From: Shin, David <david.shin_at_pearson.com>

Date: Thu 06 Jul 2006 - 03:14:23 EST

[10,] -0.3905455 0.1787059

> x.sum <- apply(x,1,sum)

*> x.sum
*

[1] 0.3311249 -0.3338355 -0.6897418 1.2792655 0.4159225 0.8223407 -2.0350183 0.3354690 -0.1196116 -0.2118395

Date: Thu 06 Jul 2006 - 03:14:23 EST

Thanks for Andy's comment and help. I should have used a better example for
my question. Below is my exact question and I will appreciate a lot for any
insights.

I generate a bi-variate normal distribution with mean = c(0, 0.2) and variance covariance matrix = matrix(c(1, .025, .025, .0025), nrow = 2):

> x <- rmvnorm(10, c(0, 0.2), matrix(c(1, .025, .025, .0025), nrow = 2))

*> x
*

[,1] [,2]

[1,] 0.1595351 0.1715898 [2,] -0.5177577 0.1839222 [3,] -0.8794011 0.1896593 [4,] 1.0584185 0.2208470 [5,] 0.1960055 0.2199169 [6,] 0.6450406 0.1773001 [7,] -2.2160986 0.1810803 [8,] 0.2131569 0.1223121 [9,] -0.3598349 0.2402232

[10,] -0.3905455 0.1787059

> x.sum <- apply(x,1,sum)

[1] 0.3311249 -0.3338355 -0.6897418 1.2792655 0.4159225 0.8223407 -2.0350183 0.3354690 -0.1196116 -0.2118395

if I call x[,1] as theta.year1 and x[,2] as growth.year1 then the mean of x.sum is 0+0.2 = -.2 and the standard deviation of x.sum is sqrt(1+.0025+2*.5*1*.05) = 1.0259

<<...OLE_Obj...>>

assume the correlation is again 0.5, I would like to generate another bi-variate normal distribution with the fixed first column that equals to x.sum. If the mean and SD of the second column is 0.2 and 0.05, respectively, the mean of this bi-variate normal distribution is c(0.2, 0.2) and the variance-covariance matrix is: matrix(c(1.0259^2, 0.5*1.0259*0.05, 0.5*1.0259*0.05, 0.0025), 2)

Thanks again for helping me.

David

-----Original Message-----

From: Liaw, Andy [mailto:andy_liaw@merck.com]
Sent: Wednesday, July 05, 2006 11:50 AM

To: Shin, David; 'r-help@stat.math.ethz.ch'
Subject: RE: [R] generate bi-variate normal data

From: Shin, David

*>
*

> Dear all,

*>
**> I would like to generate bi-variate normal data given that
**> the first column of the data is known. for example:
**> I first generate a set of data using the command, x <-
**> rmvnorm(10, c(0, 0), matrix(c(1, 0, 0, 1), 2))
**>
**> then I would like to sum up the two columns of x:
**> x.sum <- apply(x, 1, sum)
**>
**> now with x.sum I would like to generate another column of
**> data, say y, that makes cbind(x.sum, y) follow a bi-variate
**> normal distribution with mean = c(0, 0) and sigma =
**> matrix(c(1, 0, 0, 1),2)
*

x.sum as you described would be distributed as normal with mean=0 and variance=2 (so you might as well just use x.sum <- rnorm(10, 0, sqrt(2))), so I don't see how you can get to the second step where you want x.sum to have variance=1. Also, since the covariances are 0, you could just generate the columns separately using rnorm() and cbind() them together.

It might be helpful for you to get some basic understanding of math stat. I only say that because most likely there are other steps to whatever task you are doing (people are unlikely to be generating random numbers just for kicks), and there's no telling what other things you are doing inefficiently, or even erroneously.

Andy

> I will appreciate for all insights.

*>
**> David s.
**>
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