# 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.

• 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. May 23, 2021 at 16:37

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: $$(I - B) \sum_{j = 0}^{k-1} B^j + B^k = \sum_{j = 0}^{k-1} B^j - \sum_{j = 1}^{k} B^j + B^{k} = B^0 - B^{k} + B^{k} = I$$