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

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