_1 X_{t-1} + Z_t$

it is possible to demonstrate that

$E[X_t] = \phi_0 / (1 - \phi_1)$

than if $\phi_0 = 0$ we have that $E[X_t] = 0$


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