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The length of the mean vector must match the number of rows and columns of the sigma matrix. Once you give 3 entries in the mean vector you will run into the problem that the sigma you are using is not positive (semi-)definite - a variance must be the product of a matrix and its transpose. -Bill On Tue, Mar 16, 2021 at 10:55 AM hatice g?rdil
<haticegurdil1985 at gmail.com> wrote:
Code a is working. But code b is given error like given below. How can I write code b?
a<-rmvnorm(750, mean=c(0, 0),
+ sigma=matrix(c(1, .3, .3, 1), ncol=2))
head(a)
[,1] [,2] [1,] -0.97622921 -0.87129405 [2,] 0.54763494 0.16080131 [3,] -1.16627647 0.31225125 [4,] 1.72541168 2.06513939 [5,] 0.05372489 -0.07525197 [6,] -0.85062230 -1.02188473
b<-rmvnorm(round(500,0), mean=c(0,-1),
+ sigma=matrix(c(.3, 1,1,1,.3, 1, 1,1, .3), ncol=3))
Error in rmvnorm(round(500, 0), mean = c(0, -1), sigma = matrix(c(0.3, :
mean and sigma have non-conforming size
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