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Chapter 6: Q.6.16 (page 276)

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

Short Answer

Expert verified

X has conditionally Hypergeometric distribution

Step by step solution

01

Content Introduction

We are given, X and Y are independent binomial random variables with identical parameters nandp.

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

Y the number in the second n flips

02

Content Explanation

Suppose that X+Y=m鈭�{0,.....,2n}take any k鈭�{0,.....n},k鈮�m

We have,

P(X=k|X+Y=m)=P(X=k,X+Y=m)P(X+Y=m)P(X=k|X+Y=m)=P(X=k)P(Y=m-k)P(X+Y=m)

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

P(X=k)P(Y=m-k)P(X+Y=m)=nkpk(1-p)n-k脳nm-kpm-k(1-p)n-m+k2nmpm(1-p)2n-mP(X=k)P(Y=m-k)P(X+Y=m)=nk.nm-k2nm

We have prved that X has hypergeometric distribution.

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