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Re: st: Transforming Inflation


From   Nick Cox <[email protected]>
To   [email protected]
Subject   Re: st: Transforming Inflation
Date   Fri, 25 Mar 2011 19:06:04 +0000

Note also a family resemblance to another Stata function:

This is also cond(x < 0, -reldif(0, x), reldif(0, x))

When exploring such functions, I often fire up -twoway function-, e.g.

twoway function cond(x < 0, -reldif(0, x), reldif(0, x)) , ra(-100 100)

My last comment in my previous was wrong.

On Fri, Mar 25, 2011 at 6:57 PM, Nick Cox <[email protected]> wrote:
> Deflating inflation! Can you do it for real too?
>
> Applying that transform is indeed quite absurd for negative values;
> note that it is undefined for x = -1. I don't think there is any
> singularity at 1% deflation.
>
> A generalisation that makes the transform symmetric about zero is
>
> sign(x) * abs(x)/(1 +abs(x))
>
> By intent, this is monotonic and preserves sign, which seem
> economically reasonable too.
>
> I discuss a bundle of related issues in
>
> Cox, N.J. 2011. Stata tip 96: Cube roots.
> The Stata Journal 11(1): 149-154.
>
> I make a case for cube roots being the simplest shape-changing
> transformation that preserves sign and is applicable to negative, zero
> and positive values alike.
>
> Shouldn't the second formula be
>
> transformed_inflation=(inflation/100)/((100 +inflation)/100)
>
> On Fri, Mar 25, 2011 at 6:40 PM, ajjee <[email protected]> wrote:
>
>> My question is not related to Stata but I have a technical problem in
>> computing a variable. In estimation, normally we transform our inflation
>> variable to reduce the influence of extreme observations by the formula:
>>
>> transformed_inflation=(inflation)/(1+inflation)  or sometime
>> transformed_inflation=(inflation/100)/((1+inflation)/100)
>>
>> But if the value of the variable is negative, say (-1.0666355), the
>> resultant value is (16.00701) by the first formula which is misleading. What
>> should be done in this situation?
>

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