Hello,
Well, try it:
p <- .Machine$double.eps^seq(0.5, 1, by = 0.05)
z <- qnorm(p/2)
pnorm(z)
# [1] 7.450581e-09 1.228888e-09 2.026908e-10 3.343152e-11 5.514145e-12
# [6] 9.094947e-13 1.500107e-13 2.474254e-14 4.080996e-15 6.731134e-16
#[11] 1.110223e-16
p/2
# [1] 7.450581e-09 1.228888e-09 2.026908e-10 3.343152e-11 5.514145e-12
# [6] 9.094947e-13 1.500107e-13 2.474254e-14 4.080996e-15 6.731134e-16
#[11] 1.110223e-16
exp(z*z/2)
# [1] 9.184907e+06 5.301421e+07 3.073154e+08 1.787931e+09 1.043417e+10
# [6] 6.105491e+10 3.580873e+11 2.104460e+12 1.239008e+13 7.306423e+13
#[11] 4.314798e+14
p is the smallest possible such that 1 + p != 1 and I couldn't find
anything to worry about.
R version 3.6.0 (2019-04-26)
Platform: x86_64-pc-linux-gnu (64-bit)
Running under: Ubuntu 19.04
Matrix products: default
BLAS: /usr/lib/x86_64-linux-gnu/blas/libblas.so.3.8.0
LAPACK: /usr/lib/x86_64-linux-gnu/lapack/liblapack.so.3.8.0
locale:
[1] LC_CTYPE=pt_PT.UTF-8 LC_NUMERIC=C
[3] LC_TIME=pt_PT.UTF-8 LC_COLLATE=pt_PT.UTF-8
[5] LC_MONETARY=pt_PT.UTF-8 LC_MESSAGES=pt_PT.UTF-8
[7] LC_PAPER=pt_PT.UTF-8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=pt_PT.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods
[7] base
other attached packages:
[many packages loaded]
Hope this helps,
Rui Barradas
Às 15:24 de 21/06/19, jing hua zhao escreveu:
Dear R-developers,
I am keen to calculate exp(z*z/2) with z=qnorm(p/2) and p is very small. I
wonder if anyone has experience with this?
Thanks very much in advance,
Jing Hua
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