# Resolving heteroskedasticity in linear regression models

I am working on an age estimation method using 4 types of biological measurements as age predictors. I am using RStudio. So far, I have good results when I use linear regression (lm(age~predictor)), but I am encountering heteroskedasticity, and therefore cannot build prediction intervals for my models.
I have tried transformations to normalize the predictors using ln, inverse, and square root, but to no avail.
I have found a paper explaining the wls function, and I have used it in my models with the weight: $$\frac 1 {1+\frac{\text{predictor}^2} 2}$$ This has given me better age predictions, but does not solve the heteroskedasticity problem.

I have done some research, and apparently, one of my options is to create homoscedastic groups in my data by finding the data points where the residual variances change. For that, I have used the breakpoints function of strucchange, which gave me 5 breakpoints by default. I now want to give 6 different weights (weights are $\frac 1 {\text{var(age)}}$ of each interval) to my 6 intervals of data, but I cannot find a function to do that. I would greatly appreciate any help on the subject. Thanks.

## migrated from stackoverflow.comFeb 3 '14 at 20:39

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• I know this doesn't answer your question, but robust standard errors (Huber-White or otherwise, check out the sandwich package for those) won't do? Sure, the estimation will be less efficient but at least you'll be sure the variance-covariance matrix will be consistently estimated. – RoyalTS Jan 23 '14 at 12:04
• I think you would benefit from mixed effects models. A model where you can specify the variance structure would foresee heteroskedasticity. See Zuur et al. 2009 page 75+ for more details. Another good resource on MEM is Pinheiro and Bates.ž – Roman Luštrik Jan 23 '14 at 12:05
• I second @RomanLuštrik as to the use of mixed effects models. Within that framework you can explicitly model variance, and therefore heteroscedasticity. You can do it in R as well. – Maxim.K Jan 23 '14 at 12:09
• The function is gls from nlme package. – Roman Luštrik Jan 23 '14 at 13:02
• Thank you so much for all your help. I will try all of these suggestions and let you know how it goes. Thanks again! – user3227551 Jan 24 '14 at 8:53

Perhaps you can use the rlm() function from the MASS package. I believe it's intended for datasets affected by outliers, but it's at least worth a try with your dataset.

Description: Fit a linear model by robust regression using an M estimator.

Details: Fitting is done by iterated re-weighted least squares (IWLS).

# Package examples
require(MASS)
summary(rlm(stack.loss ~ ., stackloss))
rlm(stack.loss ~ ., stackloss, psi = psi.hampel, init = "lts")
rlm(stack.loss ~ ., stackloss, psi = psi.bisquare)

# Using the diamonds data set
diam_rlm <- rlm(price ~ ., diamonds)
summary(diam_rlm)

# A plot as example
ggplot(diamonds, aes(carat,price)) +
geom_point() +
stat_smooth(method = "rlm") +
xlim(0,2.9)


For the diamonds dataset, i'm sure you could improve the model with a transformation. However, as you aren't saying this works with your data i left the diamonds data "as is".