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.