On 11/08/2011 12:01 PM, Kathie wrote:
almost forgot. In fact, I want to generate correlated Poisson random vectors.
Saying you want two random variables to be correlated doesn't specify the joint distribution, so there will be a lot of solutions. Here's one, for the case where both variables have the same mean mu, and you want a positive correlation.
We know that the sum of independent Poissons is Poisson, so we'll generate 3 variables: X with mean nu, and Y & Z with mean mu-nu, and return A = X+Y and B = X+Z. If nu=0 then A and B are independent, and if nu=mu, they have correlation 1, so you must be able to solve for a value where they have any desired correlation in between.
If the means aren't the same, this method will still work up to a point, but you won't be able to get really high correlations.
If you want negative correlations it's harder, but you could use the following trick: Generate U ~ Unif(0, 1). Calculate A by the inverse CDF method from U. Compute V to be equal to U if U < a or U > 1-a, and equal to 1-U otherwise. Calculate B by the inverse CDF method on V.
Then both U and V will have Poisson distributions (and you can choose the means as you like), and there will be some range of achievable correlations which will be quite close to [-1, 1]. The joint distribution will be very weird, but you didn't say that was a problem...
Some R code: U <- runif(10000) A <- qpois(U, 5) a <- 0.115 V <- ifelse(U < a | U > 1-a, U, 1-U) B <- qpois(V, 5) cor(A, B) This gives a correlation around 0.4. Duncan Murdoch ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
