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Suppose I have a 2nd-order vector autoregressive model VAR(2):

$$ \mathbf{y}_t=\mathbf{\Phi_1}\mathbf{y}_{t-1}+\mathbf{\Phi_2}\mathbf{y}_{t-2}+\mathbf{\varepsilon}_t $$

where $\mathbf{y}_t=(x_t,y_t)$. My question is: Is it possible to calculate the joint distribution $P(x_t,y_t)$ for large $t$ limit? If yes, how?

Possible Method 1: $\mathbf{y}_t$ can be written as a MA($\infty$) process $\mathbf{y}_t=\sum_{i=0}^\infty \Psi_i\mathbf{\varepsilon}_{t-i}$. If $\Psi_i$ is a simple series, we can calculate the pdf of $\mathbf{y}_t$ through characteristics function. However, in VAR(2), $\Psi_i$ does not seem to be simple enough to work with.

Possible Method 2: According to here, AR(2) can be written as $$ X_t=\sum_{i=0}^\infty \alpha_i\epsilon_{t-i} $$ where $$\sum_{i=0}^\infty\alpha_i^2=\dfrac{1+\lambda_1\lambda_2}{(1-\lambda_1^2)(1-\lambda_2^2)(1-\lambda_1\lambda_2)}$$ and $\lambda_{1,2}$ are the eigenvalues of the F matrix. It might be possible to work out a generalization for VAR(2) but unfortunately it is very tedious.

Edit 1: Since VAR(2) can be expressed in a linear combination of innovations, which obey serial uncorrelated multivariate Gaussian distribution, the pdf of the VAR(2) itself is also a multivariate Gaussian distribution. Therefore, all I needed is the covariance matrix of VAR(2) and it can be found in many textbooks, which makes my question a bit silly.

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    $\begingroup$ Obviously, $${\mathbf y}_t=\sum_{s\geqslant0}\Psi_s\varepsilon_{t-s},$$ where $\Psi_0=I$ and, for every $s\geqslant1$, $$\Psi_s=\sum_{|\sigma|=s}\prod_{i}\Phi_{\sigma(i)},$$ where the sum enumerates every finite sequence $\sigma$ with values in $\{1,2\}$ and $|\sigma|=\sum\limits_i\sigma(i)$. For example, $$\Psi_1=\Phi_1,\quad\Psi_2=\Phi_1^2+\Phi_2,\quad\Psi_3=\Phi_1^{3}+\Phi_1\Phi_{2}+\Phi_2\Phi_1,$$ and $$\Psi_{4}=\Phi_{1}^{4}+\Phi_{1}^{2}\Phi_{2}+\Phi_{1}\Phi_{2}\Phi_{1}+\Phi_{2}\Phi_{1}^{2}+\Phi_{2}^{2}.$$ $\endgroup$ – Did Apr 6 '15 at 11:39
  • $\begingroup$ @Did, yes you are right. I can derive the expression, but it is still an infinite series. The summation $|\sigma|$ is pretty inconvenient for practical purposes. Anyway, I realized that my question is kind of silly. $\endgroup$ – wdg Apr 6 '15 at 13:09

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