Re: st: Standard normal Depvar

[Evans Jadotte <evans.jadotte@uab.es>]

Austin Nichols wrote:
> Evans Jadotte<evans.jadotte@uab.es> :
> Your desideratum
> "How can I transform the Depvar in order to force  xb^ to take on
> positive values?"
> and comments make no sense: N.B. exponentiation preserves rank.
>
> You can make x positive by
>
> su x
> replace x=x+2*abs(r(min))
>
> for example, but then you can't take the square root and have it make sense.
>
> Maybe you want
>
> g y=sign(x)*sqrt(abs(x))
>
> or somesuch?
>
> On Thu, Aug 6, 2009 at 12:26 PM, Evans Jadotte<evans.jadotte@uab.es> wrote:
>   
> > Nick Cox wrote:
> >     
> > > Exponentiation will get you all positives. After that many options are
> > > open.
> > >
> > > Evans Jadotte wrote:
> > >
> > >       
> > > > Nick Cox wrote:
> > > >         
> > > > > This produces zero or positive values.
> > > > >
> > > > > Less pedantically, if the variable is already standard normal, why does
> > > > > it need transforming?
> > > > >
> > > > > Nick
> > > > >
> > > > > Maarten buis wrote:
> > > > >
> > > > >           
> > > > > > --- On Wed, 5/8/09, Evans Jadotte wrote:
> > > > > >             
> > > > > > > I am trying to run a regression where the dependent
> > > > > > > variable has a standard normal distribution (those of you
> > > > > > > familiar with the "wealth index based on the PCA analysis",
> > > > > > > this is my Depvar).  However, I need to have the prediction  to be all
> > > > > > > positive to use for transforming.
> > > > > > > How can I transform  the Depvar in order  to
> > > > > > > force  xb^ to take on positive values?
> > > > > > >               
> > > > > > Here is one option:
> > > > > >
> > > > > > reg y x1 x2
> > > > > > predict yhat
> > > > > > sum yhat, meanonly
> > > > > > gen yhatprime = yhat + abs(r(min))
> > > > > >             
> > > > > *
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> > > > >           
> > > > Thanks to Maarten and Nick for their insight and comment on my question.
> > > >
> > > > I have both negative values and zeros in the /depvar /(/y/)/, /so/ /the
> > > > forecast, xb,  will reflect such values. And as I will need  sqrt(xb) for
> > > > further transformation at later stages, I need to transform /y/ so that xb
> > > > takes on all 'strictly' positive values and still preserve normality of /y/.
> > > > Maarten's suggestion indeed generates a 0 and the transformation I need is
> > > > in y (not yhat = xb). I have been trying a Box-Cox power transform but
> > > > results are not satisfactory.
> > > >
> > > >         
> > > *
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> > >       
> > Thanks Nick. However, exponentiation will result in a re-ranking of
> > individuals, which I must avoid. For instance, someone with a score -5
> > compared with one whose score is 4, the former will end up being ranked
> > higher than the latter after exponentiating. I need to preserve the ranks
> > and normality after transforming.
> >
> > Thanks for the feedback,
> >
> > Evans
> > *
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> >
> >     
>
> *
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>   
Slip of mind!

I was thinking about squaring all variables and the re-ranking issue 
when I wrote exponentiation. Sorry Nick and thanks Austin for your feedback.

Evans
*
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