Maximization of R
Message posted by Lakshman Nandagiri on April 26, 2000 at 12:00 AM (ET)

: Can someone give me an analytical solution to the following problem?

: Consider R sets of observations of Y (dependent variable) corresponding to X (independent variable). Let n1,n2..nR be the number of observations in each set. Using linear regression analysis we can obtain regression coefficients a1,a2..aR (intercept) and b1,b2..bR (slope) for each set of observations with the corresponding correlation coefficients R1,R2..RR. Let aC, bC and RC be the intercept,slope and correlation coefficient obtained when X,Y data from all the sets are combined together.
: Problem: The objective is to get the maximum correlation coefficient for the combined data by transforming the Y data of each set using multiplying factors k1,k2..kR (that is each observed value of Y in a set is multiplied by a factor k). The solution must consist of the optimal set of factors (k1,k2..kR) which yield the maximum correlation coefficient for the combined data.

: Any ideas, pointers,references?
: Ofcourse i am aware that a solution may be found by brute force- trial and error! But is there an elegant analytical solution?
: Also, as a corollary, can i pose the same problem not with a linear relationship but with a nonlinear one?

: Thanks!

READERS RESPOND:
(In chronological order. Most recent at the bottom.)

Re: Maximization of R
Message posted by Phil Rosenkrantz on April 27, 2000 at 12:00 AM (ET)

Actually, calculating all possible regression models and sorting by R is a very standard approach for maximizing R. There are also other procedures for sequentially finding the best models (e.g. backward stepwise, forward stepwise, etc.). Then there are a few statistics that also helps select a model that balances the number of variables in the model with the R value (e.g. Mallows Cp).

Hopefully others will chime in with more ideas. I apologize if you already knew all this.

Re: Thanks Phil, but...
Message posted by Lakshman Nandagiri on April 28, 2000 at 12:00 AM (ET)