Q.6.16 - A First Course in Probability

Question

If X and Y are independent binomial random variables with identical parameters n and p, show analytically that the conditional distribution of X given that X + Y = m is the hypergeometric distribution. Also, give a second argument that yields the same result without any computations. Hint: Suppose that 2n coins are flipped. Let X denote the number of heads in the first n flips and Y the number in the second n flips. Argue that given a total of m heads, the number of heads in the first n flips has the same distribution as the number of white balls selected when a sample of size m is chosen from n white and n black balls 

Step-by-Step Solution

Answer

X has conditionally Hypergeometric distribution

Step 1: Content Introduction

We are given, X and Y are independent binomial random variables with identical parameters $n and p$.

X denote the number of heads in the first n flips and

 Y the number in the second n flips 

Step 2: Content Explanation

Suppose that $X+Y=m\in \{0,.....,2n\}$ take any $k\in \{0,.....n\}, k\le m$

We have,

$$\begin{gathered} P(X=k| X+Y=m)=\frac{P(X=k, X+Y=m)}{P(X+Y=m)} \\ P(X=k| X+Y=m)=\frac{P(X=k)P(Y=m-k)}{P(X+Y=m)} \end{gathered}$$

Since the sum of two binomials is also a binomial, $X+Y~Binom(2n,p)$, we have that

$$\begin{gathered} \frac{P(X=k)P(Y=m-k)}{P(X+Y=m)}=\frac{\left(\begin{array}{c}n\\ k\end{array}\right)p^k{(1-p)}^{n-k}\times \left(\begin{array}{c}n\\ m-k\end{array}\right)p^{m-k}{(1-p)}^{n-m+k}}{\left(\begin{array}{c}2n\\ m\end{array}\right)p^m{(1-p)}^{2n-m}} \\ \frac{P(X=k)P(Y=m-k)}{P(X+Y=m)}=\frac{\left(\begin{array}{c}n\\ k\end{array}\right).\left(\begin{array}{c}n\\ m-k\end{array}\right)}{\left(\begin{array}{c}2n\\ m\end{array}\right)} \end{gathered}$$

We have prved that X has hypergeometric distribution.