[R] Determining Starting Values for Model Parameters in Nonlinear Regression
John Fox
j|ox @end|ng |rom mcm@@ter@c@
Sun Aug 20 00:48:28 CEST 2023
Dear John, John, and Paul,
In this case, one can start values by just fitting
> lm(1/y ~ x1 + x2 + x3 - 1, data=mydata)
Call:
lm(formula = 1/y ~ x1 + x2 + x3 - 1, data = mydata)
Coefficients:
x1 x2 x3
0.00629 0.00868 0.00803
Of course, the errors enter this model differently, so this isn't the
same as the nonlinear model, but the regression coefficients are very
close to the estimates for the nonlinear model.
Best,
John
--
John Fox, Professor Emeritus
McMaster University
Hamilton, Ontario, Canada
web: https://www.john-fox.ca/
On 2023-08-19 6:39 p.m., Sorkin, John wrote:
> Caution: External email.
>
>
> Colleagues,
>
> At the risk of starting a forest fire, or perhaps a brush fire, while it is good to see that nlxb can find a solution from arbitrary starting values, I think Paul’s question has merit despite Professor Nash’s excellent and helpful observation.
>
> Although non-linear algorithms can converge, they can converge to a false solution if starting values are sub-optimally specified. When possible, I try to specify thought-out starting values. Would it make sense to plot y as a function of (x1, x2) at different values of x3 to get a sense of possible starting values? Or, perhaps using median values of x1, x2, and x3 as starting values. Comparing results from different starting values can give some confidence that the solution obtained using arbitrary starting values are likely “correct”.
>
> I freely admit that my experience (and thus expertise) using non-linear solutions is limited. Please do not flame me, I am simply urging caution.
>
> John
>
> John David Sorkin M.D., Ph.D.
> Professor of Medicine
> Chief, Biostatistics and Informatics
> University of Maryland School of Medicine Division of Gerontology and Geriatric Medicine
> Baltimore VA Medical Center
> 10 North Greene Street<x-apple-data-detectors://12>
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>
> On Aug 19, 2023, at 4:35 PM, J C Nash <profjcnash using gmail.com<mailto:profjcnash using gmail.com>> wrote:
>
> Why bother. nlsr can find a solution from very crude start.
>
> Mixture <- c(17, 14, 5, 1, 11, 2, 16, 7, 19, 23, 20, 6, 13, 21, 3, 18, 15, 26, 8, 22)
> x1 <- c(69.98, 72.5, 77.6, 79.98, 74.98, 80.06, 69.98, 77.34, 69.99, 67.49, 67.51, 77.63,
> 72.5, 67.5, 80.1, 69.99, 72.49, 64.99, 75.02, 67.48)
> x2 <- c(29, 25.48, 21.38, 19.85, 22, 18.91, 29.99, 19.65, 26.99, 29.49, 32.47,
> 20.35, 26.48, 31.47, 16.87, 27.99, 24.49, 31.99, 24.96, 30.5)
> x3 <- c(1, 2, 1, 0, 3, 1, 0, 2.99, 3, 3, 0, 2, 1, 1, 3, 2,
> 3, 3, 0, 2)
> y <- c(1.4287, 1.4426, 1.4677, 1.4774, 1.4565,
> 1.4807, 1.4279, 1.4684, 1.4301, 1.4188, 1.4157, 1.4686, 1.4414,
> 1.4172, 1.4829, 1.4291, 1.4438, 1.4068, 1.4524, 1.4183)
> mydata<-data.frame(Mixture, x1, x2, x3, y)
> mydata
> mymod <- y ~ 1/(Beta1*x1 + Beta2*x2 + Beta3*x3)
> library(nlsr)
> strt<-c(Beta1=1, Beta2=2, Beta3=3)
> trysol<-nlxb(formula=mymod, data=mydata, start=strt, trace=TRUE)
> trysol
> # or pshort(trysol)
>
>
> Output is
>
> residual sumsquares = 1.5412e-05 on 20 observations
> after 29 Jacobian and 43 function evaluations
> name coeff SE tstat pval gradient JSingval
> Beta1 0.00629212 5.997e-06 1049 2.425e-42 4.049e-08 721.8
> Beta2 0.00867741 1.608e-05 539.7 1.963e-37 -2.715e-08 56.05
> Beta3 0.00801948 8.809e-05 91.03 2.664e-24 1.497e-08 10.81
>
> J Nash
>
>
> On 2023-08-19 16:19, Paul Bernal wrote:
> Dear friends,
> Hope you are all doing well and having a great weekend. I have data that
> was collected on specific gravity and spectrophotometer analysis for 26
> mixtures of NG (nitroglycerine), TA (triacetin), and 2 NDPA (2 -
> nitrodiphenylamine).
> In the dataset, x1 = %NG, x2 = %TA, and x3 = %2 NDPA.
> The response variable is the specific gravity, and the rest of the
> variables are the predictors.
> This is the dataset:
> dput(mod14data_random)
> structure(list(Mixture = c(17, 14, 5, 1, 11, 2, 16, 7, 19, 23,
> 20, 6, 13, 21, 3, 18, 15, 26, 8, 22), x1 = c(69.98, 72.5, 77.6,
> 79.98, 74.98, 80.06, 69.98, 77.34, 69.99, 67.49, 67.51, 77.63,
> 72.5, 67.5, 80.1, 69.99, 72.49, 64.99, 75.02, 67.48), x2 = c(29,
> 25.48, 21.38, 19.85, 22, 18.91, 29.99, 19.65, 26.99, 29.49, 32.47,
> 20.35, 26.48, 31.47, 16.87, 27.99, 24.49, 31.99, 24.96, 30.5),
> x3 = c(1, 2, 1, 0, 3, 1, 0, 2.99, 3, 3, 0, 2, 1, 1, 3, 2,
> 3, 3, 0, 2), y = c(1.4287, 1.4426, 1.4677, 1.4774, 1.4565,
> 1.4807, 1.4279, 1.4684, 1.4301, 1.4188, 1.4157, 1.4686, 1.4414,
> 1.4172, 1.4829, 1.4291, 1.4438, 1.4068, 1.4524, 1.4183)), row.names =
> c(NA,
> -20L), class = "data.frame")
> The model is the following:
> y = 1/(Beta1x1 + Beta2x2 + Beta3x3)
> I need to determine starting (initial) values for the model parameters for
> this nonlinear regression model, any ideas on how to accomplish this using
> R?
> Cheers,
> Paul
> [[alternative HTML version deleted]]
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> and provide commented, minimal, self-contained, reproducible code.
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>
> ______________________________________________
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> PLEASE do read the posting guide http://www.r-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
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