[Rd] Calculation of e^{z^2/2} for a normal deviate z

[jing hua zhao]

Hi Peter, Rui, Chrstophe and Gabriel,

Thanks for your inputs --  the use of qnorm(., log=TRUE) is a good point in line with pnorm with which we devised log(p)  as

log(2) + pnorm(-abs(z), lower.tail = TRUE, log.p = TRUE)

that could do really really well for large z compared to Rmpfr. Maybe I am asking too much since

z <-20000
> Rmpfr::format(2*pnorm(mpfr(-abs(z),100),lower.tail=TRUE,log.p=FALSE))
[1] "1.660579603192917090365313727164e-86858901"

already gives a rarely seen small p value. I gather I also need a multiple precision exp() and their sum since exp(z^2/2) is also a Bayes Factor so I  get log(x_i )/sum_i log(x_i) instead. To this point, I am obliged to clarify, see https://statgen.github.io/gwas-credible-sets/method/locuszoom-credible-sets.pdf.

I agree many feel geneticists go to far with small p values which I would have difficulty to argue againston the other hand it is also expected to see these in a non-genetic context. For instance the Framingham study was established in 1948 just got $34m for six years on phenotypewide association which we would be interesting to see.

Best wishes,


Jing Hua


________________________________
From: peter dalgaard <pdalgd using gmail.com>
Sent: 21 June 2019 16:24
To: jing hua zhao
Cc: Rui Barradas; r-devel using r-project.org
Subject: Re: [Rd] Calculation of e^{z^2/2} for a normal deviate z

You may want to look into using the log option to qnorm

e.g., in round figures:

> log(1e-300)
[1] -690.7755
> qnorm(-691, log=TRUE)
[1] -37.05315
> exp(37^2/2)
[1] 1.881797e+297
> exp(-37^2/2)
[1] 5.314068e-298

Notice that floating point representation cuts out at 1e+/-308 or so. If you want to go outside that range, you may need explicit manipulation of the log values. qnorm() itself seems quite happy with much smaller values:

> qnorm(-5000, log=TRUE)
[1] -99.94475

-pd

> On 21 Jun 2019, at 17:11 , jing hua zhao <jinghuazhao using hotmail.com> wrote:
>
> Dear Rui,
>
> Thanks for your quick reply -- this allows me to see the bottom of this. I was hoping we could have a handle of those p in genmoics such as 1e-300 or smaller.
>
> Best wishes,
>
>
> Jing Hua
>
> ________________________________
> From: Rui Barradas <ruipbarradas using sapo.pt>
> Sent: 21 June 2019 15:03
> To: jing hua zhao; r-devel using r-project.org
> Subject: Re: [Rd] Calculation of e^{z^2/2} for a normal deviate z
>
> 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
>
> �s 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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>>
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--
Peter Dalgaard, Professor,
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