[R] A vector of normal distributed values with a sum-to-zero constraint
Marc Marí Dell'Olmo
marceivissa at gmail.com
Tue Apr 1 16:56:48 CEST 2014
Boris is right. I need this vector to include as initial values of a
MCMC process (with openbugs) and If I use this last approach sum(x)
could be a large (or extreme) value and can cause problems.
The other approach x <- c(x, -x) has the problem that only vectors
with even values are obtained.
Thank you!
2014-04-01 16:25 GMT+02:00 Boris Steipe <boris.steipe at utoronto.ca>:
> But the result is not Normal. Consider:
>
> set.seed(112358)
> N <- 100
> x <- rnorm(N-1)
> sum(x)
>
> [1] 1.759446 !!!
>
> i.e. you have an outlier at 1.7 sigma, and for larger N...
>
> set.seed(112358)
> N <- 10000
> x <- rnorm(N-1)
> sum(x)
> [1] -91.19731
>
> B.
>
>
> On 2014-04-01, at 10:14 AM, JLucke at ria.buffalo.edu wrote:
>
>> The sum-to-zero constraint imposes a loss of one degree of freedom. Of N samples, only (N-1) can be random. Thus the solution is
>> > N <- 100
>> > x <- rnorm(N-1)
>> > x <- c(x, -sum(x))
>> > sum(x)
>> [1] -7.199102e-17
>>
>> >
>>
>>
>>
>>
>>
>>
>>
>>
>> Boris Steipe <boris.steipe at utoronto.ca>
>> Sent by: r-help-bounces at r-project.org
>> 04/01/2014 09:29 AM
>>
>> To
>> Marc Marí Dell'Olmo <marceivissa at gmail.com>,
>> cc
>> "r-help at r-project.org" <r-help at r-project.org>
>> Subject
>> Re: [R] A vector of normal distributed values with a sum-to-zero constraint
>>
>>
>>
>>
>>
>> Make a copy with opposite sign. This is Normal, symmetric, but no longer random.
>>
>> set.seed(112358)
>> x <- rnorm(5000, 0, 0.5)
>> x <- c(x, -x)
>> sum(x)
>> hist(x)
>>
>> B.
>>
>> On 2014-04-01, at 8:56 AM, Marc Marí Dell'Olmo wrote:
>>
>> > Dear all,
>> >
>> > Anyone knows how to generate a vector of Normal distributed values
>> > (for example N(0,0.5)), but with a sum-to-zero constraint??
>> >
>> > The sum would be exactly zero, without decimals.
>> >
>> > I made some attempts:
>> >
>> >> l <- 1000000
>> >> aux <- rnorm(l,0,0.5)
>> >> s <- sum(aux)/l
>> >> aux2 <- aux-s
>> >> sum(aux2)
>> > [1] -0.000000000006131392
>> >>
>> >> aux[1]<- -sum(aux[2:l])
>> >> sum(aux)
>> > [1] -0.00000000000003530422
>> >
>> >
>> > but the sum is not exactly zero and not all parameters are N(0,0.5)
>> > distributed...
>> >
>> > Perhaps is obvious but I can't find the way to do it..
>> >
>> > Thank you very much!
>> >
>> > Marc
>> >
>> > ______________________________________________
>> > R-help at r-project.org mailing list
>> > https://stat.ethz.ch/mailman/listinfo/r-help
>> > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> > and provide commented, minimal, self-contained, reproducible code.
>>
>> ______________________________________________
>> R-help at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
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