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It is known that x is a continuous random variable,

E(X)=μ,D(X)=δ2

real number

ε>0

Verification:

P(∥X−μ∥≥ε)≤δ2ε2

prove:

  because:

δ2=V(X)

=∫+∞−∞(t−μ)2fX(t)dt

≥∫μ−ε−∞(t−μ)2fX(t)dt+∫+∞μ−ε(t−μ)2fX(t)dt

≥∫μ−ε−∞ε2fX(t)dt+∫+∞μ−εε2fX(t)dt

  due to:

t≤μ−ε⇒ε≤∥t−μ∥⇒ε2≤(t−μ)2

So there is

=ε2∫μ−ε−∞fX(t)dt+∫+∞μ−εfX(t)dt

=ε2P(X≤μ−εorX≥μ+ε)

=ε2P(∥X−μ∥≥ε)

Therefore, there are:

δ2≥ε2P(∥X−μ∥≥ε)

It’s true!