1

Good Morning All,
May I ask for a clue to the following problem?
I got stuck and I am now wondering if I understood the problem correctly.

Let X, Y be independent random variables bin(m, p), bin(n, p) respectively.
Show that X+Y is a binomial random variable with parameter (m+n, p)

So $\mathbb P (X = i) = {m \choose i}{p^i}{q^{m-i}}$, $\mathbb P (Y = j) = {n \choose j}{p^j}{q^{n-j}}$
Let $m \le n$, $\mathbb P (X+Y = t) = $
$ \mathbb P (X=0, Y=t) + \mathbb P (X=1, Y=t-1) ... + \mathbb P (X=m, Y=t-m) $
$ = \sum_{i=1}^m \mathbb P (X=i) \mathbb P (Y=t-i)$
$ = \sum_{1=1}^m {m \choose i}{n \choose t-i} p^{i+t-i}q^{m+n-i-t+i}$

Now, I am stuck with the combinations. I can't get them to look like $m+n \choose t$.

Andy Tam
  • 3,367

2 Answers2

2

I'm sure by some books this won't be accepted as a full proof, and it doesn't answer your question, but here's a less formal proof:

Say you have $m$ identical experiments in one room, and $n$ of the same experiment in another. Also, assume that they are all independent, and that each of them have probability of success $p$. Then $X$ is the number of successful experiments in the first room and $Y$ the number of successful experiments in the other. $X + Y$ will then intuitively be the number of successful experiments in the two rooms together. This is then $m+n$ identical, independent experiments, each with success probability $p$, and it is therefore bin$(m+n, p)$.

Arthur
  • 199,419
  • Nicely explained. It does answer the original question that led to the specific question about the identity. And I expect that starting from the the fact that $X+Y$ has binomial distribution, and manipulating, probably using derivatives, we can lift the identity from that fact. – André Nicolas Jun 13 '13 at 15:41
  • @AndréNicolas I believe that from setting $q = 1-p$ and writing $$P(Z = t) = \sum_{i = 0}^t P(X= i)\cdot P(Y = t-i)$$ will give you the identity above when you multiply both sides with $p^{m}q^n$. It is the exact opposite direction of OP's approach. – Arthur Jun 13 '13 at 16:59
  • I of course mean $p^tq^{m+n-t}$ in my last comment. – Arthur Jun 13 '13 at 17:44
1

You should in your third line only go to $X=t$, $Y=0$. So you want to show that $$\sum_{i=0}^t \binom{m}{i}\binom{n}{t-i}=\binom{m+n}{t}.$$ Here is combinatorial proof of the above identity. We have a group of $m$ boys and $n$ girls, and want to pick $t$ people.

By definition, this can be done in $\dbinom{m+n}{t}$ ways.

Let us count this another way. We can pick $0$ boys and $t$ girls. This can be done in $\binom{m}{0}\binom{n}{t}$ ways.

Or else we can pick $1$ boy and $t-1$ girls. This can be done in $\binom{m}{1}\binom{n}{t-1}$ ways.

Or else we can pick $2$ boys and $t-2$ girls. This can be done in $\binom{m}{2}\binom{n}{t-2}$ ways.

Continue, and add up. We get the desired identity.

Note: If $m$ is "small" and $t$ and $n$ are largish, some of the $i$ in the sum may be greater than $m$. The above expression is still correct, if we use the convention that $\binom{a}{b}=0$ if $a$ and $b$ are non-negative integers such that $a\lt b$.

André Nicolas
  • 507,029
  • The situation where $m \lt t$ is what I am trying to capture in the third line. But then I counted wrong. TY all for the help! – Andy Tam Jun 13 '13 at 15:47