I am running a simple multivariate regression on a panel/time-series dataset, using lm() and the underlying formula $(X'X)^{-1} X'Y$
expecting to get the same coefficient values from the two methods. However, I get completely different estimates.
Here is the R code:
return = matrix(ret.ff.zoo, ncol = 50) # y vector
data = cbind(df$EQ, df$EFF, df$SIZE, df$MOM, df$MSCR, df$SY, df$UMP) # x vector
#First method
BETA = solve(crossprod(data)) %*% crossprod(data, return)
#Second method
OLS <- lm(return ~ data)
I am not sure why the estimates are different between the two methods..
Any help is appreciated! Thank you.
Your example isn't reproducible, but if you try it with some dummy data, the matrix formula and lm produce the same results when you take out the intercept:
set.seed(1)
x <- matrix(rnorm(1000),ncol=5)
y <- rnorm(200)
solve(t(x) %*% x) %*% t(x) %*% y
[,1]
[1,] -0.0826496646
[2,] -0.0165735273
[3,] -0.0009412659
[4,] 0.0070475728
[5,] -0.0642452777
> lm(y ~ x + 0)
Call:
lm(formula = y ~ x + 0)
Coefficients:
x1 x2 x3 x4 x5
-0.0826497 -0.0165735 -0.0009413 0.0070476 -0.0642453
Related
I want to estimate an equation such as:
(where the bar denotes the mean of a variable.... Meaning, I want to automatically have interactions between Z and a demeaned version of X. So far I just demean the variables manually beforehand and estimate:
lm(Y ~ .*Z, data= sdata)
This seems to be working, but I would rather use a solution that does not require manual demeaning beforehand because I would also like to include the means of more complex terms, such as:
Edit:
As requested, a working code-sample, note that in the actual thing I have large (and varying) numbers of X- variables, so that I dont want to use a hard-coded variant:
x1 <- runif(100)
x2 <- runif(100)
Z <- runif(100)
Y <- exp(x1) + exp(x2) + exp(z)
##current way of estimating the first equation:
sdata <- data.frame(Y=Y,Z=Z,x1=x1-mean(x1),x2=x2-mean(x2))
lm(Y ~ .*Z, data= sdata)
##basically what I want is that the following terms, and their interactions with Z are also used:
# X1^2 - mean(X1^2)
# X2^2 - mean(X2^2)
# X1*X2 - mean(X1*X2)
Edit 2:
Now, what I want to achieve is basically what
lm(Y ~ .^2*Z, data= sdata)
would do. However, given prior demeaing expressions in there, such as: Z:X1:X2 would correspond to: (x1-mean(x1))*(x2-mean(x2)), while what I want to have is x1*x2-mean(x1*x2)
To show that scale works inside a formula:
lm(mpg ~ cyl + scale(disp*hp, scale=F), data=mtcars)
Call:
lm(formula = mpg ~ cyl + scale(disp * hp, scale = F), data = mtcars)
Coefficients:
(Intercept) cyl scale(disp * hp, scale = F)
3.312e+01 -2.105e+00 -4.642e-05
Now for comparison let's scale the interaction outside the formula:
mtcars$scaled_interaction <- with(mtcars, scale(disp*hp, scale=F))
lm(mpg ~ cyl + scaled_interaction, data=mtcars)
Call:
lm(formula = mpg ~ cyl + scaled_interaction, data = mtcars)
Coefficients:
(Intercept) cyl scaled_interaction
3.312e+01 -2.105e+00 -4.642e-05
At least in these examples, it seems as if scale inside formulae is working.
