Probability Mass Functions of Binomial Variables

Kalinka35
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Homework Statement


Let X and Y be independent binomial random variables with parameters n and p.
Find the PMF of X+Y.
Find the conditional PMF of X given that X+Y=m.


Homework Equations


The PMF of X is P(X=k)=(n C k)pk(1-p)n-k
The PMF for Y would be the same.

The Attempt at a Solution


I am really not sure how to go about solving this problem though I have been told that the first part can be done with no computation or calculations at all. Not sure at all on the second part.
 
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Do you know anything in general about the PMF of the sum of two independent discrete random variables? Does "convolution" ring a bell?

If not, don't worry about it. Think about what a binomial distribution with parameters n and p means. You can think of it as the number of "heads" resulting from flipping a coin n times, where the probability of "head" is p. If you do that experiment TWICE, independently, and the results are X and Y, respectively, then X + Y is equivalent to simply flipping the coin 2n times, right? What's the PMF for that experiment?
 
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Hmm, not that I know of.
I haven't heard the term convolution before.
Could you provide me with a definition?
 
Kalinka35 said:
Hmm, not that I know of.
I haven't heard the term convolution before.
Could you provide me with a definition?

You need it to answer the question in general (for arbitrary PMFs), but don't worry about it in this case - for this particular example you can solve it another way (see my edited post above).
 
Ah yes, I understand it now.
Thanks very much for your clear explanation.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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