R help - Jul 2005 - Generating correlated data from uniform distribution

Dear R users,

I want to generate two random variables (X1, X2) from uniform
distribution (-0.5, 0.5) with a specified correlation coefficient r.
Does anyone know how to do it in R?

Many thanks!

Menghui

> dat<-matrix(runif(2000),2,1000)
> rho<-.77
> R<-matrix(c(1,rho,rho,1),2,2)
> dat2<-t(ch)%*%dat
> cor(dat2[1,],dat2[2,])
[1] 0.7513892
> dat<-matrix(runif(20000),2,10000)
> rho<-.28
> R<-matrix(c(1,rho,rho,1),2,2)
> ch<-chol(R)
> dat2<-t(ch)%*%dat
> cor(dat2[1,],dat2[2,])
[1] 0.2681669
> dat<-matrix(runif(200000),2,100000)
> rho<-.28
> R<-matrix(c(1,rho,rho,1),2,2)
> ch<-chol(R)
> dat2<-t(ch)%*%dat
> cor(dat2[1,],dat2[2,])
[1] 0.2814035
>
See  ?choleski

-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch
[mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Menghui Chen
Sent: July 1, 2005 4:49 PM
To: r-help at stat.math.ethz.ch
Subject: [R] Generating correlated data from uniform distribution

Dear R users,

I want to generate two random variables (X1, X2) from uniform
distribution (-0.5, 0.5) with a specified correlation coefficient r.
Does anyone know how to do it in R?

Many thanks!

Menghui

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Dear Menghui,

You may consider looking in Luc Devroye's "Non-uniform Random Number
Generation". Despite its title, section XI.3.2 describes how to
generate bivariate uniforms. The book is out of print but Devroye
himself urges you to print it from his scanned PDFs(!):

http://cgm.cs.mcgill.ca/~luc/rnbookindex.html

Hope this helps,

alejandro

On 7/1/05, Menghui Chen <menghui at gmail.com>
wrote:
> Dear R users,
> 
> I want to generate two random variables (X1, X2) from uniform
> distribution (-0.5, 0.5) with a specified correlation coefficient r.
> Does anyone know how to do it in R?
> 
> Many thanks!
> 
> Menghui
> 
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
>

While you are looking at weird distributions, here is one that 
 we have used in experiments on noise masking to explore the 
 bandwidth of visual mechanisms 

D'Zmura, M., & Knoblauch, K. (1998). Spectral bandwidths for the
detection of
color. 
Vision Research, 20, 3117-28 and
G. Monaci, G. Menegaz, S. Susstrunk and K. Knoblauch Chromatic Contrast 
Detection in Spatial 
Chromatic Noise Visual Neuroscience, Vol. 21, No 3, pp. 291-294, 2004

N <- 10000
x <- runif(N, -.5,.5)
y <- runif(N, -abs(x), abs(x))
plot(x,y)

y is not uniform but it is conditional on x.  The plot reveals
why we called this "sectored noise".

HTH

ken

--------------------------------------------------------
"Jim Brennan" <jfbrennan at rogers.com> writes:
> Yes you are right I guess this works only for normal data. Free advice
> sometimes comes with too little consideration :-)
Worth every cent...
> Sorry about that and thanks to Spencer for the correct way.
Hmm, but is it? Or rather, what is the relation between the
correlation of the normals  and that of the transformed variables? 
Looks nontrivial to me.

Incidentally, here's a way that satisfies the criteria, but in a
rather weird way:

N <- 10000
rho <- .6
x <- runif(N, -.5,.5)
y <- x * sample(c(1,-1), N, replace=T, prob=c((1+rho)/2,(1-rho)/2))

-- 
   O__  ---- Peter Dalgaard             ??ster Farimagsgade 5, Entr.B
  c/ /'_ --- Dept. of Biostatistics     PO Box 2099, 1014 Cph. K
 (*) \(*) -- University of Copenhagen   Denmark          Ph: (+45) 35327918
~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk)                  FAX: 
(+45) 35327907

____________________
Ken Knoblauch
Inserm U371, Cerveau et Vision
Department of Cognitive Neurosciences
18 avenue du Doyen Lepine
69500 Bron
France
tel: +33 (0)4 72 91 34 77
fax: +33 (0)4 72 91 34 61
portable: 06 84 10 64 10
http://www.lyon.inserm.fr/371/

"Jim Brennan" <jfbrennan at rogers.com> writes:
> OK now I am skeptical especially when you say in a weird way:-)
> This may be OK but look at plot(x,y) and I am suspicious. Is it still
> alright with this kind of relationship?
...
> N <- 10000
> rho <- .6
> x <- runif(N, -.5,.5)
> y <- x * sample(c(1,-1), N, replace=T, prob=c((1+rho)/2,(1-rho)/2))
Well, the covariance is (everything has mean zero, of course)

E(XY) = (1+rho)/2*EX^2 + (1-rho)/2*E(X*-X) = rho*EX^2 

The marginal distribution of Y is a mixture of two identical uniforms
(X and -X) so is uniform and in particular has the same variance as X.

