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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.

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  • $\begingroup$ 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. $\endgroup$
    – Andrew M
    May 23 at 16:37
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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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  • $\begingroup$ Thank you very much. $\endgroup$
    – Charles
    May 24 at 16:21
  • $\begingroup$ @Charles, no problem. If you are happy with the answer, please accept it (so it is also indicated as answered). $\endgroup$
    – tdm
    May 24 at 18:10

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