Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(Darren Dugan) #1

31.4 SOME BASIC ESTIMATORS


exact expressions, valid for samples of any sizeN, for the expectation value and


variance of ̄x. From parts (i) and (ii) of the central limit theorem, discussed in


section 30.10, we immediately obtain


E[x ̄]=μ, V[x ̄]=

σ^2
N

. (31.40)


Thus we see that ̄xis an unbiased estimator ofμ. Moreover, we note that the


standard error in ̄xisσ/



N, and so the sampling distribution ofx ̄becomes more

tightly centred aroundμas the sample sizeNincreases. Indeed, sinceV[x ̄]→ 0


asN→∞,x ̄is also a consistent estimator ofμ.


In the limit of largeN, we may in fact obtain anapproximateform for the

full sampling distribution ofx ̄. Part (iii) of the central limit theorem (see section


30.10) tells us immediately that, for largeN, the sampling distribution ofx ̄is


given approximately by the Gaussian form


P(x ̄|μ, σ)≈

1

2 πσ^2 /N

exp

[

( ̄x−μ)^2
2 σ^2 /N

]
.

Note that this doesnotdepend on the form of the original parent population.


If, however, the parent population is in fact Gaussian then this result isexact


for samples ofanysizeN(as is immediately apparent from our discussion of


multiple Gaussian distributions in subsection 30.9.1).


31.4.2 Population varianceσ^2

An estimator for the population varianceσ^2 is not so straightforward to define


as one for the mean. Complications arise because, in many cases, the true mean


of the populationμis not known. Nevertheless, let us begin by considering the


case where in factμis known. In this event, a useful estimator is


σ̂^2 =^1
N

∑N

i=1

(xi−μ)^2 =

(
1
N

∑N

i=1

x^2 i

)

−μ^2. (31.41)

Show thatσ̂^2 is an unbiased and consistent estimator of the population varianceσ^2.

The expectation value ofσ̂^2 is given by


E[σ̂^2 ]=

1


N


E


[N



i=1

x^2 i

]


−μ^2 =E[x^2 i]−μ^2 =μ 2 −μ^2 =σ^2 ,

from which we see that the estimator is unbiased. The variance of the estimator is


V[σ̂^2 ]=

1


N^2


V


[N



i=1

x^2 i

]


+V[μ^2 ]=

1


N


V[x^2 i]=

1


N


(μ 4 −μ^22 ),

in which we have used that fact thatV[μ^2 ]=0andV[x^2 i]=E[x^4 i]−(E[x^2 i])^2 =μ 4 −μ^22 ,

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