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I am trying to simulate a dataset that matches empirical data that I have, but am unsure how to estimate the errors in the original data. The empirical data includes heteroscedasticity, but I am not interested in transforming it away, but rather using a linear model with an error term to reproduce simulations of the empirical data.

For example, let's say I have some an empirical dataset and a model:

n=rep(1:100,2)
a=0
b = 1
sigma2 = n^1.3
eps = rnorm(n,mean=0,sd=sqrt(sigma2))
y=a+b*n + eps
mod <- lm(y ~ n)

using plot(n,y) we get the following. enter image description here

However, if I try to simulate the data, simulate(mod), the heteroscedasticity is removed and not captured by the model.

I can use a generalized least squares model

VMat <- varFixed(~n)
mod2 = gls(y ~ n, weights = VMat)

that provides a better model fit based on AIC, but I don't know how to simulate data using the output.

My question is, how do I create a model that will allow me to simulate data to match the original, empirical data (n and y above). Specifically, I need a way to estimate sigma2, the error, using either using a model?

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    $\begingroup$ So the linear model won't capture conditional heteroskedasticity unless it explicitly tries to do so, using one of a few approaches. Standard econometric techniques adjust the standard errors on parameters to account for heteroskedasticity, but they don't explicitly model it. $\endgroup$ – generic_user Jan 27 '17 at 14:44
  • $\begingroup$ You're right. I am trying to use a linear model to capture the heterogeneity. I think that I should be using a generalized least squares model. If there are any other recommendations, I will try them. $\endgroup$ – user44796 Jan 27 '17 at 14:55
  • $\begingroup$ There is aN ERROR IN YOUR CODE, YOU MUST USE ` lm( y ~ n)` $\endgroup$ – kjetil b halvorsen Jan 27 '17 at 15:04
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    $\begingroup$ I do not understand your question, because your code accomplishes exactly what you seem to be asking for in its title: it simulates a linear regression with heteroscedastic errors. Are you asking for methods to estimate some kind of model for the heteroscedasticity? If so, then you need to specify a model! $\endgroup$ – whuber Jan 27 '17 at 15:22
  • $\begingroup$ Hopefully I have clarified my question with edits. In the above question, n and y represent the empirical data. I want to fit a model to the data and then use the model to generate simulated data that matches the mean and residuals of the original data. $\endgroup$ – user44796 Jan 27 '17 at 15:38
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To simulate data with a varying error variance, you need to specify the data generating process for the error variance. As has been pointed out in the comments, you did that when you generated your original data. If you have real data and want to try this, you just need to identify the function that specifies how the residual variance depends on your covariates. The standard way to do that is to fit your model, check that it is reasonable (other than the heteroscedasticity), and save the residuals. Those residuals become the Y variable of a new model. Below I have done that for your data generating process. (I don't see where you set the random seed, so these won't literally be the same data, but should be similar, and you can reproduce mine exactly by using my seed.)

set.seed(568)  # this makes the example exactly reproducible

n      = rep(1:100,2)
a      = 0
b      = 1
sigma2 = n^1.3
eps    = rnorm(n,mean=0,sd=sqrt(sigma2))
y      = a+b*n + eps
mod    = lm(y ~ n)
res    = residuals(mod)

windows()
  layout(matrix(1:2, nrow=2))
  plot(n,y)
  abline(coef(mod), col="red")
  plot(mod, which=3)

enter image description here

Note that R's ?plot.lm will give you a plot (cf., here) of the square root of the absolute values of the residuals, helpfully overlaid with a lowess fit, which is just what you need. (If you have multiple covariates, you might want to assess this against each covariate separately.) There is the slightest hint of a curve, but this looks like a straight line does a good job of fitting the data. So let's explicitly fit that model:

