Mathematical Methods for Physics and Engineering : A Comprehensive Guide

STATISTICS

whereμris therth population moment. Sinceσ̂^2 is unbiased andV[σ̂^2 ]→0asN→∞,
showing that it is also a consistent estimator ofσ^2 , the result is established.

If the true mean of the population is unknown, however, a natural alternative

is to replaceμby ̄xin (31.41), so that our estimator is simply the sample variance

s^2 given by

s^2 =

1

N

∑N

i=1

x^2 i−

(

1

N

∑N

i=1

xi

) 2

.

In order to determine the properties of this estimator, we must calculateE[s^2 ]

andV[s^2 ]. This task is straightforward but lengthy. However, for the investigation

of the properties of acentralmoment of the sample, there exists a useful trick

that simplifies the calculation. We can assume, with no loss of generality, that

the meanμ 1 of the population from which the sample is drawn is equal to zero.

With this assumption, the population central moments,νr,areidenticaltothe

corresponding momentsμr, and we may perform our calculation in terms of the

latter. At the end, however, we replaceμrbyνrin the final result and so obtain

a general expression that is valid even in cases whereμ 1 =0.

CalculateE[s^2 ]andV[s^2 ]for a sample of sizeN.

The expectation value of the sample variances^2 for a sample of sizeNis given by

E[s^2 ]=

1

N

E

[

∑

i

x^2 i

]

−

1

N^2

E





(

∑

i

xi

) 2 



=

1

N

NE[x^2 i]−

1

N^2

E







∑

i

x^2 i+

∑

i,j

j=i

xixj





. (31.42)

The number of terms in the double summation in (31.42) isN(N−1), so we find

E[s^2 ]=E[x^2 i]−

1

N^2

(NE[x^2 i]+N(N−1)E[xixj]).

Now, since the sample elementsxiandxjare independent,E[xixj]=E[xi]E[xj]=0,
assuming the meanμ 1 of the parent population to be zero. Denoting therth moment of
the population byμr, we thus obtain

E[s^2 ]=μ 2 −

μ 2

N

=

N− 1

N

μ 2 =

N− 1

N

σ^2 , (31.43)

where in the last line we have used the fact that the population mean is zero, and so
μ 2 =ν 2 =σ^2. However, the final result is also valid in the case whereμ 1 =0.
Using the above method, we can also find the variance ofs^2 , although the algebra is
rather heavy going. The variance ofs^2 is given by

V[s^2 ]=E[s^4 ]−(E[s^2 ])^2 , (31.44)

whereE[s^2 ] is given by (31.43). We therefore need only consider how to calculateE[s^4 ],