How to adjust the regression line so that 90% of the lines are below the line?

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I have the following data set in R: 

x <- c(0.1, 3, 4, 5, 9, 12, 13, 19, 22, 25)
y <- c(5, 12, 17, 23, 28, 39, 26, 31, 38, 40)
bd <- data.frame(x, y)

My question is how I do in R to generate a regression model that best fits so that 90% of the data is below the regression line and the model estimates the origin (zero).

Looks like the geometric model fits this case better. I tried to use the exponential as follows, it starts at the source but the data is not 90% below the curve.

library(ggplot2)

ggplot(bd,aes(x = x, y = y)) + 
  geom_point() + 
  stat_smooth(method = 'nls', formula = 'y~a*x^b', 
              method.args = list(start = list(a = 1, b = 1)), 
              se = FALSE)

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r regression
  • 1

    Hello! What have you tried to do to solve this problem? Edit your question and share with us your code, your attempts, what errors/problems have occurred, etc.

  • Assignment arrows are backwards.

  • Are you looking for fit0 <- quantreg::rq(y ~ 0 + x, tau = 0.90)?

  • Thanks Rui, I tried to do as you put it, but I think the line does not fit well to the data, I tried to use the exponential as follows, it starts at the source but the data is not 90% below the curve: ggplot(bd,aes(x = x,y = y)) + geom_point() + stat_smooth(method = 'nls', formula = 'y~a*x b', start = list(a = 1,b=1),tau =0.9, if=FALSE)

  • ggplot(bd, aes(x = log(x), y = log(y))) + geom_point() + stat_smooth(method = 'quantreg::rq', formula = 'y ~ x', method.args = list(tau = 0.9), se = FALSE)

  • @Noisy question opened, can be answered in the corresponding field now.

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1 answer

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Here are two ways to solve the problem using the package quantreg.

The formula y = a*x^b can be transformed by applying logarithms and adjusting the resulting model, i.e., a robust regression line at quantile 0.90.

1. This can be done automatically with the function stat_smooth package ggplot2.

library(ggplot2)

ggplot(bd, aes(x = log(x), y = log(y))) + 
  geom_point() + 
  stat_smooth(method = quantreg::rq, formula = 'y ~ x', 
              method.args = list(tau = 0.9), se = FALSE)

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2. You can also make an adjustment and calculate the adjusted values.

fit90 <- quantreg::rq(log(y) ~ log(x), tau = 0.90, data = bd)
xnew <- seq(min(x), max(x), length.out = 50)
y90 <- exp(predict(fit90, newdata = data.frame(x = xnew)))
pred90 <- data.frame(x = xnew, y = y90)

ggplot(bd, aes(x, y)) +
  geom_point() +
  geom_line(data = pred90, aes(x, y), colour = "blue")

inserir a descrição da imagem aqui

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