<p>I'm not a VAR specialist, but it could go along the following lines.</p>
<p>First write the VAR in companion form:
<span class="math-container">$$
X_{t+k} = A + B X_{t+k-1} + E_{t+k-1}.
$$</span>
Where <span class="math-container">$t$</span> is the last observation, <span class="math-container">$E_j$</span> are the errors and <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are fixed coefficients.</p>
<p>Then solving recursively:
<span class="math-container">$$
\begin{align*}
X_{t+k} &amp;= A + B X_{t+k-1}+ E_{t+k-1},\\
&amp;= A + B(A + B X_{t+k-2} + E_{t+k-2}) + E_{t+k-1},\\
&amp;= (I + B)A + B^2 X_{t+k-2} + B E_{t+k-2} + E_{t+k-1},\\
&amp;= (I + B)A + B^2(A + BX_{t+k-3} + E_{t+k-3}) + B E_{t+k-2} + E_{t+k-1},\\
&amp;= (I + B + B^2)A + B^3 X_{t+k-3} + B^2 E_{t+k-3} + B E_{t+k-2} + E_{t+k-1},\\
&amp;= \ldots\\
&amp;= A \sum_{j = 0}^{k-1} B^j + B^{k} X_t + \sum_{j = 0}^{k-1} B^j E_{t+k-j-1}
\end{align*}
$$</span>
Then taking conditional expectations and using <span class="math-container">$\mathbb{E}(E_j|X_t) = 0$</span>, gives:
<span class="math-container">$$
\mathbb{E}(X_{t+k}|X_t) = A \sum_{j = 0}^{k-1} B^j + B^{k} X_t.
$$</span>
If we let <span class="math-container">$k \to \infty$</span> and if <span class="math-container">$B^k \to 0$</span> (i.e. the process is stable) we have that:
<span class="math-container">$$
\mathbb{E}(X_{t+k}|X_t)\to \mathbb{E}(X_\infty) =  A \sum_{j = 0}^\infty B^j = A(1 - B)^{-1}
$$</span>
Substituting <span class="math-container">$A = \mathbb{E}(X_\infty) (1- B)$</span> gives:
<span class="math-container">$$
\mathbb{E}(X_{t+k}) = \mathbb{E}(X_\infty)(1- B) \sum_{j = 0}^{k-1} B^j + B^{k} X_t.
$$</span>
This is a weighted sum of <span class="math-container">$\mathbb{E}(X_\infty)$</span> and <span class="math-container">$X_t$</span>. In fact a weighted average as the weights add to the unit matrix:
<span class="math-container">$$
(I - B) \sum_{j = 0}^{k-1} B^j + B^k = \sum_{j = 0}^{k-1} B^j - \sum_{j = 1}^{k} B^j + B^{k} = B^0 - B^{k} + B^{k} = I
$$</span></p>

CC BY-SA 4.0
2021-05-24T10:04:22.130Z
2021-05-24T10:04:22.130Z
tdm
<p>I am trying to determine how to write the K-step ahead forecast of a VAR(2), with two variables, as a weighted average of its mean and last observations. I understand that one must use companion form, but am unsure of how to derive the weights from it. Any guidance is much appreciated. Thank you.</p>

CC BY-SA 4.0
2021-05-23T16:37:20.037Z
It would be useful to see an attempt to see where you need help. Where are you getting stuck? Probably just useful to start with a simple AR(1) model and then work your way up to the more complicated case.
Andrew M
<econometrics><time-series>
K-step ahead forecast of VAR(2)
Charles
# K-step ahead forecast of VAR(2)

(Asked by: username_0 on 2021-05-22)

I am trying to determine how to write the K-step ahead forecast of a VAR(2), with two variables, as a weighted average of its mean and last observations. I understand that one must use companion form, but am unsure of how to derive the weights from it. Any guidance is much appreciated. Thank you.

**Tags:** econometrics, time-series

> **Comments:**
> **username_1**: It would be useful to see an attempt to see where you need help. Where are you getting stuck? Probably just useful to start with a simple AR(1) model and then work your way up to the more complicated case.  

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## 1 answers

### Answer by username_2 on 2021-05-24. Score: 1 | Accepted answer

I'm not a VAR specialist, but it could go along the following lines.

First write the VAR in companion form:
$$
X\_{t+k} = A + B X\_{t+k-1} + E\_{t+k-1}.
$$
Where $t$ is the last observation, $E\_j$ are the errors and $A$ and $B$ are fixed coefficients.

Then solving recursively:
$$
\begin{align\*}
X\_{t+k} &= A + B X\_{t+k-1}+ E\_{t+k-1},\\
&= A + B(A + B X\_{t+k-2} + E\_{t+k-2}) + E\_{t+k-1},\\
&= (I + B)A + B^2 X\_{t+k-2} + B E\_{t+k-2} + E\_{t+k-1},\\
&= (I + B)A + B^2(A + BX\_{t+k-3} + E\_{t+k-3}) + B E\_{t+k-2} + E\_{t+k-1},\\
&= (I + B + B^2)A + B^3 X\_{t+k-3} + B^2 E\_{t+k-3} + B E\_{t+k-2} + E\_{t+k-1},\\
&= \ldots\\
&= A \sum\_{j = 0}^{k-1} B^j + B^{k} X\_t + \sum\_{j = 0}^{k-1} B^j E\_{t+k-j-1}
\end{align\*}
$$
Then taking conditional expectations and using $\mathbb{E}(E\_j|X\_t) = 0$, gives:
$$
\mathbb{E}(X\_{t+k}|X\_t) = A \sum\_{j = 0}^{k-1} B^j + B^{k} X\_t.
$$
If we let $k \to \infty$ and if $B^k \to 0$ (i.e. the process is stable) we have that:
$$
\mathbb{E}(X\_{t+k}|X\_t)\to \mathbb{E}(X\_\infty) = A \sum\_{j = 0}^\infty B^j = A(1 - B)^{-1}
$$
Substituting $A = \mathbb{E}(X\_\infty) (1- B)$ gives:
$$
\mathbb{E}(X\_{t+k}) = \mathbb{E}(X\_\infty)(1- B) \sum\_{j = 0}^{k-1} B^j + B^{k} X\_t.
$$
This is a weighted sum of $\mathbb{E}(X\_\infty)$ and $X\_t$. In fact a weighted average as the weights add to the unit matrix:
$$
(I - B) \sum\_{j = 0}^{k-1} B^j + B^k = \sum\_{j = 0}^{k-1} B^j - \sum\_{j = 1}^{k} B^j + B^{k} = B^0 - B^{k} + B^{k} = I
$$


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