Dear all, I believe this should be a simple question but after trying and reading I still have no solutions.
Basically what I want to do is to run a linear model in R fixing the intercept to a specific value (which in my case is 0.69, i.e. ln(2)).
Specifically, I would like to generate a linear model with interaction, but fixing the intercept in order to have the same intercept for the different groups:
lm(y ~ day*group, data = d.df)
When I fit the data, fixing the intercept using the approaches I will soon explain, I expect to see changes in the slope coefficients between the groups but I do not: In R, I've tried using the I() as well as offset approaches and the ne library(restriktor) but when I call the summary() of the model I am still seeing intercept and slope estimates consistent with the model without fixing intercepts.
This means that the I() is not fixing the intercept as I wish.
Do you have any other solutions I might follow? Thank you very much in advance.
lm( y-intercept ~ 0+day*group, data = d.df)
There are two things important here:
1) shifting the y-values by the intercept value you want to fix, and
2) use a model without fitting an intercept term
"intercept" on the LHS of the formula is the value of the intercept you want to fix, in your case ln(2). You can also calculate the shifted y values and uses those instead.
The "0+" at the RHS of the formula tells R not to add an intercept term.
---
Alternatively, instead of shifting y you can add an offset term to the RHS:
lm(y ~ 0+day*group + offset(rep(intercept, length(y))), data = d.df)
The offset term needs to be a vector of the same length as y (you you could use a different "intercept" value for each y, if you'd want).
Dear Jochen,
Thank you for your answer. As I mentioned I tried this approach also (exactly adding the 0+ and the "i" with offset (intercept value repeated for the same length of y) term but the slopes do not change at all compared to the original model, which is surprising as setting the intercept on a specific value I would definitely expect the slopes of the different groups to change.
I have just tried again with your first suggestion but unfortunately I still do not see the intercept I set or any change in the slopes.
Thank you.
Update: I tried in Excel and the slopes slightly changed. I followed the same procedure used in Excel in R, i.e generic regression equation: y=mx+b constraint: b=ln(2); subtract b from both sides of the equation and then run the lm function. Incredible but still, the slope does not change and even if data are the same they do not match those I calculated in Excel... I really do not get what I am doing wrong.
Maybe you need this approach:
lm(y - log(2) ~ 0 + day*group, data = d.df)
As I understand, this is where you "calculate the shifted y values and uses those instead", as suggested by Jochen Wilhelm.
Regards,
Andrey
Dear Andrey and Michele,
thanks to both of you for your answers! I found a solution which is now probably working and is not considering the intercept as I want. First I shifted the y-values by the intercept value as Jochen suggested, then I used this model:
lm(y ~ 0+day:group, data = d.df). Basically I changed the "*" sign with ":" and now the output of the summary is different and is not considering the intercept at all. I did the same calculation manually to verify the slopes are correct and it seems they actually are. What do you think?
Unfortunately with all of the other "options" you proposed, in the output the intercepts between groups are not fixed but they change.
Thanks again to all of you for your comments!
Chiara
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