Ramani,
Stas Kolenikov has written a Shapley value decomposition program (-findit
shapley-), which you may find helpful.
Scott
----- Original Message -----
From: "Ramani Gunatilaka" <[email protected]>
To: <[email protected]>
Sent: Sunday, January 25, 2004 2:56 AM
Subject: st: Loop for Shapley Decomposition sub-routine
> Hi all,
> I am trying to apply the Shapley decomposition method to household consumption
data.
> I am wondering if there is a short cut to do the following subroutine.
>
> Consumption data is x, and assume there are 4 explanatory variables,
var1-var4.
>
> I have performed the following:
>
> regress x var1 var2 var3 var4
>
> saved the betas as B1, B2 etc. upto B5 (coeff on the constant) and generated
them as variables alongside var1 var2 etc.
>
> I next need to generate means of variables, vbar1, vbar2, and generate new
consumption variables x1 x2 x3 x4 as follows.
>
> Using a loop,
>
> local i=$j/*where $j is total no of vars, that is 4*/
> while `i'>=1{
> qui sum var`i'
> qui gen vbar`i'=r(mean)
> qui gen XXXXXX/*see below*/
> local i =`i'-1
> }
>
> The XXXXXX part has to read as follows if i=4
>
> x4=exp((B1*var1)+(B2*var2)+(B3*var3)+(B4*vbar4) +(B($j+1))/*constant*/
>
> if i=3, then XXXX must read,
>
> x3=exp((B1*var1)+(B2*var2)+(B3*vbar3)+(B4*var4) +(B($j+1))/*constant*/
>
> if i=2, then XXXXX goes,
> x2=exp((B1*var1)+(B2*vbar2)+(B3*var3)+(B4*var4) +(B($j+1))/*constant*/
>
> etc.
>
> Can anyone please suggest how I may write an expression for XXXXX which
includes all the above possibilities, depending on the value that `i' takes and
even if I have 25 variables?
>
> Thanks so much,
> Ramani
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