Re: Multiple Regression w/ Polynomial-in-Y?
From: Paul Victor Birke (nonlinear_at_rogers.com)
Date: 09/25/04
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Date: Sat, 25 Sep 2004 09:31:31 -0400 To: Frank Iannarilli <frankeye@cox.net>
Dear Frank
Furthermore, something like MAPLE here would be good to find the roots.
BTW MAPLE is lots of $ but handy to have around. Maybe you can download
an older version from the net.
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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