res.mod = lm(sqrt(abs(res))~fitted(mod))
summary(res.mod)
# Call:
# lm(formula = sqrt(abs(res)) ~ fitted(mod))
# 
# Residuals:
#     Min      1Q  Median      3Q     Max 
# -3.3912 -0.7640  0.0794  0.8764  3.2726 
# 
# Coefficients:
#             Estimate Std. Error t value Pr(>|t|)    
# (Intercept) 1.669571   0.181361   9.206  < 2e-16 ***
# fitted(mod) 0.023558   0.003157   7.461 2.64e-12 ***
# ---
# Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
# 
# Residual standard error: 1.285 on 198 degrees of freedom
# Multiple R-squared:  0.2195,  Adjusted R-squared:  0.2155 
# F-statistic: 55.67 on 1 and 198 DF,  p-value: 2.641e-12
windows()
  layout(matrix(1:4, nrow=2, ncol=2, byrow=TRUE))
  plot(res.mod, which=1)
  plot(res.mod, which=2)
  plot(res.mod, which=3)
  plot(res.mod, which=5)

enter image description here

We needn't be concerned that the residual variance seems to be increasing in the scale-location plot for this model as well—that essentially has to happen. There is again the slightest hint of a curve, so we can try to fit a squared term and see if that helps (but it doesn't):

res.mod2 = lm(sqrt(abs(res))~poly(fitted(mod), 2))
summary(res.mod2)
# output omitted
anova(res.mod, res.mod2)
# Analysis of Variance Table
# 
# Model 1: sqrt(abs(res)) ~ fitted(mod)
# Model 2: sqrt(abs(res)) ~ poly(fitted(mod), 2)
#   Res.Df    RSS Df Sum of Sq     F Pr(>F)
# 1    198 326.87                          
# 2    197 326.85  1  0.011564 0.007 0.9336

If we're satisfied with this, we can now use this process as an add-on to simulate data.

set.seed(4396)  # this makes the example exactly reproducible
x = n
expected.y = coef(mod)[1] + coef(mod)[2]*x
sim.errors = rnorm(length(x), mean=0,
                   sd=(coef(res.mod)[1] + coef(res.mod)[2]*expected.y)^2)
observed.y = expected.y + sim.errors

Note that this process is no more guaranteed to find the true data generating process than any other statistical method. You used a non-linear function to generate the error SDs, and we approximated it with a linear function. If you actually know the true data generating process a-priori (as in this case, because you simulated the original data), you might as well use it. You can decide if the approximation here is good enough for your purposes. We typically don't know the true data generating process, however, and based on Occam's razor, go with the simplest function that adequately fits the data we have given the amount of information available. You can also try splines or fancier approaches if you prefer. The bivariate distributions look reasonably similar to me, but we can see that while the estimated function largely parallels the true function, they do not overlap:

enter image description here

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  • $\begingroup$ This was actually a conclusion I was starting to come to, but would never have arrived at so elegant an answer. $\endgroup$ – user44796 Jan 27 '17 at 19:33
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You need to model the heteroskedasticity. One approach is via the R package (CRAN) dglm, dispersion generalized linear model. This is an extension of glm's which, in addition to the usual glm, fits a second glm for dispersion from the residuals from the first glm. I have no experience with such models, but they seem promising ... Here is some code:

n <- rep(1:100,2)
a <- 0
b <- 1
sigma2 <- n^1.3
eps <- rnorm(n,mean=0,sd=sqrt(sigma2))
y <- a+b*n + eps
mod <- lm(y ~ n)

library(dglm)  ### double glm's

mod2   <-  dglm(y ~ n, ~ n, gaussian,ykeep=TRUE,xkeep=TRUE,zkeep=TRUE)
### This uses log link for the dispersion part, should also try identity link ..

y2 <-  simulate(mod2)

plot(n, y2$sim_1)

mod3  <-  dglm(y ~ n, ~ n, gaussian, dlink="identity", ykeep=TRUE,xkeep=TRUE,zkeep=TRUE)  ### This do not work because it leads to negative weights!

The simulated plot is shown below:

enter image description here

The plot do look like the simulation have used the estimated variance, but I'm unsure, as the simulate() function do not have methods for dglm's ...

(Another possibility to look into, is using the R package gamlss, which uses another approach to modelling the variance as a function of covariables.)

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    $\begingroup$ the double generalized linear model appears to model the original data adequately. I am unclear as to how the residual error is modeled using predict(). I will have to look into that. $\endgroup$ – user44796 Jan 27 '17 at 19:08

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