Re: st: Standard normal Depvar

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

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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