Box-cox transformation

2 messages · Joshua Wiley, Ravi Varadhan

Mon, Jul 7, 2014 1:15 PM #

Dear Ravi,

In my previous example, I used the residuals, so:

sum [ (r_i / scaling)^2 ]

If you want to use the deviance from glm, that gives you:

sum [ r_i^2 ]

and since the scaling factor is just a constant for any given lambda,
then the modification would be:

sum [ r_i^2 ] / ( scaling^2 )

and is given in the modified code below (posted back to R-help in case
any else has this question).

Hope this helps,

Josh


##########################################
require(MASS)

myp <- function(y, lambda) (y^lambda-1)/lambda


lambda <- seq(-0.05, 0.45, len = 20)
N <- nrow(quine)
res <- matrix(numeric(0), nrow = length(lambda), 2, dimnames =
list(NULL, c("Lambda", "LL")))

# scaling contant
C <- exp(mean(log(quine$Days+1)))

for(i in seq_along(lambda)) {
  SS <- deviance(glm(myp(Days + 1, lambda[i]) ~ Eth*Sex*Age*Lrn, data = quine))
  LL <- (- (N/2) * log(SS/((C^lambda[i])^2)))
  res[i, ] <- c(lambda[i], LL)
}

# box cox
boxcox(Days+1 ~ Eth*Sex*Age*Lrn, data = quine, lambda = lambda)
# add our points on top to verify match
points(res[, 1], res[,2], pch = 16)

##########################################

On Mon, Jul 7, 2014 at 11:57 PM, Ravi Varadhan <ravi.varadhan at jhu.edu> wrote:

Dear Josh,
Thank you very much.  I knew that the scaling had to be adjusted, but was not sure on how to do this.

Can you please show me how to do this scaling with `glm'? In other words, how would I scale the deviance from glm?

Thanks,
Ravi

-----Original Message-----
From: Joshua Wiley [mailto:jwiley.psych at gmail.com]
Sent: Sunday, July 06, 2014 11:34 PM
To: Ravi Varadhan
Cc: r-help at r-project.org
Subject: Re: [R] Box-cox transformation

Hi Ravi,

Deviance is the SS in this case, but you need a normalizing constant adjusted by the lambda to put them on the same scale.  I modified your example below to simplify slightly and use the normalization (see the LL line).

Cheers,

Josh

######################################

require(MASS)

myp <- function(y, lambda) (y^lambda-1)/lambda


lambda <- seq(-0.05, 0.45, len = 20)
N <- nrow(quine)
res <- matrix(numeric(0), nrow = length(lambda), 2, dimnames = list(NULL, c("Lambda", "LL")))

# scaling contant
C <- exp(mean(log(quine$Days+1)))

for(i in seq_along(lambda)) {
  r <- resid(lm(myp(Days + 1, lambda[i]) ~ Eth*Sex*Age*Lrn, data = quine))
  LL <- (- (N/2) * log(sum((r/(C^lambda[i]))^2)))
  res[i, ] <- c(lambda[i], LL)
}

# box cox
boxcox(Days+1 ~ Eth*Sex*Age*Lrn, data = quine, lambda = lambda) # add our points on top to verify match points(res[, 1], res[,2], pch = 16)

######################################



On Mon, Jul 7, 2014 at 11:33 AM, Ravi Varadhan <ravi.varadhan at jhu.edu> wrote:

Hi,

I am trying to do Box-Cox transformation, but I am not sure how to do it correctly.  Here is an example showing what I am trying:



# example from MASS

require(MASS)
boxcox(Days+1 ~ Eth*Sex*Age*Lrn, data = quine,
       lambda = seq(-0.05, 0.45, len = 20))

# Here is My attempt at getting the profile likelihood for the Box-Cox
parameter lam <- seq(-0.05, 0.45, len = 20) dev <- rep(NA, length=20)

for (i in 1:20) {
a <- lam[i]
ans <- glm(((Days+1)^a-1)/a ~ Eth*Sex*Age*Lrn, family=gaussian, data =
quine) dev[i] <- ans$deviance }

plot(lam, dev, type="b", xlab="lambda", ylab="deviance")

I am trying to create the profile likelihood for the Box-Cox parameter, but obviously I am not getting it right.  I am not sure that ans$deviance is the right thing to do.

I would appreciate any guidance.

Thanks & Best,
Ravi




        [[alternative HTML version deleted]]

______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Joshua F. Wiley
Ph.D. Student, UCLA Department of Psychology
http://joshuawiley.com/
Senior Analyst, Elkhart Group Ltd.
http://elkhartgroup.com
Office: 260.673.5518

Ravi Varadhan

Mon, Jul 7, 2014 1:50 PM #

Thank you.  It is very helpful.

