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Chapter 8: Q. 8.21 (page 392)

Let Xbe a non-negative random variable. Prove that

E[X]≤EX21/2≤EX31/3≤⋯

Short Answer

Expert verified

Apply Lyapunov's inequality (proof is given inside) to a random variable0<1<2<3<⋯.

Step by step solution

01

Given Information.

LetX be a nonnegative random variable.

02

Explanation.

Suppose that Xis a nonnegative random variable. The relationship

E[X]≤EX21/2≤EX31/3≤⋯

we need to prove the consequence of Lyapunov's inequality, so let's prove this general result.

Lyapunov's inequality: If 0<s<t, thenE|X|s1/s≤E|X|t1/t.

Proof: Consider the functionf(y)=|y|r,r>1. The function f(y)is a convex function. Then, to Jensen's inequality,

E[f(Y)]≥f(E[Y])⇔E|Y|r≥|E[Y]|rr=ts>1,letY=|X|sE|X|t≥E|X|st/s

| taking the troot of each side we get:

E|X|t1/t≥E|X|s1/s.

Since X≥0we have that|X|=X. Apply Lyapunov's inequality to a random variable X=localid="1649893415983" 0<1<2<3<⋯.

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Most popular questions from this chapter

A tobacco company claims that the amount of nicotine in one of its cigarettes is a random variable with a mean of 2.2mg and a standard deviation of 0.3mg. However, the average nicotine content of 100randomly chosen cigarettes was 3.1mg. Approximate the probability that the average would have been as high as or higher than 3.1 if the company’s claims were true

The strong law of large numbers states that with probability 1, the successive arithmetic averages of a sequence of independent and identically distributed random variables converge to their common mean . What do the successive geometric averages converge to? That is, what is

limn→∞∏i=1nXi1/n

We have 100components that we will put to use in a sequential fashion. That is, the component 1is initially put in use, and upon failure, it is replaced by a component2, which is itself replaced upon failure by a componentlocalid="1649784865723" 3, and so on. If the lifetime of component i is exponentially distributed with a mean 10+i/10,i=1,...,100estimate the probability that the total life of all components will exceed1200. Now repeat when the life distribution of component i is uniformly distributed over(0,20+i/5),i=1,...,100.

8.5 The amount of time that a certain type of component functions before failing is a random variable with probability density function

f(x)=2x0<x<1

Once the component fails, it is immediately replaced by
another one of the same type. If we let denote the life-time of the ith component to be put in use, then Sn=∑i=1nXirepresents the time of the nth failure. The long-term rate at which failures occur, call itr, is defined by
r=limn→∞nSn

Assuming that the random variables Xi,i≥1,are independent, determine r.

If, Xis a nonnegative random variable with a mean of25, what can be said about

role="math" localid="1649836663727" (a)E[X3]?

(b)E[√X]?

(c)E[logX]?

role="math" localid="1649836643047" (d)E[e-X]?
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