5
$\begingroup$

Let's say $X$ and $Y$ are binomial random variables with parameters $n$ and $p$ and $X+Y=m$. I want to show that the conditional distribution of $X$ if $X+Y=m$ is a hypergeometric distribution.

I'm thinking about putting these in terms of coin flips with the $m^\text{th}$ head occurring after $2n$ tosses. $X$ and $Y$ have $n$ tosses with probability of heads $p$. I'm not sure how to relate this to the hypergeometric distribution however.

$\endgroup$
6
$\begingroup$

We prove a somewhat more general result. Let $X$ and $Y$ be independent binomially distributed random variables with the same probability $p$ of "success" and number of trials equal to $a$ and $b$ respectively. (In your problem, we have $a=b=n$.)

Let us find the probability that $X=k$ given that $X+Y=m$. By the usual formula for the conditional probability, this is the probability that $X=k$ and $Y=m-k$ divided by the probability that $X+Y=m$.

The probability that $X=k$ and $Y=m-k$ is $\binom{a}{k}p^k(1-p)^{a-k}\binom{b}{m-k}p^{m-k} (1-p)^{b-(m-k)}$. This is $\binom{a}{k}\binom{b}{m-k}p^m(1-p)^{a+b-m}$.

The probability that $X+Y=m$ is $\binom{a+b}{m} p^m(1-p)^{a+b-m}$.

Divide. We get $$\frac{\binom{a}{k}\binom{b}{m-k}}{\binom{a+b}{m}}.\tag{1}$$

This is precisely the probability of getting $k$ red balls if we choose, without replacement, $m$ balls from $a+b$ balls $a$ of which are red and $b$ of which are blue. Formula (1) is a version of the familiar hypergeometric distribution function.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.