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I am estimating the following panel data twoways fixed effect model:

y = alfa*y.lag + beta1*z + beta2*z^2 + theta*id + gamma*t     (1)

where id is the individual effect, t is the time effect. I am using the {plm} package in r, therefore the code goes like this

require(plm)
fe.full <- plm(y ~ lag(y, 1) + z +z2, data = mdt, model= "within", effect="twoways")

Now I need to "extract" the predicted value for y. I know it can be done by doing predict(fe.full), however I need to do that by hand for comparing it with other models specifications. Summing alfa*y.lag + beta1*z + beta2*z^2 is not enough since I have to consider the coefficients for id and t. I know that the former can be obtained as fixef(fe.full). However I dont know how to extract the value for the time effect. In Stata coefficients for time are displayed in the summary of the model. In R I was not able to manage that. I also tried to sum the coefficients for time taken from Stata to the value of X*coeff + fixef(fe.full) obtained in R, but still the output is not equal to the value given by predict(fe.full). So I am doing something wrong, I guess.

Therefore the question is : how can I compute manually the sum of the right hand side of equation (1) starting from a plm object ?

Bob
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1 Answers1

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Well I finally managed it out, applying the suggestions by the authors of the plm package here http://r.789695.n4.nabble.com/fitted-from-plm-td3003924.html

So what I did is just to apply

fitted <- as.numeric(fe.full$model[[1]] - fe.full$residuals)

where I need the as.numeric function since I need to use it as a vector to plug in for further manipulations. I also want to point out that if your model has a lagged dependent variable on the right hand side, the solution above with as.numeric provides a vector already NET of the missing values because of the lag. For me this is exactly what I needed to.

Cheers

Bob
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    for doing that "by hand" (as someone requested), see here http://stats.stackexchange.com/questions/166531/hand-computation-for-plm-package-in-r-for-predicted-values/181450#181450 – Helix123 Nov 12 '15 at 14:22
  • does not answer how to deal with time effect – Bob Nov 13 '15 at 14:52