Could anyone guide me to a document where they derive the distribution of the difference between two binomial random variables. So $X \sim \mathrm{Bin}(n_1, p_1) $ and $Y \sim \mathrm{Bin}(n_2, p_2) $, what is the distribution of $|X-Y|$.
thank you.
(Also $X$ and $Y$ are independent)
Best Answer
I can give you an answer for the pmf of X-Y. From there |X - Y| is straightforward.
So we start with
$X \sim Bin(n_1, p_1)$
$Y \sim Bin(n_2, p_2)$
We are looking for the probability mass function of $Z=X-Y$
First note that the min and max of the support of Z must be $(-n_2, n_1)$ since that covers the most extreme cases ($X=0$ and $Y=n_2$) and ($X=n_1$ and $Y=0$).
Then we need a modification of the binomial pmf so that it can cope with values outside of its support.
$m(k, n, p) = \binom {n} {k} p^k (1-p)^{n-k}$ when $k \leq n$ and 0 otherwise.
Then we need to define two cases
In the first case
$p(z) = \sum_{i=0}^{n_1} m(i+z, n_1, p_1) m(i, n_2, p_2)$
since this covers all the ways in which X-Y could equal z. For example when z=1 this is reached when X=1 and Y=0 and X=2 and Y=1 and X=4 and Y=3 and so on. It also deals with cases that could not happen because of the values of $n_1$ and $n_2$. For example if $n_1 = 4$ then we cannot get Z=1 as a combination of X=5 and Y=4. In this case thanks to our modified binomial pmf the probablity is zero.
For the second case we just reverse the roles. For example if z=-1 then this is reached when X=0 and Y=1, X=1 and Y=2 etc.
$p(z) = \sum_{i=0}^{n_2} m(i, n_1, p_1) m(i+z, n_2, p_2)$
Put them together and that's your pmf.
$ f(z)= \begin{cases} \sum_{i=0}^{n_1} m(i+z, n_1, p_1) m(i, n_2, p_2),& \text{if } z\geq 0\\ \sum_{i=0}^{n_2} m(i, n_1, p_1) m(i+z, n_2, p_2), & \text{otherwise} \end{cases} $
Here's the function in R and a simulation to check it's right (and it does work.)
https://gist.github.com/coppeliaMLA/9681819