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Chapter 4: Q. 4.29 (page 172)

For a hypergeometric random variable, determine

P{X=k+1}/P{X=k}

Short Answer

Expert verified

(K-k)(n-k)(k+1)(N-K-n+k+1)

Step by step solution

01

Given information

For Hypergeometric random variable with parameters N,K,nwe have that

P(X=k)=Kk路N-Kn-kNn

02

Calculation

So we have that

P(X=k+1)P(X=k)=Kk+1路N-Kn-(k+1)NnKk路N-Kn-kNn=Kk+1路N-Kn-(k+1)Kk路N-Kn-k

03

Continue Calculation

When we write out these binomial coefficients, we get that the expression above is equal to

K!(k+1)!(K-k-1)!路(N-K)!(n-k-1)!(N-K-n+k+1)!K!k!(K-k)!路(N-K)!(n-k)!(N-K-n+k)!

04

Final answer

If we cancel out everything we can we left with.

(K-k)(n-k)(k+1)(N-K-n+k+1)

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Most popular questions from this chapter

A total of 2npeople, consisting of nmarried couples, are randomly divided into npairs. Arbitrarily number the women, and let Widenote the event that woman iis paired with her husband.

  1. FindP(Wi).
  2. For i鈮�j,find role="math" localid="1646662043709" PWi鈭�Wj.
  3. When nis large, approximate the probability that no wife is paired with her husband.
  4. If each pairing must consist of a man and a woman, what does the problem reduce to?

Consider n independent sequential trials, each of which is successful with probability p. If there is a total of k successes, show that each of the n!/[k!(n 鈭� k)!] possible arrangements of the k successes and n 鈭� k failures is equally likely.

Here is another way to obtain a set of recursive equations for determining Pn, the probability that there is a string of kconsecutive heads in a sequence of nflips of a fair coin that comes up heads with probability p:

(a) Argue that for k<n, there will be a string of kconsecutive heads if either

1. there is a string of kconsecutive heads within the first n-1flips, or

2. there is no string of kconsecutive heads within the first n-k-1flips, flip n-kis a tail, and flips n-k+1,鈥�,nare all heads.

(b) Using the preceding, relate PntoPn-1. Starting with Pk=pk, the recursion can be used to obtain Pk+1, thenPk+2, and so on, up to Pn.

Let X be a binomial random variable with parameters (n, p). What value of p maximizes P{X = k}, k = 0, 1, ... , n? This is an example of a statistical method used to estimate p when a binomial (n, p) random variable is observed to equal k. If we assume that n is known, then we estimate p by choosing that value of p that maximizes P{X = k}. This is known as the method of maximum likelihood estimation.

From a set of n elements, a nonempty subset is chosen at random in the sense that all of the nonempty subsets are equally likely to be selected. Let X denote the number of elements in the chosen subset. Using the identities given in Theoretical Exercise 12of Chapter1, show that

E[X]=n2鈭�12n鈭�1

Var(X)=n鈰�22n鈭�2鈭�n(n+1)2n鈭�22n鈭�12

Show also that for n large,

Var(X)~n4

in the sense that the ratio Var(X) ton/4approaches 1as n approaches q. Compare this formula with the limiting form of Var(Y) when P{Y =i}=1/n,i=1,...,n.
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