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We know that

$X$ follows $\mathrm{Bin}(n_1,p_1)$, $Y$ follows $\mathrm{Bin}(n_2,p_2)$, $X$ and $Y$ are independent. What does $X+Y$ follow?

I know the answer, that if $p_1=p_2=p$ then $X+Y$ follow $\mathrm{Bin}(n_1+n_2,p)$ but I do not quite get it. I tried to match $X+Y$ to the criterion of Bin distribution:

  • Success/fail
  • Independence, yes since Y and X are independent, and all $X$'s are independent with themselves, and all Y's are independent with themselves.
  • same p. yes, If we are allowed to assume that.

Also, what more can we say about $X+Y$?

Glen_b
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user1603548
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  • Do you mean: you're throwing a die and flipping a coin. You then look at the sum of 'one's + 'head's? – cbeleites unhappy with SX Feb 14 '14 at 11:51
  • The situation at hand is not given, although yes your situation seem to be accurate since X: # of 1 follows Bin and Y: # of heads follow Bin – user1603548 Feb 14 '14 at 12:28
  • [This thread](http://math.stackexchange.com/questions/29998/sum-of-independent-binomial-random-variables-with-different-probabilities) is related to this question. – COOLSerdash Feb 14 '14 at 13:03
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    A very closely related thread at http://stats.stackexchange.com/questions/5347 addresses this question for the case of more than two binomial variables (with a focus on a much larger number of them). Of course all the exact solutions offered there apply to this simpler case! – whuber Feb 14 '14 at 15:49

1 Answers1

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This is the Poisson-binomial distribution.

A number of answers here discuss this distribution.

We can say a number of things about it. Here are a few examples:

  • we can compute its mean and variance using basic properties of mean and of variance under independence.

  • It is unimodal*

* with the proviso that if two adjacent outcomes are equally maximally probable, we still call that 'unimodal'.

  • if both $p_i$ values are very small, $X+Y$ may be well approximated by a Poisson distribution

  • Le Cam's theorem applies to it

we might say many more things.

Community
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Glen_b
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  • Can you provide a proof for this more simple case? – user1603548 Feb 16 '14 at 15:22
  • A proof of what? – Glen_b Feb 16 '14 at 23:10
  • Proce that if $X$ follows $\mathrm{Bin}(n_1,p_1)$, $Y$ follows $\mathrm{Bin}(n_2,p_2)$, $X$ and $Y$ are independent, and $p_1=p_2=p$ then $X+Y$ follow $\mathrm{Bin}(n_1+n_2,p)$. – user1603548 Feb 17 '14 at 14:29
  • That would definitely fall under the 'routine question' part of the `self-study` [tag wiki info](http://stats.stackexchange.com/tags/self-study/info) (*q.v.*), to which helpful hints would be given if you show us where you had difficulty. The deleted answer by @Zen was the start of one way to approach that question. (I presume it got deleted because you asked about the Poisson binomial, a more complicated question to which his answer wasn't as directly useful.) -- what have you tried? – Glen_b Feb 17 '14 at 18:11