Re: Re: st: sample splitting problem

 From dist@interfree.it To statalist@hsphsun2.harvard.edu Subject Re: Re: st: sample splitting problem Date 21 Nov 2007 07:14:24 -0000

```Thank you for you reply but it is not clear to me if your suggested procedure is a solution. The restore command gives me back an "error" (nothing to restore r(622);).  Besides, after posting my request I have also searched for a code. Do you (or someone else) think the following one is a way to solve my problem?

> gen ratio = x/y
> egen m1=mean(ratio), by(id)
> egen medm1=median(m1)
> gen hi1=m1<medm1 if !mi(m1, medm1)

( uncertain if it should be: by(id) or by (firm) )

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>Probably the simplest thing is to collapse the dataset and then
>obtaining the median. Something like this (change the varnames where
>appropiate):
>
>preserve
>gen ratio = x/y
>collapse ratio, by(firm)
>su ratio, detail
>local threshold = r(p50)
>restore
>
>gen more_x = (x/y > `threshold')
>
>In the end, we restore the data and create a dummy variable that indicates
>whether each firm has more of x or not (not having more of x
>is the same as having more of y).
>
>HTH,
>Sergio
>
>PS: I'm assuming your panel is in -long- format. If not, set it as long.
>
>On 21 Nov 2007 06:00:21 -0000,  <dist@interfree.it> wrote: > Dear all, here I'm
>again for a new request of help. I would like to replicate a sample splitting
>into my research. > Given an unbalanced panel (ts-cs data) and two variables, x
>and y, the sample splitting criterion is the following: > a firm is identified
>as having more "x" (high_x) if "x/y" is larger than the median accross firms of
>the firms' time averages. Conversely, a firm is identified as having less "x"
>(low_x) if "x/y" is less than (or equal to) the median accross firms of these
>time averages. I would be grateful if someone can help me with the commands
>(just an example to use for my dataset). >
>> Thank you very much.
>>
>>

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