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I am working on a structural vector autoregression that requires imposing constraints on a matrix factorization.

In particular, I have an N-dimensional positive definite matrix $\Sigma$ that I need to factor with some N-dimensional square matrix $R$, such that $RR'=\Sigma$. Moreover, given some N-dimensional square matrix $C$, let $D \equiv CR$ and assume constraints are imposed such that some elements of $D$ are zero. In particular, enough of these constraints are imposed such that $R$ is identified. I believe this requires $n^2-n*(n+1)/2$

Any idea how to recover $R$?

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  • $\begingroup$ Do we know anything about $C$ and $D$ (are they full rank, symmetric, etc)? How many constraints do you have (i.e. how many elements of $D$ must be zero)? $\endgroup$ Commented Nov 20, 2021 at 15:36
  • $\begingroup$ @cwindolf The matrix $C$ represents the sum of the coefficients in the vector moving average representation. This means that it represents the permanent impact of the error terms on the variables of interest. It can be assumed to be full rank for our purposes, but not necessarily symmetric. The companion matrix that can be used to build $C$ is assumed to be stationary (so $C$ is finite). $\endgroup$
    – John
    Commented Nov 22, 2021 at 14:48
  • $\begingroup$ @cwindolf In terms of number of constraints, as said above, $R * R' \equiv \Sigma$ imposes $n*(n+1)/2$ constraints, so then I need to impose $n^2 - n*(n+1)/2$ additional constraints, which I believe is actually $n*(n-1)/2$. So when $n=2$, you need one additional constraint (for D to be 0), for $n=3$ you need 3, etc. $\endgroup$
    – John
    Commented Nov 22, 2021 at 14:49
  • $\begingroup$ This $RR'=\Sigma$ is the Cholesky decomposition, where $R$ is a lower triangular matrix. $\Sigma$ is symmetric and it has $n*(n+1)/2$ free elements. $R$ also has $n*(n+1)/2$ free elements. $\endgroup$
    – papgeo
    Commented Dec 6, 2021 at 16:01
  • $\begingroup$ @papgeo Cholesky is commonly used in this case, but there is still leftover restrictions that need to be imposed. Nevertheless, I did find a way to resolve it using a simplified version of what is in the paper King, Plosser, Stock, & Watson (1991), available here: ideas.repec.org/a/aea/aecrev/v81y1991i4p819-40.html $\endgroup$
    – John
    Commented Dec 6, 2021 at 19:47

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