On 10/28/25 08:26, Michael Dewey wrote:


On 28/10/2025 10:13, Peter Dalgaard wrote:
I suspect this is more like a relic from times when people would do (say) 1 - pchisq(x,f) instead of pchisq(x, f, lower=FALSE) and intended to avoid the embarrassment of printing 0 for things that weren’t actually impossible.

People have been known to have unexpected uses for the tiny probabilities (one case came from theoretical physics - I think it got recorded as a fortune() entry) but rarely as low as 10^-16 in actual significance testing. Things like whole genome scans may suggest some hefty Bonferroni multipliers, but the numer of tests are not (yet?) in the trillions (US).

A paper in Nature https://www.nature.com/articles/nature14151 by stensola and colleagues (sorry there does not seem to be a DOI but it is also vol 518, pages 207–212 (2015)) reports a p-value of Z = 405, P = 2.2 * 10^{−226} which is believed to be the current record. To give credit where it is due this was posted in a comment by user amoeba on CrossValidated.

Michael


> pnorm(-405)
[1] 0
> pnorm(-405, log=TRUE)
[1] -82019.42
> pt(-405, 146)
[1] 6.299392e-225
> pt(-405, 147)
[1] 3.090621e-226

They must not be using R ;-) sg>

- pd

On 26 Oct 2025, at 23.34, Ben Bolker <[email protected]> wrote:

  One possible source of confusion is that the `print.Coefmat` function uses .Machine$double.eps as its threshold for printing "< [minimum value]" rather than the precise computed p-value (presumably on the grounds that a number smaller than this is likely to be unrealistic as an accurate statement of the unlikeliness of an outcome in the real world).

On 10/26/25 10:41, Richard O'Keefe wrote:
No, 0 and 5-19 are not "equalled".  THey are quite distinct.
As for pt() returning something smaller than double.eps, why wouldn't it? If I calculate 10^-30, I get 1e-30, which is much smaller than double.eps, but is still correct.  It would be a serious error to return 0 for 10^-30.
Welcome to the wonderful world of floating-point arithmetic.
This really has nothing to do with R.
On Sun, 26 Oct 2025 at 09:38, Christophe Dutang <[email protected]> wrote:

Thanks for your answers.

I was not aware of the R function expm1().

I’m completely aware that 1 == 1 - 5e-19. But I was wondering why pt() returns something smaller than double.eps.

For students who will use this exercise, it is disturbing to find 0 or 5e-19 : yet it will be a good exercise to find that these quantities are equalled.

Regards, Christophe

Le 25 oct. 2025 à 12:14, Ivan Krylov <[email protected]> a écrit :

В Sat, 25 Oct 2025 11:45:42 +0200
Christophe Dutang <[email protected]> пишет:

Indeed, the p-value is lower than the epsilon machine

pt(t_score, df = n-2, lower=FALSE) < .Machine$double.eps
[1] TRUE

Which means that for lower=TRUE, there will not be enough digits in R's
numeric() type to represent the 5*10^-19 subtracted from 1 and
approximately 16 zeroes.

Instead, you can verify your answer by asking for the logarithm of the number that is too close to 1, thus retaining more significant digits:

print(
-expm1(pt(t_score, df = n-2, lower=TRUE, log.p = TRUE)),
digits=16
)
# [1] 2.539746620181249e-19
print(pt(t_score, df = n-2, lower=FALSE), digits=16)
# [1] 2.539746620181248e-19

expm1(.) computes exp(.)-1 while retaining precision for numbers that
are too close to 0, for which exp() would otherwise return 1.

See the links in
https://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R- think-these-numbers-are-equal_003f
for a more detailed explanation.

--
Best regards,
Ivan
(flipping the "days since referring to R FAQ 7.31" sign back to 0)

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--
Dr. Benjamin Bolker
Professor, Mathematics & Statistics and Biology, McMaster University
Associate chair (graduate), Mathematics & Statistics
Director, School of Computational Science and Engineering
* E-mail is sent at my convenience; I don't expect replies outside of working hours.

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