As I wrote in my first response, lower triangular.
My last post was in answer to your reply implying that the interpretation of the variance decomposition differed from the interpretation of the impulse responses. To quote:
So, for impulse responses Variant B is appropriate. But regarding Variance decomposition which one is true?
I'd like to suppose that this is the same, but citing User's guide II (p.356):
"As with the impulse responses, the variance decomposition ... . For example,
the first period decomposition for the first variable in the VAR ordering is completely due to
its own innovation."
I understood that Variant A is appropriate, since only its own shock affects the first variable.
So, is it true? Maybe I misunderstood something ..
For those following at home, I think we got mixed up on our A's and B's. Variant A is the lower and variant B is the upper. I'm not certain why my original response was interpreted as "Variant B is appropriate," since I said "lower" it should have been A, but given that it was, I now understand why we have the confusion about the implications for variance decomposition.
And for anyone who wants context regarding what we're debating. The variance decomposition is essentially a rotation of the entire system to triangularity and diagonal errors. If you do so by pre-multiplying the estimated reduced form VAR coefficients by the inverse of the
lower triangular Cholesky of the residual covariance, the first equation has only it's own contemporaneous error, the second equation has it's own error and the implicit error from the contemporaneous first equation, etc... This matches the interpretation of the impulse responses and what is written in the manual.
If the Cholesky were upper, then the last equation would only have it's own errors, the second to last would have its own and the last equation's errors, etc...
I hope this clarifies things.