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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
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>
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