On 15-Dec-05 Lisa Wang wrote: > Hello there, > > I would like to simulate X --Normal (20, 5) > Y-- Normal (40, 10) > > and the correlation between X and Y is 0.6. How do I do it in R?
... and, as well as using mvrnorm (MASS) or rmvnorm (mvtnorm), as have been suggested, you could simply do it "by hand": If U, V are independent and N(0,1), then E(U + a*V)*(U - a*V) = 1 - a^2 E(U+a*V)^2 = E(U - a*V) = 1 + a*2 so the correlation between (U + a*V) and U - a*V) is r = (1 - a^2)/(1 + a^2) Hence, for -1 < r < 1, choose a = sqrt((1 - r)/(1 + r)) which, for r = 0.6, gives a = sqrt(0.4/1.6) = sqrt(1/4) = 1/2 (how nice! ... ). Then Var(U + a*V) = 1 + a^2 = 1 + 1/4 = 5/4 (I smell more smooth numbers coming ... ). Then, since the correlation between two variables is unchanged if you add a constant to either, or multiply either by a constant, you can give (U + a*V) variance 5 by multiplying it by 2, and give (U - a*V) variance 10 by multiplying by 2*sqrt(2), both still having expectation 0. So finally add 10 and 20: X = 10 + 2*(U + V/2) ; Y = 20 + 2*sqrt(2)*(U - V/2) So you can get U and V by sampling from rnorm(), and then X and Y as described. (Which is how I used to do it before starting to use R, e.g. in matlab/octave). Best wishes, Ted. -------------------------------------------------------------------- E-Mail: (Ted Harding) <[EMAIL PROTECTED]> Fax-to-email: +44 (0)870 094 0861 Date: 15-Dec-05 Time: 17:04:18 ------------------------------ XFMail ------------------------------ ______________________________________________ [email protected] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
