st: RE: how to calculate the accumulative work experience for panel data

[Nick Cox <n.j.cox@durham.ac.uk>]

I think we need -- correction, I need -- your definition of accumulated or cumulative work experience. (By the way, I suggest either word, not accumulative.)

Do you or do you not count time served simultaneously in two or more jobs with that much weight or you just count time served in employment? 

Nick 
n.j.cox@durham.ac.uk 

Hey Sky


yesterday I post a question on how to calculate the mean wage under  panel data. 
thanks for David valuable suggestions. the tested code are  as above.

now I have another question about panel data: how to calculate the  accumulative 
work experience while getting rid off the overlapping  period? 


example data as follows, the j_sta/j_sto represents job starting/stoping  time 
and id, as before, represents the people who participates in  survey.  


as you can see, the various overlapping situation make it hard to  entangle and 
get the right work experience. for id=1, a perfect  situation that he start his 
2nd job after finished his 1st one. for  id=2, even there is an overlapping 
period between his 1st and 2nd job,  it is not hard to get accumulative exp 
right. 


now consider a little more complicated case for 3 or more jobs. such as   id=3, 
overlapping between 1st and 2nd but not 3rd job. for id=7, his  all 3 jobs are 
overlapped. now take a look at id=5, though there is no  overlapping between his 
1st and 3rd job, his 2nd job are totally covered  by his 3rd job. id=6 is in a 
similar situation with overlapping  between 1st and 3rd job.

any suggestions are appreciated.



**** Tested code to get mean wage under panel data ****
clear
input id jobid wage1 wage2 wage3 j_sta j_sto
1 1 1 2 . 1000 1050
1 2 . 3 . 1055 1260
2 1 1 . . 1000 1050
2 2 . 2 2 1022 1240

3 1 1 2 . 1000 1050
3 2 . 3 . 1030 1150
3 3 . . 4 1155 1280

4 1 2 . . 1000 1050
4 2 . 3 4 1070 1150
4 3 . . 5 1160 1240

5 1 1 . . 1000 1050
5 2 . 2 3 1080 1100
5 3 . . 5 1040 1290

6 1 1 . . 1000 1050
6 2 . 2 3 1080 1100
6 3 . . 5 1060 1290

7 1 2 2 . 1000 1050
7 2 . 2 3 1040 1110
7 3 . . 5 1100 1250

7 1 2 4 6 1020 1270
8 1 2 3 . 1001 1210
9 1 2 . 4 1033 1244

end


 format  j_sta j_sto %td

**** suggested by David  ****
reshape long wage, i(id jobid) j(wave)

drop if mi(wage)
sort id wave jobid
by id wave (jobid): keep if _n==_N

by id: egen meanwage = mean(wage)
***************

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