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

P{D}11r22.

P{D}11r22

Short Answer

Expert verified

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

P{|X-|>||}222.

Therefore,

P{D}=P{|X-|||}1-222=r=|u|=1-1r22.

Step by step solution

01

Given Information.

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

02

Explanation.

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

P{D}=PX-=P{|X-|||}

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

P{|X-|>||}222.

Since

P{|X-|>||}=1-P{|X-|||},

therefore

P{D}=P{|X-|||}1-222r=||P{D}1-1r22.

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