Introduction to Probability and Statistics for Engineers and Scientists

30 Chapter 2:Descriptive Statistics

Proof

Letyi=xi− ̄x,i=1,...,n. For anyb>0, we have that

∑n

i= 1

(yi+b)^2 ≥

∑

i:yi≥ks

(yi+b)^2

≥

∑

i:yi≥ks

(ks+b)^2

=N(k)(ks+b)^2 (2.4.1)

where the first inequality follows because (yi+b)^2 ≥0, and the second because bothks
andbare positive. However,

∑n

i= 1

(yi+b)^2 =

∑n

i= 1

(yi^2 + 2 byi+b^2 )

=

∑n

i= 1

yi^2 + 2 b

∑n

i= 1

yi+nb^2

=(n−1)s^2 +nb^2

where the final equation used that

∑n

i= 1 yi =

∑n

i= 1 (xi− ̄x)=

∑n
i= 1 xi−nx ̄ =0.
Therefore, we obtain from Equation (2.4.1) that

N(k)≤

(n−1)s^2 +nb^2

(ks+b)^2

implying that
N(k)
n

≤

s^2 +b^2

(ks+b)^2

Because the preceding is valid for allb>0, we can setb =s/k(which is the value ofb
that minimizes the right-hand side of the preceding) to obtain that

N(k)

n

≤

s^2 +s^2 /k^2

(ks+s/k)^2

Multiplying the numerator and the denominator of the right side of the preceding byk^2 /s^2
gives

N(k)

n

≤

k^2 + 1

(k^2 +1)^2

=

1
k^2 + 1
and the result is proven. Thus, for instance, where the usual Chebyshev inequality shows
that at most 25 percent of data values are at least 2 standard deviations greater than
the sample mean, the one-sided Chebyshev inequality lowers the bound to “at most
20 percent.” ■