Proof of the Chebyshev inequality (continuous case):
Given: X a real continuous random variables with E(X) = µ, V (X) =
σ 2 , real number > 0.
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To show: P (|X − µ| ≥ ) ≤ σ2 .
Then
σ 2 = V (X)
Z ∞
(t − µ)2 fX (t) dt
=
−∞
Z µ−
Z
2
≥
(t − µ) fX (t) dt +
−∞
∞
(t − µ)2 fX (t) dt,
µ+
where the last line is by restricting the region over which we integrate
a positive function. Then this is
Z ∞
Z µ−
2
2 fX (t) dt,
fX (t) dt +
≥
−∞
µ+
2
since t ≤ µ − =⇒ ≤ |t − µ| =⇒ ≤ (t − µ)2 . But we rearrange
and use the definition of the density function to get
Z µ−
Z ∞
2
fX (t) dt
= fX (t) dt +
−∞
µ+
2
= P (X ≤ µ − or X ≥ µ + )
= 2 P (|X − µ| ≥ ).
Thus,
σ 2 ≥ 2 P (|X − µ| ≥ ),
and dividing through by 2 gives the desired.
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