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Re: st: Standard normal Depvar


From   Evans Jadotte <evans.jadotte@uab.es>
To   statalist@hsphsun2.harvard.edu
Subject   Re: st: Standard normal Depvar
Date   Thu, 06 Aug 2009 19:10:25 +0200

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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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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