How to compute p-values from z-scores in R when the Z score is large (pvalue much below zero)?

问题

In genetics very small p-values are common (for example 10^-400), and I am looking for a way to get very small p-values (two-tailed) when the z-score is large in R, for example:

z=40
pvalue = 2*pnorm(abs(z), lower.tail = F)

This gives me a zero instead of a very small value which is very significant.

回答1:

The inability to handle p-values less than about 10^(-308) (.Machine$double.xmin) is not really R's fault, but is rather a generic limitation of any computational system that uses double precision (64-bit) floats to store numeric information.

It's not hard to solve the problem by computing on the log scale, but you can't store the result as a numeric value in R; instead, you need to store (or print) the result as a mantissa plus exponent.

pvalue.extreme <- function(z) {
   log.pvalue <- log(2) + pnorm(abs(z), lower.tail = FALSE, log.p = TRUE)
   log10.pvalue <- log.pvalue/log(10) ## from natural log to log10
   mantissa <- 10^(log10.pvalue %% 1)
   exponent <- log10.pvalue %/% 1
   ## or return(c(mantissa,exponent))
   return(sprintf("p value is %1.2f times 10^(%d)",mantissa,exponent))
}

Test with a not-too-extreme case:

pvalue.extreme(5)
## [1] "p value is 5.73 times 10^(-7)"
2*pnorm(5,lower.tail=FALSE)
## [1] 5.733031e-07

More extreme:

pvalue.extreme(40)
## [1] "p value is 7.31 times 10^(-350)"

There are a variety of packages that handle extremely large/small numbers with extended precision in R (Brobdingnag, Rmpfr, ...) For example,

2*Rmpfr::pnorm(mpfr(40, precBits=100), lower.tail=FALSE, log.p = FALSE)
## 1 'mpfr' number of precision  100   bits 
## [1] 7.3117870818300594074979715966414e-350

However, you will pay a big cost in computational efficiency and convenience for working with an arbitrary-precision system.

来源：https://stackoverflow.com/questions/46416027/how-to-compute-p-values-from-z-scores-in-r-when-the-z-score-is-large-pvalue-muc

标签

statistics

normal-distribution

genetics