Ravi

-----Original Message-----
From: Joshua Wiley [mailto:jwiley.psych at gmail.com] 
Sent: Monday, July 07, 2014 4:15 PM
To: Ravi Varadhan
Cc: r-help at r-project.org
Subject: Re: [R] Box-cox transformation

Dear Ravi,

In my previous example, I used the residuals, so:

sum [ (r_i / scaling)^2 ]

If you want to use the deviance from glm, that gives you:

sum [ r_i^2 ]

and since the scaling factor is just a constant for any given lambda, then the modification would be:

sum [ r_i^2 ] / ( scaling^2 )

and is given in the modified code below (posted back to R-help in case any else has this question).

Hope this helps,

Josh


##########################################
require(MASS)

myp <- function(y, lambda) (y^lambda-1)/lambda


lambda <- seq(-0.05, 0.45, len = 20)
N <- nrow(quine)
res <- matrix(numeric(0), nrow = length(lambda), 2, dimnames = list(NULL, c("Lambda", "LL")))

# scaling contant
C <- exp(mean(log(quine$Days+1)))

for(i in seq_along(lambda)) {
  SS <- deviance(glm(myp(Days + 1, lambda[i]) ~ Eth*Sex*Age*Lrn, data = quine))
  LL <- (- (N/2) * log(SS/((C^lambda[i])^2)))
  res[i, ] <- c(lambda[i], LL)
}

# box cox
boxcox(Days+1 ~ Eth*Sex*Age*Lrn, data = quine, lambda = lambda) # add our points on top to verify match points(res[, 1], res[,2], pch = 16)

##########################################

On Mon, Jul 7, 2014 at 11:57 PM, Ravi Varadhan <ravi.varadhan at jhu.edu> wrote:

Dear Josh,
Thank you very much.  I knew that the scaling had to be adjusted, but was not sure on how to do this.

Can you please show me how to do this scaling with `glm'? In other words, how would I scale the deviance from glm?

Thanks,
Ravi

-----Original Message-----
From: Joshua Wiley [mailto:jwiley.psych at gmail.com]
Sent: Sunday, July 06, 2014 11:34 PM
To: Ravi Varadhan
Cc: r-help at r-project.org
Subject: Re: [R] Box-cox transformation

Hi Ravi,

Deviance is the SS in this case, but you need a normalizing constant adjusted by the lambda to put them on the same scale.  I modified your example below to simplify slightly and use the normalization (see the LL line).

Cheers,

Josh

######################################

require(MASS)

myp <- function(y, lambda) (y^lambda-1)/lambda


lambda <- seq(-0.05, 0.45, len = 20)
N <- nrow(quine)
res <- matrix(numeric(0), nrow = length(lambda), 2, dimnames = 
list(NULL, c("Lambda", "LL")))

# scaling contant
C <- exp(mean(log(quine$Days+1)))

for(i in seq_along(lambda)) {
  r <- resid(lm(myp(Days + 1, lambda[i]) ~ Eth*Sex*Age*Lrn, data = quine))
  LL <- (- (N/2) * log(sum((r/(C^lambda[i]))^2)))
  res[i, ] <- c(lambda[i], LL)
}

# box cox
boxcox(Days+1 ~ Eth*Sex*Age*Lrn, data = quine, lambda = lambda) # add 
our points on top to verify match points(res[, 1], res[,2], pch = 16)

######################################



On Mon, Jul 7, 2014 at 11:33 AM, Ravi Varadhan <ravi.varadhan at jhu.edu> wrote:

Hi,

I am trying to do Box-Cox transformation, but I am not sure how to do it correctly.  Here is an example showing what I am trying:



# example from MASS

require(MASS)
boxcox(Days+1 ~ Eth*Sex*Age*Lrn, data = quine,
       lambda = seq(-0.05, 0.45, len = 20))

# Here is My attempt at getting the profile likelihood for the 
Box-Cox parameter lam <- seq(-0.05, 0.45, len = 20) dev <- rep(NA, 
length=20)

for (i in 1:20) {
a <- lam[i]
ans <- glm(((Days+1)^a-1)/a ~ Eth*Sex*Age*Lrn, family=gaussian, data 
=
quine) dev[i] <- ans$deviance }

plot(lam, dev, type="b", xlab="lambda", ylab="deviance")

I am trying to create the profile likelihood for the Box-Cox parameter, but obviously I am not getting it right.  I am not sure that ans$deviance is the right thing to do.

I would appreciate any guidance.

Thanks & Best,
Ravi




        [[alternative HTML version deleted]]

______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

--
Joshua F. Wiley
Ph.D. Student, UCLA Department of Psychology http://joshuawiley.com/ Senior Analyst, Elkhart Group Ltd.
http://elkhartgroup.com
Office: 260.673.5518