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# binomial distribution: Xi~B(n, p) Yj~B(m, p) Xi+Yj~B(n+m, p) Tweet

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1. binomial distribution: Xi~B(n, p) Yj~B(m, p) Xi+Yj~B(n+m, p)
Where can I find out (the prove of)this topic?

Xi has binomial distribution B(n, p)
Yj has binomial distribution B(m, p)

Xi+Yj has binomial distribution B(n+m, p)

Combine poison distribution
P() and P()

= P()
Last edited by BookInquiry; 17-07-2012 at 06:13.
2. Re: binomial distribution: Xi~B(n, p) Yj~B(m, p) Xi+Yj~B(n+m, p)
Write the Binomials as a sum of Bernoulli's and its obvious

Poisson one can be proved from the MGF, see http://www.proofwiki.org/wiki/Sum_of...les_is_Poisson. You could use the same method to prove the Binomial one. You could also use the univariate transformation theorem for both.
Last edited by poohat; 17-07-2012 at 06:36.