I have a VAR(2) model:


where $\textbf{y}_t$ is a 2x1 vector, $\textbf{A}_1, \textbf{A}_2$ are two 2x2 matrices and $\textbf{u}_t$ is a 2x1 vector of white noise distrubuted as $N(0,\Sigma)$. Notice that $\Sigma$ is not diagonal, i.e. the noise are contemporaneous correlated. Moreover: the two matrices $\textbf{A}_1, \textbf{A}_2$ are respectively:

\begin{array}{cc} 0.6 & 0.2 \\ 0.2 & 0.4 \\ \end{array}


\begin{array}{cc} -0.2 & 0.2 \\ 0.4 & -0.4 \\ \end{array}

and $\Sigma$ \begin{array}{cc} -0.2 & 0.3 \\ 0.1 & -0.4 \\ \end{array}

Can someone help me in writing $\textbf{y}_{t+h}$ foh $h=1,2$ as a function of $\textbf{y}_0$ and $\textbf{y}_{-1}$ only? Notice that I have that $\textbf{y}_0=(1,1)$ and $\textbf{y}_{-1}=(0,0)$.

This will be needed to calculat the generalized impulse response function:

$IR_{it}(h)=E_t[\textbf{y}_{t+h}|u_{it=k},\textbf{y}_{t-1}.. ]-E_t[\textbf{y}_{t+h}|\textbf{y}_{t-1}.. ]$


You only need to substitute t+h into your original equation:


So you would have:


When h=1,2 respectievly

  • $๐ฒ_{๐‘ก+1}=๐€_1๐ฒ_{๐‘ก}+๐€_2๐ฒ_{๐‘ก-1}+๐ฎ_{๐‘ก+1}$
  • $๐ฒ_{๐‘ก+2}=๐€_1๐ฒ_{๐‘ก+1}+๐€_2๐ฒ_{๐‘ก}+๐ฎ_{๐‘ก+2}$

When t=-1 and h=1,2 respectievly

  • $๐ฒ_{0}=๐€_1๐ฒ_{-1}+๐€_2๐ฒ_{-2}+๐ฎ_{0}$
  • $๐ฒ_{1}=๐€_1๐ฒ_{0}+๐€_2๐ฒ_{-1}+๐ฎ_{1}$

When t=0 and h=1,2 respectievly

  • $๐ฒ_{1}=๐€_1๐ฒ_{0}+๐€_2๐ฒ_{-1}+๐ฎ_{1}$
  • $๐ฒ_{2}=๐€_1๐ฒ_{1}+๐€_2๐ฒ_{0}+๐ฎ_{2}$

Now you have your ingredeints for you complete solution.

$๐ฒ_{1}$ is a function of $๐ฒ_{0}$ and $๐ฒ_{-1}$ and $๐ฒ_{2}$ is a function of $๐ฒ_{1}$ and $๐ฒ_{0}$

  • $\begingroup$ Unofortunately this is not what I meant. My aim is to express $\textbf{y}_{t+1}$ and $\textbf{y}_{t+1}$ as functions of $\textbf{y}_{-1}$ and $\textbf{y}_{0}$ only leaving the $t$ index generic, i.e. I am not allowed to arbitrary fix $t$. Essentially, my problem is to derive a \enquote{nice} representation of $\textbf{y}_{t+1}$ by iteratively substituting backwards until time 0 and 1, but I have troubles in seeing the right pattern $\endgroup$ – giorgio Jul 14 '19 at 17:05
  • $\begingroup$ You should look carefully at the answer. What you seek is there. Even though I didn't spell it out for you. $\endgroup$ – grldsndrs Jul 14 '19 at 22:00

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