Re: [Gretl-users] svar inquiry
On Wed, 2 Jul 2014, Sven Schreiber wrote:
> The problem is unlikely to be the rank of your contraint matrix.
If you
> fill a 9x9 matrix with 36 zeros at random, the probability if it being
> singular is zero (it shouldn't be difficult to prove this formally)
> although of course the event is not impossible (nice example of an event
> with 0 Lebesgue measure).
Jack, I don't think this is true, for example the probability of getting
a column with all (9) zeros by chance may be small, but certainly not
zero if you allocate 36 zeros randomly. If I were good in combinatorics
I could give you the exact numbers, but I'm not.
Apart from that the constraints are not chosen at random. A line has
measure zero in a plane, but still I can pick exactly points on the line.
So I'm not saying the matrix is singular, but without further
investigation/thinking I cannot rule out that it is.
You're absolutely right, my friend. After all, the number of ways you can
put n*(n-1)/2 zeros into a square matrix with n rows is finite, and some
of those do give rise to singular matrices, so yes, it isn't a Lebesgue 0.
I stand corrected.
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Riccardo (Jack) Lucchetti
Dipartimento di Scienze Economiche e Sociali (DiSES)
Università Politecnica delle Marche
(formerly known as Università di Ancona)
r.lucchetti(a)univpm.it
http://www2.econ.univpm.it/servizi/hpp/lucchetti
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