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Re: st: Modelling Relative Risks with -fracpoly-


From   Nick Cox <njcoxstata@gmail.com>
To   statalist@hsphsun2.harvard.edu
Subject   Re: st: Modelling Relative Risks with -fracpoly-
Date   Wed, 20 Mar 2013 15:12:51 +0000

Sorry, that was too hasty. You said much more than I noticed in a
brisk reading.

However, forcing through the origin here still seems more problematic
than usual.

In the easiest applications, (0, 0) is unattainable but a sensible
limit on physical (biological, economic, ...) grounds. (Mundane
example: length and area of objects.) In the best applications,
forcing a function through the origin is also consistent with the data
say.

Here it seems that RR < 1 and RR > 1 could be something you observe
even for exposure at or near 0, just as a matter of empirical
fluctuation. If they are about equally common, your curve should
reflect that any way. If they aren't equally common, force is not
nice.

Nick

On Wed, Mar 20, 2013 at 3:03 PM, Nick Cox <njcoxstata@gmail.com> wrote:
> What is the origin here?
>
> Normally something we should all have been able to answer at age 13 or
> so, but please bear with me.
>
> If logRR = 0 then RR = 1.
>
> If RR = 0 then logRR is indeterminate.
>
> Do you want either limiting behaviour?
>
> If so, why? If not, what else?
>
> Either way, you could try choosing a set of powers that had the
> behaviour you want, but that might get in the way of -fracpoly-'s
> scope for adjusting to the data.
>
> Nick
>
> On Wed, Mar 20, 2013 at 2:52 PM, Colin Angus <c.r.angus@sheffield.ac.uk> wrote:
>
>> I'm using the -fracpoly- command to model the log relative risk of an
>> event as a function of a single continuous exposure variable, where
>> the reference category for my relative risk is those with an exposure
>> of 0 (i.e. my log RR at 0 exposure is 0). So my command is:
>>
>> -fracpoly: regress logRR exposure [weight=weight]-
>>
>> I cannot see how to force the fitted fractional polynomial function
>> through the origin. Even if I use the -nocon- command to supress the
>> constant term, the transformations of the exposure variable mean that
>> the fitted value at 0 isn't 0.
>>
>> Can anybody help me?
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