Explanation Calculation: Consider the random variable Z = | g ( Y ) | . This is a random variable as Y itself is a random variable, and the support of Z is a subset of [ 0 , ∞ ) , with a distribution function F . The indicator variable is defined as follows: I A = { 1 , if event A occurs 0 , otherwise Now consider, P ( Z > k ) = E ( I ( Z > k ) ) = ∫ 0 ∞ I ( Z > k ) d F ( z ) Consider the case, if z > k , then z k > 1 = I { z > k } and if 0 < z < k implies that z k > 0 = I { z > k } . Therefore, it can be observed that z k ≥ 1 { z > k } ∀ z ≥ 0

Mathematical Statistics with Applications

7th Edition

ISBN: 9780495110811

Author: Dennis Wackerly, William Mendenhall, Richard L. Scheaffer

Publisher: Cengage Learning

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Chapter 4, Problem 198SE

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Prove that for every constant k, P(|g(Y)|≤k)≥1−E(|g(Y)|)k

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Chapter 4, Problem 197SE

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Chapter 4 Solutions

Mathematical Statistics with Applications

Ch. 4.2 - Prob. 1E Ch. 4.2 - A box contains five keys, only one of which will...Ch. 4.2 - A Bernoulli random variable is one that assumes...Ch. 4.2 - Let Y be a binomial random variable with n = 1 and...Ch. 4.2 - Suppose that Y is a random variable that takes on...Ch. 4.2 - Consider a random variable with a geometric...Ch. 4.2 - Let Y be a binomial random variable with n=10 and...Ch. 4.2 - Prob. 8E Ch. 4.2 - A random variable Y has the following distribution...Ch. 4.2 - Refer to the density function given in Exercise...

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Prove that for every constant k , P ( | g ( Y ) | ≤ k ) ≥ 1 − E ( | g ( Y ) | ) k

Solution Summary: The author explains that the random variable Z is a subset of left[0,inftyright], and the indicator variable is defined as follows:

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Prove that for every constant k, P(|g(Y)|≤k)≥1−E(|g(Y)|)k

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