Re: Multiple Regression w/ Polynomial-in-Y?
From: Paul Victor Birke (nonlinear_at_rogers.com)
Date: 09/26/04
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Date: Sun, 26 Sep 2004 08:57:18 -0400 To: Frank Iannarilli <frankeye@cox.net>
Dear Frank
Doing more poking around and remembering the notion of a stationary
value, it might be that your problem is related to Stationary Point
Solutions of Multivariate Regression.
Here is a paper where we are after the centre of y contours
http://www.stat.rutgers.edu/~buyske/591/lect07.pdf
Stationary Points are denoted as those points where differentials = zero
but there is likely much more than that to this approach.
I briefly looked at is some years ago in context of Neural Networks.
Again one seems to get the dependent variable "y" on both sides of the
equation.
thanks again for bringing up this problem.
I will try to find the papers I know I have upstairs to look at this
problem again wrt the Stationary Point.
Paul
Frank Iannarilli wrote:
> Hi,
>
> Is what I'm tempted to call "Polynomial Root Regression" so obvious
> that nobody talks/writes about it?
>
> Here's the model, which is readily solved with multiple regression
> methods (also tried it successfully with shrinkage regression
> technique such as Partial Least Squares) - note well, the coefficients
> of the dependent-variable polynomial are unknown, and are estimated by
> the (standard) regression (in addition to the usual weights for the
> independent variables).
>
> Here, a quadratic-in-Y polynomial (extension to higher-order
> obvious):
>
> x.w = y + c*y^2 + error
>
> M indep. vars; N samples; (intercept term suppressed)
> Both w vector and c coefficient unknown/to be estimated
>
>
> |x(1,1), x(1,2)...x(1,M); -y(1)^2| |w(1)| |y(1)|
> |x(2,1), x(2,2)...x(2,M); -y(2)^2| |w(2)| |y(2)|
> |....... | |... | = |... |
> |x(N,1), x(N,2)...x(N,M); -y(N)^2| |w(M)| |y(N)|
> | c |
>
> Solve above with appropriate linear least-squares solver (e.g., OLS,
> or RR or PLS if multicollinearity). Yields w-vector and c coefficient
> for quadratic term for y.
>
> To predict a new y-value, just find the root y of the determined
> polynomial
>
> c*y^2 + y - x.w = 0
>
>
> I tried this successfully for a problem whose non-linearity was best
> modeled on the dependent-variable side, rather than forcing polynomial
> terms of independent variables (which I suspect would not be so good
> for my application). One drawback is needing to decide which root is
> the actual solution, but for many situations this is probably "easy".
>
> Thoughts? Does this have a name?
>
> Thanks!
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