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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
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