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I'm reading the paper Bayesian Vector Autoregressions by Thomas Wozniak. He considers the model $$y_t = \mu + A_1 y_{t-1} + \cdots A_k y_{t-k} + u_t$$ where each $y_i$ is a $N$-vector, each $A_j$ is a $N\times N$ matrix and $u_t \sim N(0, \Sigma)$

He assigns the prior for $\Sigma$ to be inverse Wishart. I think that the errors are independent, which would mean that $\Sigma$ is diagonal. If this is the case, then $N^2 - N$ of the entries are zero (i.e. non-random). If this is not the case, I am working with a similar model where the covariance matrix is diagonal. In either case, I am led me to the following question:

Is it okay to assign the inverse Wishart distribution to diagonal matrix?

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  • $\begingroup$ What makes you think that the error covariance matrix is diagonal? I have never seen that assumed in the context of VARs. $\endgroup$ – hejseb May 15 at 4:02
  • $\begingroup$ I believe the errors are independent. $\endgroup$ – grxxvytony May 15 at 5:14
  • $\begingroup$ Over time yes, but literally no one assumes that $\Sigma$ is diagonal. Maybe you are conflicting two different notions of independence. You typically assume time independence ($u_t$ and $u_s$ are independent for $t\neq s$), but that is different from cross-sectional independence which you do not assume. The latter would entail assuming that $u_{i,t}$ and $u_{j,t}$ are independent for $i\neq j$ and would yield a diagonal $\Sigma$. But no one makes that assumption, so $\Sigma$ is never diagonal in this context. $\endgroup$ – hejseb May 15 at 10:25
  • $\begingroup$ That makes sense. I've never heard the term cross-sectional independence before. $\endgroup$ – grxxvytony May 15 at 17:42
  • $\begingroup$ Time independence: $\text{Cov}(u_t, u_s) = 0$ if $t \neq s$. Cross-sectional independence: $\text{Var}(u_t)$ is a diagonal matrix. $\endgroup$ – grxxvytony May 15 at 17:43

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