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$\newcommand{\Bin}{\operatorname{Bin}}$ My text doesn't define $X\sim \Bin(n,p)$ but after mentioning it, in the next few lines it writes that $f(x)$=$ {n\choose x}p^xq^{n-x}$ is the p.m.f. of the binomial distribution. ($p+q=1$)

It goes on to state that $X_1+X_2$ follows $\Bin(n_1+n_2,p)$ where $X_1$ and $X_2$ are independent random variables. I do not really understand how we can prove this. Can anyone please tell me how to do it?

Did
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    There are lots of things that your book does not seem to mention or, more likely, you think are not important enough to include in your question. Does it say somewhere that the result you want to prove holds when $X_1$ and $X_2$ are independent random variables? Also, though most people reading this question will know what $q$ means, could you tell us what the value of $q$ is in your formula for the p.m.f.? – Dilip Sarwate Jan 14 '13 at 15:03
  • @DilipSarwate I have edited my question. –  Jan 14 '13 at 15:13

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Hint: You are told that $\mathbb P(X_1=k)={n_1\choose k}p^k(1-p)^{n_1-k}$ for every $0\leqslant k\leqslant n_1$ and that $\mathbb P(X_2=k)={n_2\choose k}p^k(1-p)^{n_2-k}$ for every $0\leqslant k\leqslant n_2$, and you are interested in the value of $\mathbb P(X_1+X_2=k)$. Obviously, $$ [X_1+X_2=k]=\bigcup_{i=0}^k[X_1=i]\cap[X_2=k-i]. $$ Using the hypothesis that $X_1$ and $X_2$ are independent, can you deduce $\mathbb P(X_1+X_2=k)$ from this decomposition?

Did
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  • There are more direct ways to prove the result but since you said nothing about the level of your knowledge, this one will do. – Did Jan 14 '13 at 15:22