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
- Re: please solve