Sum of Independent Binomials

Question

Let $X \sim \text{Bin}(m,p)$ and $Y \sim \text{Bin}(n,p)$. I want to find the distribution of $W = X + Y$ using a convolution theorem that says: $$f_W(w) = \sum\limits_x f_X (x) f_Y(w-x).$$ So using the binomial distributions for $X$ and $Y$: $$\Rightarrow \sum\limits_{x=0}^w {m \choose x} p^x(1-p)^{m-x} {n \choose w-x}p^{w-x}(1-p)^{n-w+x}$$

$$\Rightarrow \sum\limits_{x=0}^w {m \choose x} {n \choose w-x} p^w(1-p)^{m+n-w}.$$

And it's the sum over $x$, so:

$$ \Rightarrow p^w(1-p)^{m+n-w} \sum\limits_{x=0}^w {m \choose x} {n \choose w-x}.$$

My question is, what is the best approach to evaluate $ \sum\limits_{x=0}^w {m \choose x} {n \choose w-x}$?? Kinda stuck on this part, but I know it should equal ${m+n \choose w}$ somehow...

Answer 1

This is a known result called Vandermonde's identity after the French mathematician.

If you don't like that, an alterantive approach to your problem would be to use moment generating functions. Since the probability is constant, it is much easier to arrive at the result.

Answer 2

If I have $m$ things in my left hand, and $n$ things in my right hand, and you choose $w$ of them,you must have chosen x for my left, and w-x from my right, for some $x$ between 0 and $w$. This is essentially a bijective proof.