How do you get ${X_i}+{X_j}{\sim}Binom(m,{p_i}+{p_j}) $

Question

How do you get$$ {X_i}+{X_j}{\sim}Binom(m,{p_i}+{p_j}) \\ from \\f(x_i,x_j)={m!\over{x_i!x_j!(m-x_i-x_j)!}}(p_i)^{x_i}(p_j)^{x_j}(1-p_i-p_j)^{m-x_i-x_j}\large? $$

$ My \; best \; try:$ The only thing I thought of was manipulating the formula to $f(x_i,x_j)=({m \choose x_i+x_j}(p_i+p_j)^{x_i+x_j}(1-(p_i+p_j)^{m-(x_i+x_j)})({ {x_i+x_j}\choose{x_i}}({p_i\over p_i+p_j})^{x_i}(1-{p_i\over p_i+p_j})^{(x_i+x_j)-x_i})=f(x_i+x_j)f(x_i|x_{k\neq i,j}) \\ with \\ f(x_i+x_j)\sim Binom(m,x_i+x_j) \; and \; f(x_i|x_{k\neq i,j}) \sim Binom(x_i+x_j,({p_i \over p_i+p_j})) \\$

Though I feel like this is wrong and I need something more to show the distribution for $X_i+X_j$

Answer 1

To do it by algebra; We use begin with a convolution formula and later use the Binomial Theorem: $$\begin{align}f_{X_i+X_j}(k) ~&=~ \sum_{x_i=0}^{k} f_{X_I,X_j}(x_i, k-x_i) \\[1ex] ~&= \mathop{\sum\qquad}_{x_I=\max(0,k-m)}^{\min(m,k)} \frac{m!\,p_i^{x_i}\,p_j^{k-x_i}\,(1-p_i-p_j)^{m-k}}{x_i!\,(k-x_i)!\,(m-k)!}\\[1ex] &~~\vdots \\[1ex] &= \dfrac{m!\,(1-p_i-p_j)^{m-k}\,(p_i+p_j)^k }{(m-k)!~k!} \end{align}$$

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To do it by words: We argue that as $X_1,X_2$ are multinomial distributed random variables; the counts of type-1 and type-2 successes, respectively, among $m$ independent trials with mutually exclusive success rates(in any trial) of $p_1, p_2$ respectively; then $X_1+X_2$ will be the total count of success among the trials, which occur at rate $p_1+p_2$.   Hence $X_1+X_2$ is binomially distributed.

$$(X_1+X_2)\sim\mathcal{Bin}(m, p_1+p_2)$$