-devel@r-project.org
Subject: Re: [Rd] Calculation of e^{z^2/2} for a normal deviate z
>>>>> jing hua zhao
>>>>> on Mon, 24 Jun 2019 08:51:43 + writes:
> Hi All,
> Thanks for all your comments which allows me to appreciate more of these
in Pyt
each=nrow(lxabs
if(anyNA(sum.) || any(sum. <= 0))
(if(strict) stop else warning)("lssum found non-positive sums")
l.off + log(sum.)
}
----------
> Best wishes,
> Jing Hua
> ___
r close?
Best wishes,
Jing Hua
From: R-devel on behalf of Martin Maechler
Sent: 24 June 2019 08:37
To: William Dunlap
Cc: r-devel@r-project.org
Subject: Re: [Rd] Calculation of e^{z^2/2} for a normal deviate z
>>>>> William Dunlap via R-devel
>
lty 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.
>> >
e association which we would be
interesting to see.
Best wishes,
Jing Hua
From: peter dalgaard
Sent: 21 June 2019 16:24
To: jing hua zhao
Cc: Rui Barradas; r-devel@r-project.org
Subject: Re: [Rd] Calculation of e^{z^2/2} for a normal deviate z
You may
hich we would be interesting to see.
> >
> > Best wishes,
> >
> >
> > Jing Hua
> >
> >
> >
> > From: peter dalgaard
> > Sent: 21 June 2019 16:24
> > To: jing hua zhao
> > Cc: Rui Barradas; r-deve
r instance the Framingham study was
> established in 1948 just got $34m for six years on phenotypewide association
> which we would be interesting to see.
>
> Best wishes,
>
>
> Jing Hua
>
>
> ________
> From: peter dalgaard
> Sen
15:03
*To:* jing hua zhao; r-devel@r-project.org
*Subject:* Re: [Rd] Calculation of e^{z^2/2} for a normal deviate z
Hello,
Well, try it:
p <- .Machine$double.eps^seq(0.5, 1, by = 0.05)
z <- qnorm(p/2)
pnorm(z)
# [1] 7.450581e-09 1.22e-09 2.026908e-10 3.343152e-11 5.514145e-12
# [6]
Hello,
Well, try it:
p <- .Machine$double.eps^seq(0.5, 1, by = 0.05)
z <- qnorm(p/2)
pnorm(z)
# [1] 7.450581e-09 1.22e-09 2.026908e-10 3.343152e-11 5.514145e-12
# [6] 9.094947e-13 1.500107e-13 2.474254e-14 4.080996e-15 6.731134e-16
#[11] 1.110223e-16
p/2
# [1] 7.450581e-09 1.22e-09
: peter dalgaard
Sent: 21 June 2019 16:24
To: jing hua zhao
Cc: Rui Barradas; r-devel@r-project.org
Subject: Re: [Rd] Calculation of e^{z^2/2} for a normal deviate z
You may want to look into using the log option to qnorm
e.g., in round figures:
> log(1e-300)
[1] -690.7755
> qnorm(-691, log=TRU
Rui,
> >
> > Thanks for your quick reply -- this allows me to see the bottom of this.
> I was hoping we could have a handle of those p in genmoics such as 1e-300
> or smaller.
> >
> > Best wishes,
> >
> >
> > Jing Hua
> >
> > ___
15:03
> To: jing hua zhao; r-devel@r-project.org
> Subject: Re: [Rd] Calculation of e^{z^2/2} for a normal deviate z
>
> Hello,
>
> Well, try it:
>
> p <- .Machine$double.eps^seq(0.5, 1, by = 0.05)
> z <- qnorm(p/2)
>
> pnorm(z)
> # [1] 7.450581e-09 1.2
st wishes,
>
>
> Jing Hua
>
>
> From: Rui Barradas
> Sent: 21 June 2019 15:03
> To: jing hua zhao; r-devel@r-project.org
> Subject: Re: [Rd] Calculation of e^{z^2/2} for a normal deviate z
>
> Hello,
>
> Well, try it:
>
> p <
-devel@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 1.22e-09 2.026908e-10 3.343152e-11 5.514145e-12
# [6] 9.094947e-13 1.500107e-13 2.4742
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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