# Sum of Binomial distribution when the success rate is different.

Is there any easy way to calculate the probability of the sum of two binomial random variable if the success rates of them are different each other?

I mean that $X \sim Bin(n,p_0)%$, $Y \sim Bin(m, p_1)$, $Z = X+Y$, $p_0 \neq p_1$ and hope to calculate the distribution function of $Z$.

I know the distribution of sum of random variables is calculated by convolution but I wonder if there is more easy way to get this. For example, $Z \sim Bin(m+n, p)$ if $p = p_0 = p_1$. Is there any similar formula in the case of $p_0 \neq p_1$?

We look at the probability generating functions. \begin{align} \Pi_X(s) & = \mathsf E(s^X) \\[0ex] & = (sp_0-(1-p_0))^n \\[2ex] \Pi_Y(s) & = (sp_1-(1-p_1))^m \\[2ex] \Pi_{X+Y}(s) & = \mathsf E(s^{X+Y}) \\[0ex] & = \Pi_X(s)\Pi_Y(s) \\[0ex] & = (sp_0-(1-p_0))^n(sp_1-(1-p_1))^m \end{align} Now, if $p=p_0=p_1$ then we would immediately have $\;\Pi_{X+Y}(s) = (sp+(1-p))^{m+n}\;$ which would indicate that $X+Y\sim\mathcal{Bin}(m+n, p)$ .
Unfortunately we clearly don't have as nice a result for $p_0\neq p_1$.
We have to consider the moment generating function. \begin{align} M_Z(t) & = \mathbb E(e^{t(X+Y)}) = \\[0ex] & = \mathbb E(e^{tX}e^{tY}) \iff X indep. Y= \\[0ex] & = M_X(t)M_Y(t)= \\[2ex] & = (1-p_0+pe^t)^n (1-p_1+pe^t)^m \\ \end{align}