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Re: st: Standard normal Depvar

From   Nick Cox <>
Subject   Re: st: Standard normal Depvar
Date   Fri, 07 Aug 2009 08:38:41 -0500

I am glad we agree that exponentiation, meaning exp(), preserves ranks. Sorry, but I don't understand what you want otherwise. Nor it seems do Maarten Buis, Austin Nichols or Martin Weiss.

I suggest that you try again with a fuller explanation, together with examples.


Evans Jadotte wrote:

> You did not read my earlier mail apologizing for my mind lapsus. I was
> thinking about squaring all variables when I mentioned the re-ranking
> issue. Evidently exponentiation preserves ranks. Also, as Austin
> mentioned that the exercise I want to carry out does not make sense, it
> has been applied many time in papers published from refereed journals
> and was proposed the time by: Amemiya (1977) The ML
> Estimator............ /Econometrica 45:955-68/.
> Many thanks again for the feedback.

Nick Cox wrote:

To echo Austin Nichols, your assertion about exponentiation and change of rank is quite incorrect. Think of a plot of exp(x) and you will see that it is a monotonic function with any real as argument and so rank reversal will not occur.

Also, to expand on my earlier comment, only linear transforms will map normals to normals.

Evans Jadotte wrote:

> 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.

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?

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