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


From   Hey Sky <[email protected]>
To   [email protected]
Subject   Re: st: RE: how to calculate the accumulative work experience for panel data
Date   Thu, 17 Feb 2011 11:25:39 -0800 (PST)

hey Nick

thanks for your reply. 

sorry I am not familiar with the following command you suggested. could you plz 
explain what is the meaning and what can get from this? 

bysort id (j_) : gen nojobs = sum((_j == "sta") - (_j == "sto"))
especially this part: sum((_j == "sta") - (_j == "sto"))

and any hint for the next step--subtract off the gaps when there was no job from 
the total length of employment? I cannot imagine how to do it due to not 
understanding the upper code.  thanks for your time and kind reply in advance.

Nan
from Montreal






----- Original Message ----
From: Nick Cox <[email protected]>
To: [email protected]
Sent: Thu, February 17, 2011 1:49:46 PM
Subject: Re: st: RE: how to calculate the accumulative work experience for panel 
data

OK. Here's a start.

I note first that in your sample data you have two observations with
id == 7, jobid == 1.
First I drop one

drop in 20

Then I -reshape long-

reshape long j_, i(id jobid) string

The number of jobs currently held is

bysort id (j_) : gen nojobs = sum((_j == "sta") - (_j == "sto"))

You now need to subtract off the gaps when there was no job from the
total length of employment.

On Thu, Feb 17, 2011 at 5:11 PM, Hey Sky <[email protected]> wrote:

> for the cumulative work exp, no matter how many jobs they hold simutaneously, 
I
> only count them served in employment. that is, only count the overlapping time
> once.
>
> From: Nick Cox <[email protected]>
>
> 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?

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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