# [R] nonlinear least squares, multiresponse

Douglas Bates bates at stat.wisc.edu
Thu Apr 25 17:04:57 CEST 2002

Bill Simpson <W.Simpson at gcal.ac.uk> writes:

> Here is a solution that is fairly easy to do. (Not sure how statistically
> reasonable it is--seems OK to me)
>
> Use nlm() or optimize() to minimise the sum of the squared residuals
> across all DVs simultaneously. If your IVs are a,b,c , your parameters
> are b0,b1,b2, and your DVs are x,y,z, then nlm() will iteratively minimize
>
> (x-fnx(a,b,c,b0,b1,b2))^2+
> (y-fny(a,b,c,b0,b1,b2))^2+
> (z-fnz(a,b,c,b0,b1,b2))^2
>
> Or I suppose you could write a function that returns a vector
> xhat, yhat, zhat
>
> Here is a simple 1 IV 1 DV example to base your code on:
>
> x<-c(0.02, 0.02, 0.06, 0.06, 0.11, 0.11, 0.22, 0.22, 0.56, 0.56, 1.10,
> 1.10)
> y<-c(76, 47, 97, 107, 123, 139, 159, 152, 191, 201, 207, 200)
> fn <- function(p) sum((y - (p[1] * x)/(p[2] + x))^2)
> out<-nlm(fn,p=c(200,.1),hessian=TRUE)
> out
> #SSE = out\$minimum; MSE = out\$minimum/(n-p), n is #pts, p is #params
> #estimates=out\$estimate
> se<-sqrt(diag(2*out\$minimum/(length(y)-2)*solve(out\$hessian)))  #SEs

You are making assumptions in doing that.  You are assuming that the
"noise terms" on your responses are independent and have comparable
variances.  This may not be the case.

The Box-Draper criterion is to minimize the square of the product of
the singular values of the matrix of the residuals.  I would recommend

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