K-step ahead forecast of VAR(2) 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.
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:
$$
\begin{align*}
X_{t+k} &amp;= A + B X_{t+k-1}+ E_{t+k-1},\\
&amp;= A + B(A + B X_{t+k-2} + E_{t+k-2}) + E_{t+k-1},\\
&amp;= (I + B)A + B^2 X_{t+k-2} + B E_{t+k-2} + E_{t+k-1},\\
&amp;= (I + B)A + B^2(A + BX_{t+k-3} + E_{t+k-3}) + B E_{t+k-2} + E_{t+k-1},\\
&amp;= (I + B + B^2)A + B^3 X_{t+k-3} + B^2 E_{t+k-3} + B E_{t+k-2} + E_{t+k-1},\\
&amp;= \ldots\\
&amp;= A \sum_{j = 0}^{k-1} B^j + B^{k} X_t + \sum_{j = 0}^{k-1} B^j E_{t+k-j-1}
\end{align*}
$$
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:
$$
(I - B) \sum_{j = 0}^{k-1} B^j + B^k = \sum_{j = 0}^{k-1} B^j - \sum_{j = 1}^{k} B^j + B^{k} = B^0 - B^{k} + B^{k} = I
$$