/*! 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} Problem 21 Let \(X\) be a nonnegative rando... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 21

Let \(X\) be a nonnegative random variable. Prove that $$E[X] \leq\left(E\left[X^{2}\right]\right)^{1 / 2} \leq\left(E\left[X^{3}\right]\right)^{1 / 3} \leq \dots$$

Short Answer

Expert verified
Using Jensen's inequality with a reverse concave square root function, we first prove that \(E[X] \leq (E[X^{2}])^{1 / 2}\). Then, by applying H枚lder's inequality for each pair of successive powers, we prove the generalized inequality: \(\left( E[X^{k}] \right)^{1/k} \leq \left( E[X^{k+1}] \right)^{1/(k+1)}\). Combining these results, we conclude that \(E[X] \leq\left( E[X^{2}]\right)^{1 / 2} \leq\left(E[X^{3}]\right)^{1 / 3} \leq \dots\).

Step by step solution

01

Recall that Jensen's inequality states that for any convex function \(f\) and a random variable \(Y\), we have \[E[f(Y)] \geq f(E[Y]).\] And H枚lder's inequality states that for any nonnegative random variable \(Z\) and positive integer powers \(p\) and \(q\), such that \(\frac{1}{p}+\frac{1}{q}=1\), we have \[\left( E[Z^p] \right)^{1/p} \leq \left( E[Z^q] \right)^{1/q}.\] #Step 2: Apply Jensen's inequality with the square root function#

To show that \(E[X] \leq \left(E\left[X^{2}\right]\right)^{1 / 2}\), set \(f(Y) = \sqrt{Y}\) and \(Y = X^2\). Since the square root function is concave, we can apply the reverse Jensen's inequality, which states that for any concave function \(f\) and a random variable \(Y\), we have \[E[f(Y)] \leq f(E[Y]).\] Therefore, now we have \[E[\sqrt{X^2}] \leq \sqrt{E[X^2]}.\] Since \(X\) is nonnegative, we know that \(\sqrt{X^2} = X\). Thus, we get \[E[X] \leq\left(E\left[X^{2}\right]\right)^{1 / 2}.\] #Step 3: Apply H枚lder's inequality for each pair of successive powers#
02

To prove the generalized inequality \(\left(E\left[X^{k}\right]\right)^{1 / k} \leq \left(E\left[X^{k+1}\right]\right)^{1 / (k+1)}\), apply H枚lder's inequality with \(p=k+1\) and \(q=k+1-k=k/(k+1)\) so that we can satisfy the criteria of the inequality \(\frac{1}{p}+\frac{1}{q}=1\). By H枚lder's inequality, \[\left( E[X^{k+1}] \right)^{1/(k+1)} \leq \left( E[X^k] \right)^{1/k}.\] Since the inequality holds for any arbitrary \(k\), we have \[\left(E\left[X^{2}\right]\right)^{1 / 2} \leq \left(E\left[X^{3}\right]\right)^{1 / 3} \leq \left(E\left[X^{4}\right]\right)^{1 / 4} \leq \dots\] #Step 4: Combine the results#

Combining the results from steps 2 and 3, we conclude that \[E[X] \leq\left(E\left[X^{2}\right]\right)^{1 / 2} \leq\left(E\left[X^{3}\right]\right)^{1 / 3} \leq \dots\] And thus, we have proven the inequality.

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

A person has 100 light bulbs whose lifetimes are independent exponentials with mean 5 hours. If the bulbs are used one at a time, with a failed bulb being replaced immediately by a new one, approximate the probability that there is still a working bulb after 525 hours.

A.J. has 20 jobs that she must do in sequence, with the times required to do each of these jobs being independent random variables with mean 50 minutes and standard deviation 10 minutes. M.J. has 20 jobs that he must do in sequence, with the times required to do each of these jobs being independent random variables with mean 52 minutes and standard deviation 15 minutes. (a) Find the probability that A.J. finishes in less than 900 minutes. (b) Find the probability that M.J. finishes in less than 900 minutes. (c) Find the probability that A.J. finishes before M.J.

Civil engineers believe that \(W\), the amount of weight (in units of 1000 pounds) that a certain span of a bridge can withstand without structural damage resulting, is normally distributed with mean 400 and standard deviation \(40 .\) Suppose that the weight (again, in units of 1000 pounds) of a car is a random variable with mean 3 and standard deviation \(3 .\) Approximately how many cars would have to be on the bridge span for the probability of structural damage to exceed \(.1 ?\)

Let \(X_{1}, \ldots, X_{20}\) be independent Poisson random variablcs with mean 1. (a) Use the Markov incquality to obtain a bound on $$P\left\\{\sum_{1}^{20} X_{i}>15\right\\}$$ (b) Use the central limit theorem to approximate $$P\left\\{\sum_{1}^{20} X_{i}>15\right\\}$$

We have 100 components that we will put in use in a sequential fashion. That is, component 1 is initially put in use, and upon failure, it is replaced by component \(2,\) which is itself replaced upon failure by component \(3,\) and so on. If the lifetime of component \(i\) is exponcntially distributed with mean \(10+i / 10, i=1, \ldots, 100,\) estimate the probability that the total life of all components will exceed \(1200 .\) Now repeat when the life distribution of componcnt \(i\) is uniformly distributcd over \((0,20+i / 5), i=1, \dots, 100\)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation
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.