How to define initial kicks for nls function for potential regression model?

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I would like to know how to define the initial "kicks" to use the power nonlinear regression model. I am working with estimation of data, testing various regression models, even if there is no adjustment, I would like to quote this model, but I have difficulty defining the values initial, below follows only a fragment of data for testing, I would like to know a method that allows me to apply in any data set, for obtaining the initial values of A and B for the function nls(y~B*x^A,start=list(A=1, B=1.7))

dados<-structure(list(y = c(44.42, 77.9, 95.72, 40.24, 63.7, 46.62, 
84.6, 52.49, 88.53, 56.52, 71.21, 65.16, 72.24, 53.81, 67.02), 
    x = c(11.26, 14.78, 17.56, 10.37, 13.27, 10.3, 14.07, 12.26, 
    13.3, 12.84, 13.72, 12.8, 14.86, 11.47, 15.06)), .Names = c("y", 
"x"), row.names = c(NA, -15L), class = "data.frame")

attach(dados)

nls(y~B*x^A,start=list(A=1, B=1.7))
r

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That answer Cross-Validated seems a good solution.

The suggestion here is to take the log on both sides and adjust a linear model. That would look something like this:

y = b*(x^a)
log(y) = log(b) + a*log(x)

Therefore, by making the linear model you will have an initial estimate for log(b) and to a.

Adjust the linear model:

linear <- lm(log(y) ~ log(x), data = dados)
linear
# Call:
#   lm(formula = log(y) ~ log(x), data = dados)
# 
# Coefficients:
#   (Intercept)       log(x)  
#        0.1371       1.5611

Thus you will have 0.1371 the estimate of log(b) and 1.5611 estimation of a. A direct estimate of b can be exp(0.1371) = 1.146943. These values can be used as starting kicks.

modelo <- nls(y~B*x^A,start=list(A=coef(linear)[2], B = exp(coef(linear)[1])), data = dados)
modelo

# Nonlinear regression model
# model: y ~ B * x^A
# data: dados
# A     B 
# 1.436 1.598 
# residual sum-of-squares: 1040
# 
# Number of iterations to convergence: 5 
# Achieved convergence tolerance: 3.365e-07

In the link has several other answers that may help!

  • 1

    Excellent answer. Logging on both sides was my idea to linearize the problem and answer it. However, is this question not beyond the scope of the site? I looked for questions on meta and it seems to me that there is no consensus among users as to what to do in cases like this. Personally, I think you were right to answer, but there’s no risk of anyone coming here and closing the question?

  • 1

    @Marcusnunes Thanks! For me this question is on the frontier of "inside" and "outside" the scope... There is a risk of someone closing, but as it is at the border, you will find it difficult to get the 5 votes needed. The answer from Carlos reflects well my opinion. I mean, I just don’t answer if the question has nothing to do with programming. Here the problem of AP is more: I want to run nls and I can not pq the convergence problem, how to correct?. The way I see it, he didn’t come to me for help in statistics.

  • I will take this into account in my next responses. I need to learn to answer what is asked and not go too far in statistical theory on this site.

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