Re: Related to Poisson

[Daniel Murphy <chiefmurphy@gmail.com>]

So the new random variable can take values 0, A, 2A, ... ? You might want to look at the overdispersed poisson distribution. Your new random variable X=A*n will have mean A*lambda, where lambda is the mean of the poisson r.v. n. The variance of X will be A^2*Var(n)=A^2*lambda=A*(mean(X)). I.e, the variance of X will not equal its mean (as is the case with a poisson), but will be a multiple of its mean, with multiple A, called the "(over)dispersion parameter" (sometimes denoted with the greek letter phi). If you are given n (and therefore know lambda) and you know A, then P(X=x)=P(n=x/A) which you can calculate from the formula for n.

-Dan 

On Tue, May 12, 2009 at 4:41 AM, Jewel Bright <jwlbright@yahoo.com> wrote:

Folks:

 

I have a seemingly simple question, but cannot resolve it (at least without much of thinking and digging.

 

Suppose that "n" is a Poisson random variable drawn from the distribution with Poisson lambda "lambda". What is the distribution of the random variable A*n, where A is an arbitrary real number?

 

Please, note, I am not asking how to generate this random variable, I still remember how to multiply a set of numbers by a constant. I am asking about analytical form of this distribution, and about how to derive the distribution function (or density) in their analytical form.

 

A standard approach through the characteristic functions did not bring immediate success. I am sure there there are a lot of smart people in the list who would consider this problem very simple. Please, help.

 

Thanks in advance.

 

Jewel