In summary,  EXY/sqrt(EX^2EY^2) == rho

So as I said, it satisfies the formal requirements. X and Y are
uniformly distributed and their correlation is rho. 

If for nothing else, I suppose that this example is good for
demonstrating that independence and uncorrelatedness is not the same
thing. 

-- 
   O__  ---- Peter Dalgaard             ??ster Farimagsgade 5, Entr.B
  c/ /'_ --- Dept. of Biostatistics     PO Box 2099, 1014 Cph. K
 (*) \(*) -- University of Copenhagen   Denmark          Ph: (+45) 35327918
~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk)                  FAX: (+45) 35327907

Here is an approach using 'optim' and simulated annealing:

x <- sort(runif(1000))
y <- sort(runif(1000))

ord <- 1:1000
target <- function(ord){ ( cor(x, y[ord]) - 0.6 ) ^2 }
new.point <- function(ord){
	tmp <- sample(length(ord), 2)
	ord[tmp] <- ord[rev(tmp)]
	ord
}

new.point2 <- function(ord){
	tmp <- sample(length(ord) -100, 1)
	tmp2 <- sample(100, 1)
	ord[ c(tmp, tmp+tmp2) ] <- ord[ c(tmp+tmp2, tmp) ]
	ord
}

res <- optim(ord, target, new.point, method="SANN",
	control = list(maxit=6000, temp=2000, trace=TRUE))

res2 <- optim(ord, target, new.point2, method="SANN",
	control = list(maxit=60000, temp=200, trace=TRUE))

y <- y[res$par]

par(mfrow=c(2,2))
hist(x)
hist(y)
plot(x,y)
cor(x,y)


y <- sort(y)[res2$par]

par(mfrow=c(2,2))
hist(x)
hist(y)
plot(x,y)
cor(x,y)

Hope this helps,

Greg Snow, Ph.D.
Statistical Data Center, LDS Hospital
Intermountain Health Care
greg.snow at ihc.com
(801) 408-8111
>>> "Jim Brennan" <jfbrennan at rogers.com> 07/01/05
05:25PM >>>
OK now I am skeptical especially when you say in a weird way:-)
This may be OK but look at plot(x,y) and I am suspicious. Is it still
alright with this kind of relationship?

For large N it appears Spencer's method is returning slightly lower
correlation for the uniforms as compared to the normals so maybe there is a
problem!?!

Hope we are all learning something and Menghui gets/has what he wants . :-)

-----Original Message-----
From: pd at pubhealth.ku.dk [mailto:pd at pubhealth.ku.dk] On Behalf Of Peter
Dalgaard
Sent: July 1, 2005 6:59 PM
To: Jim Brennan
Cc: 'Tony Plate'; 'Menghui Chen'; r-help at stat.math.ethz.ch 
Subject: Re: [R] Generating correlated data from uniform distribution

"Jim Brennan" <jfbrennan at rogers.com> writes:
> Yes you are right I guess this works only for normal data. Free advice
> sometimes comes with too little consideration :-)
Worth every cent...
> Sorry about that and thanks to Spencer for the correct way.
Hmm, but is it? Or rather, what is the relation between the
correlation of the normals  and that of the transformed variables? 
Looks nontrivial to me.

Incidentally, here's a way that satisfies the criteria, but in a
rather weird way:

N <- 10000
rho <- .6
x <- runif(N, -.5,.5)
y <- x * sample(c(1,-1), N, replace=T, prob=c((1+rho)/2,(1-rho)/2))

-- 
   O__  ---- Peter Dalgaard             ??ster Farimagsgade 5, Entr.B
  c/ /'_ --- Dept. of Biostatistics     PO Box 2099, 1014 Cph. K
 (*) \(*) -- University of Copenhagen   Denmark          Ph: (+45) 35327918
~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk)                  FAX: (+45) 35327907

______________________________________________
R-help at stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help 
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html