Similarly,
> .Machine$double.xmin
[1] 2.225074e-308
is NOT really the smallest number currently available on my computer,
but it's moderately close, e.g.,
> .Machine$double.xmin/2
[1] 1.112537e-308
> .Machine$double.xmin^1.1
[1] 0
To maximize likelihood, I routinely use log or log.p = TRUE in functions
like dt and pt, because nonlinear optimization too often throws and
error with a likelihood of 0, whose logarithm is -Inf but does not have
a problem on a log scale:
> log(.Machine$double.xmin)
[1] -708.3964
> log(.Machine$double.xmin)*2
[1] -1416.793
Spencer Graves
On 10/26/25 09:41, Richard O'Keefe wrote:
No, 0 and 5-19 are not "equalled". THey are quite distinct.
As for pt() returning something smaller than double.eps, why wouldn't it?
If I calculate 10^-30, I get 1e-30, which is much smaller than double.eps,
but is still correct. It would be a serious error to return 0 for 10^-30.
Welcome to the wonderful world of floating-point arithmetic.
This really has nothing to do with R.
On Sun, 26 Oct 2025 at 09:38, Christophe Dutang <[email protected]> wrote:
Thanks for your answers.
I was not aware of the R function expm1().
I’m completely aware that 1 == 1 - 5e-19. But I was wondering why pt() returns
something smaller than double.eps.
For students who will use this exercise, it is disturbing to find 0 or 5e-19 :
yet it will be a good exercise to find that these quantities are equalled.
Regards, Christophe
Le 25 oct. 2025 à 12:14, Ivan Krylov <[email protected]> a écrit :
В Sat, 25 Oct 2025 11:45:42 +0200
Christophe Dutang <[email protected]> пишет:
Indeed, the p-value is lower than the epsilon machine
pt(t_score, df = n-2, lower=FALSE) < .Machine$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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