<p>There are three things going on (slightly rewriting the expectation operator):</p>
<ol>
<li>Backward recursion</li>
</ol>
<p><span class="math-container">\begin{equation}
\begin{split}
x_{t+2} &amp;= \phi_1 x_{t+1} + \phi_2 x_{t} + \phi x_{t-1} + \epsilon_{t+2}\\
&amp;= \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}\\
&amp;= \cdots
\end{split}
\end{equation}</span></p>
<ol start="2">
<li><p>Application / Rules of the expectation operator:
<span class="math-container">$$
E[a + bX] = a + bE[X]
$$</span>
for constants <span class="math-container">$a,b$</span> and a random variable <span class="math-container">$X$</span>.</p>
</li>
<li><p><span class="math-container">$$E_t[\epsilon_{t+h}] = 0 \quad \textrm{for all} \quad h&gt;0$$</span></p>
</li>
</ol>
<p>Applying these and collecting coefficients on <span class="math-container">$x_t, x_{t-1}, x_{t-2}$</span> gives you the result.</p>

CC BY-SA 4.0
2021-08-07T09:08:25.803Z
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.
EB3112
CC BY-SA 4.0
2021-08-07T10:05:37.593Z
Edited my response accordingly.
BrsG
CC BY-SA 4.0
2021-08-07T07:30:02.107Z
2021-08-07T10:05:16.287Z
2021-08-07T10:05:16.287Z
BrsG
<p>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):</p>
<p><span class="math-container">$$x_t = \phi x_{t-1} + \phi_2 x_{t-2} + \phi x_{t-3} + \epsilon_t\qquad \epsilon_t \sim(0,\sigma^2)$$</span></p>
<p>In particular, could someone point out why the substitution (highlighted by substitution of the first underbrace into the second underbrace) is legitimate?</p>
<p>The <span class="math-container">$t+1$</span> forecast is given:</p>
<p><span class="math-container">\begin{equation}
    \begin{split}
\underbrace{E(x_{t+1}|x_t,x_{t-1},...)} &amp; = E(\phi_1x_t + \phi_2 x_{t-1} + \phi_3 x_{t-2} + \epsilon_{t+1}|x_t, x_{t-1},...)\\ &amp; =\phi_1x_t + \phi_2 x_{t-1} + \phi_3 x_{t-2} 
\end{split}
\end{equation}</span></p>
<p>And the <span class="math-container">$t+2$</span> forecast</p>
<p><span class="math-container">\begin{equation}
\begin{split}
E(x_{t+2}|x_t,x_{t-1},...) &amp; = \underbrace{E(\phi_1 x_{t+1}} + \phi_2 x_t + \phi_3 x_{t-1} + \epsilon_{t+2}|x_t, x_{t-1},...) \\
&amp; = \underbrace{\phi_1E(x_{t+1}|x_t,x_{t-1},...)} + \phi_2x_t +\phi_3 x_{t-1},\\
&amp; = \underbrace{\phi_1(\phi_1x_t +\phi_2 x_{t-1} + \phi_3 x_{t-2})} + \phi_2x_t + \phi_3x_{t-1}\\
&amp; = (\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}</span></p>
<p>Would be appreciated.</p>

<mathematical-economics><time-series>
Algebra for two period forecasting in AR (3) Model
EB3112
# Algebra for two period forecasting in AR (3) Model

(Asked by: username_0 on 2021-08-06)

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.

**Tags:** mathematical-economics, time-series


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## 1 answers

### Answer by username_1 on 2021-08-07. Score: 3 | Accepted answer

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}

2. Application / Rules of the expectation operator:
   $$
   E[a + bX] = a + bE[X]
   $$
   for constants $a,b$ and a random variable $X$.
3. $$E\_t[\epsilon\_{t+h}] = 0 \quad \textrm{for all} \quad h>0$$

Applying these and collecting coefficients on $x\_t, x\_{t-1}, x\_{t-2}$ gives you the result.

> **Comments:**
> **username_0**: Thanks @ username_1, 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.  
> **username_1**: Edited my response accordingly.  

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