To provide a solution to your specific issue:
Alternative 1: Use formulae
# fit without Z
mod <- lm(Y ~ (.)^2, data= sdata[, names(sdata) != "Z" ])
vars <- attr(mod$terms, "term.labels")
vars <- gsub(":", "*", vars) # needed so that scale works later
vars <- paste0("scale(", vars, ", scale=F)")
newf <- as.formula(paste0("Y ~ ", paste0(vars, collapse = "+")))
# now interact with Z
f2 <- update.formula(newf, . ~ .*Z)
# This fives the following formula:
f2
Y ~ scale(x1, scale = F) + scale(x2, scale = F) + scale(x1*x2, scale = F) +
Z + scale(x1, scale = F):Z + scale(x2, scale = F):Z + scale(x1*x2, scale = F):Z
Alternative 2: Use Model Matrices
# again fit without Z and get model matrix
mod <- lm(Y ~ (.)^2, data= sdata[, names(sdata) != "Z" ])
modmat <- apply(model.matrix(mod), 2, function(x) scale(x, scale=F))
Here, all x's and the interactions are demeaned:
> head(modmat)
(Intercept) x1 x2 x1:x2
[1,] 0 0.1042908 -0.08989091 -0.01095459
[2,] 0 0.1611867 -0.32677059 -0.05425087
[3,] 0 0.2206845 0.29820499 0.06422944
[4,] 0 0.3462069 -0.15636463 -0.05571430
[5,] 0 0.3194451 -0.38668844 -0.12510551
[6,] 0 -0.4708222 -0.32502269 0.15144812
> round(colMeans(modmat), 2)
(Intercept) x1 x2 x1:x2
0 0 0 0
You can use the model matrix as follows:
modmat <- modmat[, -1] # remove intercept
lm(sdata$Y ~ modmat*sdata$Z)
It is not beautiful, but should do the work with any number of explanatory variables. You can also add Y and Z to the matrix so that the output looks prettier if this is a concern. Note that you can also create the model matrix directly without fitting the model. I took it from the fitted model directly since it have already fitted it for the first approach.
As a sidenote, it may be that this is not implemented in a more straight forward fashion because it is difficult to imagine situations in which demeaning the interaction is more desirable compared to the interaction of demeaned variables.
Comparing both approaches:
Here the output of both approaches for comparison. As you can see, apart from the coefficient names everything is identical.
> lm(sdata$Y ~ modmat*sdata$Z)
Call:
lm(formula = sdata$Y ~ modmat * sdata$Z)
Coefficients:
(Intercept) modmatx1 modmatx2 modmatx1:x2 sdata$Z
4.33105 1.56455 1.43979 -0.09206 1.72901
modmatx1:sdata$Z modmatx2:sdata$Z modmatx1:x2:sdata$Z
0.25332 0.38155 -0.66292
> lm(f2, data=sdata)
Call:
lm(formula = f2, data = sdata)
Coefficients:
(Intercept) scale(x1, scale = F) scale(x2, scale = F)
4.33105 1.56455 1.43979
scale(x1 * x2, scale = F) Z scale(x1, scale = F):Z
-0.09206 1.72901 0.25332
scale(x2, scale = F):Z scale(x1 * x2, scale = F):Z
0.38155 -0.66292
I am writing a custom script to bootstrap standard errors in a GLM in R and receive the following error:
Error in eval(predvars, data, env) : numeric 'envir' arg not of length one
Can someone explain what I am doing wrong? My code:
#Number of simulations
sims<-numbersimsdesired
#Set up place to store data
saved.se<-matrix(NA,sims,numberofcolumnsdesired)
y<-matrix(NA,realdata.rownumber)
x1<-matrix(NA,realdata.rownumber)
x2<-matrix(NA,realdata.rownumber)
#Resample entire dataset with replacement
for (sim in 1:sims) {
fake.data<-sample(1:nrow(data5),nrow(data5),replace=TRUE)
#Define variables for GLM using fake data
y<-realdata$y[fake.data]
x1<-realdata$x1[fake.data]
x2<-realdata$x2[fake.data]
#Run GLM on fake data, extract SEs, save SE into matrix
glm.output<-glm(y ~ x1 + x2, family = "poisson", data = fake.data)
saved.se[sim,]<-summary(glm.output)$coefficients[0,2]
}
An example: if we suppose sims = 1000 and we want 10 columns (suppose instead of x1 and x2, we have x1...x10) the goal is a dataset with 1,000 rows and 10 columns containing each explanatory variable's SEs.
