"Jim Brennan" <[EMAIL PROTECTED]> 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 Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~~~~~~~~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 ______________________________________________ R-help@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
