Quick probability generating function question?

Question

Using probability generating functions, find

P(X+Y=n) if X and Y are independent poisson distibutions with parameters a and b respectively.

Hence find P(X=k | X + Y = n) and E(X=k | X + Y = n)

Answer 1

I really don't think you need to use generating functions, but it is one way to do this. The work here is a little difficult to read because of the limited formating I can do on Y! Answers. I would suggest you write it out in a better formate with pen and paper to help see how this all works.

Generating functions

Let ξ be an integer valued random variable.

We define the probability mass function for ξ as:

P( ξ = k ) = pk for k = 0, 1, 2, 3, 4, ....

The generating function is φ(s) = E(s ^ ξ) = ∑ pk s^k where the sum has limits k = 0 to ∞.

the generating function if related to the mass. The good news is that the relationship is one - to - one, i.e., if you know one then you know the other. the relationship between the generating function and the probability mass function is:

pk = 1/k! * d^k / ds^k φ(s) | s=0

(it's difficult to format this on Y! Answers. in word the above equation says the P( ξ = k ) = pk is equal to 1 over k factorial multiplied by the kth derivative of φ(s) evaluated at s = 0

if you have ξ1, ξ2, ..., ξn independent random variables then the sum of these random variables have a generating function equal to the product of the individual generating functions.

for a Poisson random variable with parameter λ the generating function φ(s) = exp( - λ * (1 - s) ) for |s| < 1

You have X + Y ~ Poisson(a + b)

the generating function is: φ(s) = exp( (a + b) (s - 1) )

P( X + Y = n) = 1/n! * (a + b) ^ n * exp( (a + b) (s - 1) ) | s = 0
P( X + Y = n) = 1/n! * (a + b) ^ n * exp( -(a + b) )

this is the same as you would find by just using the knowledge that the sum of independent Poisson random variables has the Poisson distribution with the parameter for the sum equal to the sum of the parameters for the individual Poissons.

to find P(X = k | X + Y = n) do the following:

P(X = k | X + Y = n)
= P(X = k ∩ X + Y = n) / P(X + Y = n)
= P(X = k ∩ Y = n - k) / P(X + Y = n)
= (a ^ k exp( -a ) / k! ) ( b ^ (n - k) exp( -b) ) / ( (a+b) ^ n exp(-(a+b)) / n!)
= n! / (k! (n - k)!) * (a / (a + b)) ^ k * ( b / (a + b)) ^ (n - k)

this is the binomial mass function for a binomial with n trials and success probability a / (a + b)

Answer 2

Th foundation of Agnosticism is that a probability or even a preponderance can not be logically or scientifically asserted by man, we know nothing of what may be beyond phenomena. Agnostics simply don't have enough information, to assert if existence has intention or not, so we admit our ignorance. We are not 50-50 or any other number we are ignorant, you are too, but you just cant bring yourself to see it. Those that try to assert some probability have only to provide the scientific grounds, to convince Agnostics everywhere. It's easy to see from the answers given here that both Theist and Atheist alike willingly disregard logic, reason and science, when it comes to the question - does existence have intention or not? You may as well deal with the fact that every last one of us will take your last breath not knowing what's about to happen next. What an adventure! TJ Bradders