Re: Asymmetric quadratic coefficient problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg49909] Re: Asymmetric quadratic coefficient problem*From*: "Dr. Wolfgang Hintze" <weh at snafu.de>*Date*: Thu, 5 Aug 2004 09:22:36 -0400 (EDT)*References*: <cequfb$k3m$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

As to your first question: You can easily convice yourself that the asymmetric part of the matrix A just drops out of the term z'Az. Let S=(A+Transpose[A])/2 (symmetric part of A) N=(A-Transpose[A])/2 (asymmetric part of A) then, obviously A = S+N and we have z'Az = z'Sz + z'Nz but the term x=z'Nz vanishes because (not in strict Mathematica style) x=z'Nz = Sum[ z[i] N[i,k] z[k] ] = Sum[ z[k] N[k,i] z[i] ] (just renaming summation indices) = - Sum [ z[i] N[i,k] z[k] ] (because of N[k,i] = - N[i,k] ) = - x Hences x=0. Hope this helps. Regards, Wolfgang Xiao Huang wrote: > Dear All, > > Does any one know how to deal with asymmetric quadratic coefficient in > QuadraticFormDistribution[ {A,b,c}, {mu, sigma}] for z'Az+b'z+c? > Mathematica assumes A is a symmetric matrix, but I now have an asymmetric > and singular matrix A. > > Also, I'd really appreciate if any expert here can give me some > clue/reference about the following problem: how can I calculate different > quantiles of a mixture distribution? One way is to do the simulation, but is > it possible to use the function Quantile[.] in Mathematica? if yes, how? > > Thank you! > > Xiao > > _________________________________________________________________ > Overwhelmed by debt? Find out how to 'Dig Yourself Out of Debt' from MSN > Money. http://special.msn.com/money/0407debt.armx > >