There isn't a reason to reinvent the wheel. Here is an example of bootstrapping the standard error of the intercept with the boot package:
set.seed(42)
counts <- c(18,17,15,20,10,20,25,13,12)
x1 <- 1:9
x2 <- sample(9)
DF <- data.frame(counts, x1, x2)
glm1 <- glm(counts ~ x1 + x2, family = poisson(), data=DF)
summary(glm1)$coef
# Estimate Std. Error z value Pr(>|z|)
#(Intercept) 2.08416378 0.42561333 4.896848 9.738611e-07
#x1 0.04838210 0.04370521 1.107010 2.682897e-01
#x2 0.09418791 0.04446747 2.118131 3.416400e-02
library(boot)
intercept.se <- function(d, i) {
glm1.b <- glm(counts ~ x1 + x2, family = poisson(), data=d[i,])
summary(glm1.b)$coef[1,2]
}
set.seed(42)
boot.intercept.se <- boot(DF, intercept.se, R=999)
#ORDINARY NONPARAMETRIC BOOTSTRAP
#
#
#Call:
#boot(data = DF, statistic = intercept.se, R = 999)
#
#
#Bootstrap Statistics :
# original bias std. error
#t1* 0.4256133 0.103114 0.2994377
Edit:
If you prefer doing it without a package:
n <- 999
set.seed(42)
ind <- matrix(sample(nrow(DF), nrow(DF)*n, replace=TRUE), nrow=n)
boot.values <- apply(ind, 1, function(...) {
i <- c(...)
intercept.se(DF, i)
})
sd(boot.values)
#[1] 0.2994377
Is it possible to replace coefficients in lm object?
I thought the following would work
# sample data
set.seed(2157010)
x1 <- 1998:2011
x2 <- x1 + rnorm(length(x1))
y <- 3*x1 + rnorm(length(x1))
fit <- lm( y ~ x1 + x2)
# view origional coefficeints
coef(fit)
# replace coefficent with new values
fit$coef(fit$coef[2:3]) <- c(5, 1)
# view new coefficents
coef(fit)
Any assistance would be greatly appreciated
Your code is not reproducible, as there's few errors in your code. Here's corrected version which shows also your mistake:
set.seed(2157010) #forgot set.
x1 <- 1998:2011
x2 <- x1 + rnorm(length(x1))
y <- 3*x2 + rnorm(length(x1)) #you had x, not x1 or x2
fit <- lm( y ~ x1 + x2)
# view original coefficients
coef(fit)
(Intercept) x1 x2
260.55645444 -0.04276353 2.91272272
# replace coefficients with new values, use whole name which is coefficients:
fit$coefficients[2:3] <- c(5, 1)
# view new coefficents
coef(fit)
(Intercept) x1 x2
260.5565 5.0000 1.0000
So the problem was that you were using fit$coef, although the name of the component in lm output is really coefficients. The abbreviated version works for getting the values, but not for setting, as it made new component named coef, and the coef function extracted the values of fit$coefficient.
I'm looking for suggestions on how to deal with NA's in linear regressions when all occurrences of an independent/explanatory variable are NA (i.e. x3 below).
I know the obvious solution would be to exclude the independent/explanatory variable in question from the model but I am looping through multiple regions and would prefer not to have a different functional forms for each region.
Below is some sample data:
set.seed(23409)
n <- 100
time <- seq(1,n, 1)
x1 <- cumsum(runif(n))
y <- .8*x1 + rnorm(n, mean=0, sd=2)
x2 <- seq(1,n, 1)
x3 <- rep(NA, n)
df <- data.frame(y=y, time=time, x1=x1, x2=x2, x3=x3)
# Quick plot of data
library(ggplot2)
library(reshape2)
df.melt <-melt(df, id=c("time"))
p <- ggplot(df.melt, aes(x=time, y=value)) +
geom_line() + facet_grid(variable ~ .)
p
I have read the documentation for lm and tried various na.action settings without success:
lm(y~x1+x2+x3, data=df, singular.ok=TRUE)
lm(y~x1+x2+x3, data=df, na.action=na.omit)
lm(y~x1+x2+x3, data=df, na.action=na.exclude)
lm(y~x1+x2+x3, data=df, singular.ok=TRUE, na.exclude=na.omit)
lm(y~x1+x2+x3, data=df, singular.ok=TRUE, na.exclude=na.exclude)
Is there a way to get lm to run without error and simply return a coefficient for the explanatory reflective of the lack of explanatory power (i.e. either zero or NA) from the variable in question?
