R help - Jun 2005 - Linear Models with mean as Intercept.

Dear advanced statisticians,

*******Objectif********

I try to set up linear models with mean as intercept:
Answer: y
Variable: x, as factor of two modalities: x(1), x(2).

I would like to have a model as:
y = mean(y)+A(i)+residuals,
with i in (1,2) and A(1) coefficient for x(1) and A(2) coefficient for x(2).

*******Trials in R*******

## Firstly:

I write in R:
 >Model<-lm(y~x,Data)
 >summary(Model)
...
I've got the coefficients for:
- the intercept (x(1) as been choosen) that we can call B(1)
- the second modality: x(2) that we can call B(2)

If I have well understood we have for the model and predictions:
if x(1): y=B(1)
if x(2): y=B(1)+B(2)
which is quite different as y=mean(y)+A(i)

## Secondly
I tried to skip the intercept
 >Model2<-lm(y~0+x,Data)
 >summary(Model2)
...
I've got the coefficients for:
- the first modality: x(1) that we can call C(1)
- the second modality: x(2) that we can call C(2)

And the model and predictions, if I'm right, are:
if x(1): y=C(1)
if x(2): y=C(2)

******* Questions ***********
How can I obtain a predictive model y=mean(y)+A(i) ?
Is it possible to settle mean(y) as intercept?

Thanks for your help.

Ghislain V., retarded statistician.

you could model: (y-mean(y))~x
hth, ingmar
> From: Ghislain Vieilledent <ghislainv at gmail.com>
> Date: Thu, 30 Jun 2005 15:35:30 +0200
> To: r-help at stat.math.ethz.ch
> Subject: [R] Linear Models with mean as Intercept.
> 
> ******* Questions ***********
> How can I obtain a predictive model y=mean(y)+A(i) ?
> Is it possible to settle mean(y) as intercept?

Ghislain Vieilledent wrote:
> Dear advanced statisticians,
> 
> *******Objectif********
> 
> I try to set up linear models with mean as intercept:
> Answer: y
> Variable: x, as factor of two modalities: x(1), x(2).
> 
> I would like to have a model as:
> y = mean(y)+A(i)+residuals,
> with i in (1,2) and A(1) coefficient for x(1) and A(2) coefficient for
x(2).
> 
> *******Trials in R*******
> 
> ## Firstly:
> 
> I write in R:
>  >Model<-lm(y~x,Data)
>  >summary(Model)
> ...
> I've got the coefficients for:
> - the intercept (x(1) as been choosen) that we can call B(1)
> - the second modality: x(2) that we can call B(2)
> 
> If I have well understood we have for the model and predictions:
> if x(1): y=B(1)
> if x(2): y=B(1)+B(2)
> which is quite different as y=mean(y)+A(i)
> 
> ## Secondly
> I tried to skip the intercept
>  >Model2<-lm(y~0+x,Data)
>  >summary(Model2)
> ...
> I've got the coefficients for:
> - the first modality: x(1) that we can call C(1)
> - the second modality: x(2) that we can call C(2)
> 
> And the model and predictions, if I'm right, are:
> if x(1): y=C(1)
> if x(2): y=C(2)
> 
> ******* Questions ***********
> How can I obtain a predictive model y=mean(y)+A(i) ?
> Is it possible to settle mean(y) as intercept?
> 
> Thanks for your help.
> 
> Ghislain V., retarded statistician.

You have to set the contrasts differently:

set.seed(1)
y <- rnorm(10)
x <- factor(rep(letters[1:5], each = 2))
fit <- lm(y ~ x, contrasts = list(x = "contr.sum"))
all.equal(mean(y), coef(fit)[1])
#[1] TRUE

"contr.helmert" will work too. The default "contr.treatment"
does not
have columns that sum to zero. Be forewarned: the coefficients need to 
be interpretted differently.

HTH,

--sundar