On 11/08/2011 1:33 PM, Duncan Murdoch wrote:
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
I meant A and B in the line above...
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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