Poisson probability

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• Mar 21st 2011, 03:33 AM
ulysses123
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=n-x)

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
• Mar 21st 2011, 03:48 PM
matheagle
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/math-he...on-167960.html

it should just be

$\displaystyle P(X=k|X+Y=n)={P(X=k)P(Y=n-k)\over P(X+Y=n)}$
• Mar 21st 2011, 10:32 PM
ulysses123
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.
• Mar 21st 2011, 10:38 PM
matheagle
please show your work
• Mar 22nd 2011, 12:01 AM
ulysses123
Solving for this conditional probability:

P(X=x/X+Y=n)=[n! (at)^x (bt)^(n-x)] /(at+bt)^n (x!(n-x)!
=n choose x [(a^x)*(b^(n-x)]/(a+b)^n

since the exponential terms cancel out