Multiple Regression w/ Polynomial-in-Y?

From: Frank Iannarilli (frankeye_at_cox.net)
Date: 09/25/04 

Date: 24 Sep 2004 20:30:58 -0700

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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