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I saw the answer on this post and got confused about a couple things in its explanation.

Mainly, I am unsure of

  1. How the poster immediately knows the process $X_t = c+\phi_1 Y_{t-1} + \epsilon_t$ is weakly stationary. Or is it just an assumption that is made?
  2. How the poster goes from $E_t[X_{t+h}] = E_t[\alpha X_{t+h-1}+\epsilon_t] = \alpha^hX_t$
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1 Answer 1

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  1. This AR(1) process is WSS by definition. So, it is also weakly stationary. Of course, assuming $|\phi_1|<1$ as the poster suggests.

  2. I simply follow the definitions and the notation in the post:

    $E_t[Y_{t+h}]=E_t[\phi_1Y_{t+h-1}+\epsilon_{t+h}]=E_t[\phi_1^2Y_{t+h-2}+\epsilon_{t+h}+\phi_1\epsilon_{t+h-1}]=E_t[\phi^hY_{t}]+\sum_{m=0}^{h-1}{\phi_1^mE_t[\epsilon_{t+h-m}}]=E_t[\phi^hY_t]=\phi^hY_t$, because expectation of noise is 0.

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    $\begingroup$ What does WSS stand for? $\endgroup$ Commented Oct 7, 2018 at 20:08
  • $\begingroup$ Wide-Sense-Stationary $\endgroup$
    – gunes
    Commented Oct 27, 2018 at 15:39

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