Please forgive my sloppy notation as I am not a Mathematician, I will give my very best to explain what I mean.
Given $\quad p\in(0,1)$, $\quad n,n'\in\mathbb{N}$, $\quad m\in\{0,\dots,n\}$, $\quad m'\in\{0,\dots,n'\}\quad$ and the two Binomial distributions
$f_n(m) = \begin{pmatrix}n \\ m\end{pmatrix} p^m (1-p)^{n-m}$
and
$f_{n'}(m') = \begin{pmatrix}n' \\ m'\end{pmatrix} p^{m'} (1-p)^{n'-m'}$
I would like to find the Probability Density of $\delta=\frac{m}{n}-\frac{m'}{n'}$.
My calculation so far is as follows
$ \text{Pr}\left(\frac{m}{n}-\frac{m'}{n'}=\delta\right) = \sum_{\alpha\in I} \text{Pr}\left(r=\alpha\right) \text{Pr}\left(r'=\alpha-\delta\right), \quad I=\{\frac{m}{n}, 0\leq m \leq n\} \\ = \sum_{\alpha\in I} \left[ n \begin{pmatrix}n \\ n\alpha \end{pmatrix} p^{n\alpha} (1-p)^{n(1-\alpha)} \right] \left[ n' \begin{pmatrix}n' \\ n'(\alpha-\delta) \end{pmatrix} p^{n'(\alpha-\delta)} (1-p)^{n'(1-(\alpha-\delta))} \right] \\ = n n'\left(\frac{1-p}{p}\right)^{n'\delta} \sum_{\alpha\in I} \begin{pmatrix}n \\ n\alpha \end{pmatrix} \begin{pmatrix}n' \\ n'(\alpha-\delta)\end{pmatrix} p^{\alpha(n+n')} (1-p)^{(1-\alpha)(n+n')} $
Can anyone tell me if this is at least the right direction?