
Poisson probability
X, Y are two independant poisoon processes with parameter a and b respectively. Find the conditional distribution of X given X+y=n,
Using P(X/Y=nx)
I get something that is almost a binomial probability but i get an extra term that i am not sure should be there, specifically 1/((a+b)t)^n
I have done this three times but i am not sure if i am making a mistake anywhere, is anyone able to check whether they get a binomial distribution with this extra term, or am i stuffing up somewhere

can you post your work
it would a lot easier than me doing it,
because I think I did this problem here last year
try a search here at this site
I found this, which I'm not if it helps you
http://www.mathhelpforum.com/mathhe...on167960.html
it should just be
$\displaystyle P(X=kX+Y=n)={P(X=k)P(Y=nk)\over P(X+Y=n)}$

acording to the link you provided the distribution for the numerator is a poisson distribution for X+Y which now gives me that this probability is 1.


Solving for this conditional probability:
P(X=x/X+Y=n)=[n! (at)^x (bt)^(nx)] /(at+bt)^n (x!(nx)!
=n choose x [(a^x)*(b^(nx)]/(a+b)^n
since the exponential terms cancel out