/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 7.51 Use Table 7.2 to determine the d... [FREE SOLUTION] | 91影视

91影视

Use Table 7.2 to determine the distribution ofi=1nXi when X1,,Xnare independent and identically distributed exponential random variables, each having mean1/.

Short Answer

Expert verified

The identically distributed exponential random variables value areXi~Gamma(n,).

Step by step solution

01

Given Information

Determine the distribution ofi=1nXi,whenX1,,Xnare independent.

02

Explanation 

We have that X1,,Xnhave distribution Expo()and that they are independent.

So, they have all common MGF

M(t)=-t

Define Y=i=1nXiwe have that,

MY(t)=EetY=Eeti=1nXi=i=1nEetXi=M(t)n.

03

Explanation 

Where the third and fourth equality hold since variablesXiare independent and equally distributed.

So, we have that,

MY(t)=-tn

Now, using the calculated MGF of Y, we can recognize that Y has Gamma distribution with parameters n and .

04

Final answer

We can recognize that Y has Gamma distribution with parameters nandvalue areXi~Gamma(n,).

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 91影视!

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

For an event A, let IA equal 1 if A occurs and let it equal 0 if A does not occur. For a random variable X, show that E[X|A] = E[XIA] P(A

In the text, we noted that

Ei=1Xi=i=1EXi

when the Xiare all nonnegative random variables. Since

an integral is a limit of sums, one might expect that E0X(t)dt=0E[X(t)]dt

whenever X(t),0t<,are all nonnegative random

variables; this result is indeed true. Use it to give another proof of the result that for a nonnegative random variable X,

E[X)=0P(X>t}dt

Hint: Define, for each nonnegative t, the random variable

X(t)by

role="math" localid="1647348183162" X(t)=1ift<X\\0iftX

Now relate4q

0X(t)dttoX

The random variables X and Y have a joint density function is given by

f(x,y)={2e2x/x0x<,0yx0otherwise

ComputeCov(X,Y)

The number of accidents that a person has in a given year is a Poisson random variable with mean 蹋 However, suppose that the value of changes from person to person, being equal to 2for 60percent of the population and 3for the other 40percent. If a person is chosen at random, what is the probability that he will have

(a) 0accidents and,

(b) Exactly 3accidents in a certain year? What is the conditional probability that he will have3 accidents in a given year, given that he had no accidents the preceding year?

For Example 2i, show that the variance of the number of coupons needed to a mass a full set is equal toi=1N1iN(Ni)2

When Nis large, this can be shown to be approximately equal (in the sense that their ratio approaches 1 as N) to N22/6.

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.