The distribution for a function of Binomially distributed variables with different probabilities

Question

If (X_1,...,X_m) ~ Bernoulli(p) and (Y_1,...Y_n) ~ Bernoulli(1-p), how should I go about finding the distribution for the statistic T = (Sum of X's) - (Sum of Y's) when X and Y are distributed independently??? Thanks!

Answer 1

If $Y\sim\operatorname{Bernoulli}(1-p)$ then $1-Y\sim\operatorname{Bernoulli}(p).$

So \begin{align} & \sum_{i=1}^m X_i - \sum_{j=1}^n Y_i \\[10pt] = {} & \sum_{i=1}^m X_i + \sum_{j=1}^n (1-Y_i) - \sum_{j=1}^n 1 \\[10pt] = {} & \Big[ \text{the sum of $m+n$ indepedent Bernoulli$(p)$ random variables} \Big] - n. \end{align}

Answer 2

$\sum_{i=1}^nX_i$ is $\mbox{Bin}(n,p)$. $\sum_{i=1}^mY_i$ is $\mbox{Bin}(m,1-p)=m-\mbox{Bin}(m,p)$. Then

$$\sum_{i=1}^nX_i-\sum_{i=1}^nY_i=\mbox{Bin}(m,p)+\mbox{Bin}(m,p)-m=\mbox{Bin}(2m,p)-m,$$

where the equalities are in distribution.