Here's one idea:
set.seed(23409)
n <- 100
time <- seq(1,n, 1)
x1 <- cumsum(runif(n))
y <- .8*x1 + rnorm(n, mean=0, sd=2)
x2 <- seq(1,n, 1)
x3 <- rep(NA, n)
df <- data.frame(y=y, time=time, x1=x1, x2=x2, x3=x3)
replaceNA<-function(x){
if(all(is.na(x))){
rep(0,length(x))
} else x
}
lm(y~x1+x2+x3, data= data.frame(lapply(df,replaceNA)))
Call:
lm(formula = y ~ x1 + x2 + x3, data = data.frame(lapply(df, replaceNA)))
Coefficients:
(Intercept) x1 x2 x3
0.05467 1.01133 -0.10613 NA
lm(y~x1+x2, data=df)
Call:
lm(formula = y ~ x1 + x2, data = df)
Coefficients:
(Intercept) x1 x2
0.05467 1.01133 -0.10613
So you replace the variables which contain only NA's with variable which contains only 0's. you get the coefficient value NA, but all the relevant parts of the model fits are same (expect qr decomposition, but if information about that is needed, it can be easily modified). Note that component summary(fit)$alias (see ?alias) might be useful.
This seems to relate your other question: Replace lm coefficients in [r]
You won't be able to include a column with all NA values. It does strange things to model.matrix
x1 <- 1:5
x2 <- rep(NA,5)
model.matrix(~x1+x2)
(Intercept) x1 x2TRUE
attr(,"assign")
[1] 0 1 2
attr(,"contrasts")
attr(,"contrasts")$x2
[1] "contr.treatment"
So your alternative is to programatically create the model formula based on the data.
Something like...
make_formula <- function(variables, data, response = 'y'){
if(missing(data)){stop('data not specified')}
using <- Filter(variables,f= function(i) !all(is.na(data[[i]])))
deparse(reformulate(using, response))
}
variables <- c('x1','x2','x3')
make_formula(variables, data =df)
[1] "y ~ x1 + x2"
I've used deparse to return a character string so that there is no environment issues from creating the formula within the function. lm can happily take a character string which is a valid formula.
Let's say I have a data matrix d
pc = prcomp(d)
# pc1 and pc2 are the principal components
pc1 = pc$rotation[,1]
pc2 = pc$rotation[,2]
Then this should fit the linear regression model right?
r = lm(y ~ pc1+pc2)
But then I get this error :
Errormodel.frame.default(formula = y ~ pc1+pc2, drop.unused.levels = TRUE) :
unequal dimensions('pc1')
I guess there a packages out there who do this automatically, but this should work too?
Answer: you don't want pc$rotation, it's the rotation matrix and not the matrix of rotated values (scores).
Make up some data:
x1 = runif(100)
x2 = runif(100)
y = rnorm(2+3*x1+4*x2)
d = cbind(x1,x2)
pc = prcomp(d)
dim(pc$rotation)
## [1] 2 2
Oops. The "x" component is what we want. From ?prcomp:
x: if ‘retx’ is true the value of the rotated data (the centred (and scaled if requested) data multiplied by the ‘rotation' matrix) is returned.
dim(pc$x)
## [1] 100 2
lm(y~pc$x[,1]+pc$x[,2])
##
## Call:
## lm(formula = y ~ pc$x[, 1] + pc$x[, 2])
## Coefficients:
## (Intercept) pc$x[, 1] pc$x[, 2]
## 0.04942 0.14272 -0.13557