In Self-Test Problem 7.1, we showed how to use the value of a uniform (0,1) random variable (commonly called a random number) to obtain the value of a random variable whose mean is equal to the expected number of distinct names on a list. However, its use required that one choose a random position and then determine the number of times that the name in that position appears on the list. Another approach, which can be more efficient when there is a large amount of replication of names, is as follows: As before, start by choosing the random variable X as in Problem 7.1. Now identify the name in position X, and then go through the list, starting at the beginning, until that name appears. Let I equal 0if you encounter that name before getting to position X, and let l equal1ifyour first encounter with the name is at position X. Show that E[mI]=d.

Hint: Compute E[I] by using conditional expectation.

Short Answer

Expert verified

The position is equality likely Therefore, to be chosen as,

E[mI]=E[mn(X)]=d

Step by step solution

01

Given Information

Show thatE[ml]=d.

02

Explanation

Name chosen as n(x),

position is equality likely Therefore, to be chosen as

E[In(X)]=P{I=1n(X)}=1N(X)

E[I]=E[In(X)]

Therefore,

E[mI]=E[mn(X)]=d

03

Final Answer

E[mI]=E[mn(X)]=d

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A group of 20 people consisting of 10 men and 10 women is randomly arranged into 10 pairs of 2 each. Compute the expectation and variance of the number of pairs that consist of a man and a woman. Now suppose the 20 people consist of 10 married couples. Compute the mean and variance of the number of married couples that are paired together.

Consider a gambler who, at each gamble, either wins or loses her bet with respective probabilities pand 1-p. A popular gambling system known as the Kelley strategy is to always bet the fraction 2p-1of your current fortune when p>12. Compute the expected fortune afterngambles of a gambler who starts with xunits and employs the Kelley strategy.

AThere are n+1participants in a game. Each person independently is a winner with probability p. The winners share a total prize of 1 unit. (For instance, if 4people win, then each of them receives 14, whereas if there are no winners, then none of the participants receives anything.) Let A denote a specified one of the players, and let Xdenote the amount that is received by A.

(a) Compute the expected total prize shared by the players.

(b) Argue that role="math" localid="1647359898823" E[X]=1(1p)n+1n+1.

(c) Compute E[X] by conditioning on whether is a winner, and conclude that role="math" localid="1647360044853" E[(1+B)1]=1(1p)n+1(n+1)p when B is a binomial random variable with parameters n and p

Consider 3 trials, each having the same probability of success. Let Xdenote the total number of successes in these trials. If E[X]=1.8, what is

(a) the largest possible value of PX=3?

(b) the smallest possible value of P{X=3}}?

Suppose that X1and X2 are independent random variables having a common mean μ. Suppose also that VarX1=σ12 and VarX2=σ22. The value of μ is unknown, and it is proposed that μ be estimated by a weighted average of X1 and X2. That is, λX1+(1-λ)X2 will be used as an estimate of μ for some appropriate value of λ. Which value of λ yields the estimate having the lowest possible variance? Explain why it is desirable to use this value ofλ.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free