       • 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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