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Chapter 8: Q. 8.2 (page 392)

ItXhas, a mean渭and standard deviation蟽, the ratior=|渭|/蟽is called the measurement signal-to-noise ratioX. The idea is that Xcan be expressed asX=渭+(X鈭�渭), 渭representing the signal and X鈭�渭the noise. If we define|(X鈭�渭)/渭|=Dit as the relative deviation Xfrom its signal (or mean)渭, show that for伪>0,

P{D鈮�伪}鈮�1鈭�1r2伪2.

P{D鈮�伪}鈮�1鈭�1r2伪2

Short Answer

Expert verified

Since |渭|伪>0using Chebyshev's inequality we get:

P{|X-渭|>|渭|伪}鈮�蟽2渭2伪2.

Therefore,

P{D鈮�伪}=P{|X-渭|鈮�|渭|伪}鈮�1-蟽2渭2伪2=r=|u|=1-1r2伪2.

Step by step solution

01

Given Information.

Xhas to mean渭and standard deviation蟽, the ratiolocalid="1649850029051" r=|渭|/蟽called the measurement signal-to-noise ratio of X.

02

Explanation.

Assume that the random variable Xhas mean 渭and standard deviation蟽, and let伪>0. Then,

P{D鈮�伪}=P鈭�X-渭渭鈮�伪=P{|X-渭|鈮�|渭|伪}

Since localid="1649850016019" |渭|伪>0using Chebyshev's inequality we get:

P{|X-渭|>|渭|伪}鈮�蟽2渭2伪2.

Since

P{|X-渭|>|渭|伪}=1-P{|X-渭|鈮�|渭|伪},

therefore

P{D鈮�伪}=P{|X-渭|鈮�|渭|伪}鈮�1-蟽2渭2伪2鈬�r=|渭|蟽P{D鈮�伪}鈮�1-1r2伪2.

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Most popular questions from this chapter

Would the results of Example5fchange be if the investor were allowed to divide her money and invest the fraction伪,0<伪<1,in the risky proposition and invest the remainder in the risk-free venture? Her return for such a split investment would beR=伪X+(1鈭�伪)m.

Let Xbe a binomial random variable with parameters nandp. Show that, fori>np,

(a)the minimum e-tiEetXoccurs when tis such thatet=iq(n-i)pwhereq=1-p.

(b)P{X鈮�i}鈮�nni2(n-i)n-ipi(1-p)n-i

LetZn, n鈮�1be a sequence of random variables andca constant such that for each

蔚>0,P|Zn鈭�c|>蔚鈫�0asn鈫�q. Show that for any bounded continuous functiong,

E[g(Zn)]鈫�g(c)asn鈫�q.

A certain component is critical to the operation of an electrical system and must be replaced immediately upon failure. If the mean lifetime of this type of component is 100 hours and its standard deviation is 30 hours, how many of these components must be in stock so that the probability that the system is in continual operation for the next 2000 hours is at least .95?

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=鈭�i=1nXirepresents the time of the nth failure. The long-term rate at which failures occur, call itr, is defined by
r=limn鈫�鈭�nSn

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