R devel - Mar 2000 - d-p-q-r-tests: why is plnorm(exp(Inf)) equal pnorm(Inf)?

I stumbled across a result of d-p-q-r-tests:

d-p-q-r-tests expects pnorm(Inf) and pnorm(-Inf)  to give the same results as 
plnorm(exp(Inf))
and plnorm(exp(-Inf)), respectively.

Does this mean that all expressions result in NaN (which I expect for the 3rd
and 4th expression only) or
does it mean that R can prevent the exp(Inf) to be evaluated before being parsed
in the context of plnorm?

Thomas Hoffmann.

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On Sat, 18 Mar 2000, Thomas Hoffmann wrote:
> I stumbled across a result of d-p-q-r-tests:
> 
> d-p-q-r-tests expects pnorm(Inf) and pnorm(-Inf)  to give the same results
as  plnorm(exp(Inf))
> and plnorm(exp(-Inf)), respectively.
> 
> Does this mean that all expressions result in NaN (which I expect for
> the 3rd and 4th expression only) or does it mean that R can prevent
> the exp(Inf) to be evaluated before being parsed in the context of
> plnorm?
No, it means that exp(Inf) and exp(-Inf) are evaluated correctly (Inf and
0 respectively) and the lognormal cdf is then evaluated as 1 and 0
respectively. 

If the result were NaN then equality would not hold. Nothing is equal to
NaN, not even itself.



	-thomas

Thomas Lumley
Assistant Professor, Biostatistics
University of Washington, Seattle

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