Is there a closed form answer for absolute value of difference of two identical binomial random variables with identical distributions when $p=1/2$? In particular, what is the the distribution of $|X-Y|$ when $X$ and $Y$ are independent and both $~Bin(n,1/2)$?
Difference of two binomial random variables with identical distribution
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binomial-distribution
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0Relevant (general case): http://math.stackexchange.com/questions/562119/difference-of-two-binomial-random-variables – 2017-01-23
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0I guess that is not a closed form answer as it uses the hypergeomteric function. – 2017-01-23
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2No, but it is still relevant for future searches, and having a link here helps in that. (Another one: http://math.stackexchange.com/questions/1065487/difference-between-two-independent-binomial-random-variables-with-equal-success) – 2017-01-23
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1Note that what you want can be rephrased as follows: what is the probability distribution of $\lvert X - n\rvert$, where $X\sim\operatorname{Bin}(2n,\frac{1}{2})$? (I.e., the distribution of the "distance from the mean") – 2017-01-23
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1You need to assume your two binomial random variables are independent. – 2017-01-23
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0Thank Clement. Can you explain how these two are equivalent? – 2017-01-23
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0Yes they are also independent. – 2017-01-23
1 Answers
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From the various answers at the linked question, it looks as if $$P(|X-Y|=0)={2n \choose n} \frac{1}{2^{2n}}$$ while for positive $z$ $$P(|X-Y|=z)={2n \choose n+z} \frac{1}{2^{2n-1}}$$
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0And now I read the second linked question where [my answer](http://math.stackexchange.com/a/1065670/6460) said in passing "For the case $p=\frac12$, $m-Y$ has the same distribution as $Y$ so $X+Y$ and $X-Y+m$ have the same distribution, which is indeed binomial." $m$ there is $n$ here – 2017-01-23