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Re: Re: please solve

• To: mathgroup at smc.vnet.net
• Subject: [mg53017] Re: [mg52998] Re: please solve
• From: DrBob <drbob at bigfoot.com>
• Date: Mon, 20 Dec 2004 06:34:33 -0500 (EST)
• References: <cprk0m$r87$1@smc.vnet.net><cpuh1i$gui$1@smc.vnet.net> <200412191114.GAA17997@smc.vnet.net>
• Reply-to: drbob at bigfoot.com
• Sender: owner-wri-mathgroup at wolfram.com

As I said before, if the matrix weren't singular, what you need is

Inverse[Transpose[x].x].Transpose[x].y

There isn't any simpler expression for it, in general.

Bobby

On Sun, 19 Dec 2004 06:14:43 -0500 (EST), <rcmcll at yahoo.com> wrote:

> Maybe I should define t. t is just time 1,2,3..,n and t^2 is
> 1,4,9,...n^2 like that.
> I didn't think about singularity as this is a simple regression problem
> that works when one uses actual values. I'm trying to get simple
> formulas for beta instead of the usual ones.Sorry about that. Maybe I
> can take usual formulas and try to get a simple solution thru that.
> Thanks for all the effort
> Bob
>
> Robert M. Mazo wrote:
>> On Thu, 16 Dec 2004 09:20:22 +0000 (UTC), rcmcll at yahoo.com wrote:
>>
>> >Greetings:
>> >
>> >Could someone please solve this symbolically?
>> >This is just the ols formula for beta-hat but I need a symbolic
>> >solution for this special case.
>> >
>> >b = inv(x'x)x'y
>> >
>> >where
>> >
>> >x =  1  t  t^2
>> >1  t  t^2
>> >1  t  t^2
>> >
>> >
>> >and simplify simplify simplify!!
>> >
>> >Thanks,
>> >
>> >Bob
>>
>> If I understand the notation, x is a matrix.  Yes?  If so, x is
>> singular and doesn't have an inverse.  Hence neither does x'x, so the
>> problem has no solution.  Or, better put, the problem is ill posed.
>>
>> A dsifferent Bob
>
>
>
>



-- 
DrBob at bigfoot.com
www.eclecticdreams.net

• References:
  • Re: please solve
    • From: rcmcll@yahoo.com