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I wondered if some folks could help fill in a knowledge gap for me with some time-series algebra please regarding the following AR (3):

$$x_t = \phi x_{t-1} + \phi_2 x_{t-2} + \phi x_{t-3} + \epsilon_t\qquad \epsilon_t \sim(0,\sigma^2)$$

In particular, could someone point out why the substitution (highlighted by substitution of the first underbrace into the second underbrace) is legitimate?

The $t+1$ forecast is given:

\begin{equation} \begin{split} \underbrace{E(x_{t+1}|x_t,x_{t-1},...)} & = E(\phi_1x_t + \phi_2 x_{t-1} + \phi_3 x_{t-2} + \epsilon_{t+1}|x_t, x_{t-1},...)\\ & =\phi_1x_t + \phi_2 x_{t-1} + \phi_3 x_{t-2} \end{split} \end{equation}

And the $t+2$ forecast

\begin{equation} \begin{split} E(x_{t+2}|x_t,x_{t-1},...) & = \underbrace{E(\phi_1 x_{t+1}} + \phi_2 x_t + \phi_3 x_{t-1} + \epsilon_{t+2}|x_t, x_{t-1},...) \\ & = \underbrace{\phi_1E(x_{t+1}|x_t,x_{t-1},...)} + \phi_2x_t +\phi_3 x_{t-1},\\ & = \underbrace{\phi_1(\phi_1x_t +\phi_2 x_{t-1} + \phi_3 x_{t-2})} + \phi_2x_t + \phi_3x_{t-1}\\ & = (\phi_1^2 + \phi_2)x_t + (\phi_1 \phi_2 + \phi_3)x_{t-1} + \phi_1\phi_3x_{t-2}. \end{split} \end{equation}

Would be appreciated.

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There are three things going on (slightly rewriting the expectation operator):

  1. Backward recursion

\begin{equation} \begin{split} x_{t+2} &= \phi_1 x_{t+1} + \phi_2 x_{t} + \phi x_{t-1} + \epsilon_{t+2}\\ &= \phi_1(\phi_1 x_{t} + \phi_2 x_{t-1} + \phi x_{t-1} + \epsilon_{t+1}) + \phi_2 x_{t} + \phi x_{t-1} + \epsilon_{t+2}\\ &= \cdots \end{split} \end{equation}

  1. Application / Rules of the expectation operator: $$ E[a + bX] = a + bE[X] $$ for constants $a,b$ and a random variable $X$.

  2. $$E_t[\epsilon_{t+h}] = 0 \quad \textrm{for all} \quad h>0$$

Applying these and collecting coefficients on $x_t, x_{t-1}, x_{t-2}$ gives you the result.

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  • $\begingroup$ Thanks @ BrsG, I appreciate your help. I edited the post a little. I appreciate your feedback on the rule. It makes sense now, you've said it. Initially I was stuck on why the first and second underbraces were equivalent. $\endgroup$
    – EB3112
    Aug 7 '21 at 9:08
  • $\begingroup$ Edited my response accordingly. $\endgroup$
    – BrsG
    Aug 7 '21 at 10:05
  • $\begingroup$ Much appreciated. Thank you $\endgroup$
    – EB3112
    Aug 7 '21 at 10:47

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