Mathematical Methods for Physics and Engineering : A Comprehensive Guide

31.3 ESTIMATORS AND SAMPLING DISTRIBUTIONS

For the 90% central confidence interval, we requireα=β=0.05. From table 30.3, we
find

Φ−^1 (1−α)=Φ−^1 (0.95) = 1. 65 ,

and using (31.31) and (31.32) we obtain

a−= ̄x− 1. 65 σ ̄x=1. 11 −(1.65)(0.32) = 0. 58 ,

a+= ̄x+1. 65 σ ̄x=1.11 + (1.65)(0.32) = 1. 64.

Thus, the 90% central confidence interval onμis [0. 58 , 1 .64]. For comparison, the true
value used to create the sample wasμ=1.

In the case where the standard errorσaˆin (31.33) is not known in advance,

one must use a valueσˆaˆestimated from the sample. In principle, this complicates

somewhat the construction of confidence intervals, since properly one should

consider the two-dimensional joint sampling distributionP(a,ˆσˆˆa|a). Nevertheless,

in practice, providedσˆaˆis a fairly good estimate ofσaˆthe above procedure may

be applied with reasonable accuracy. In the special case where the sample values

xiare drawn from a Gaussian distribution with unknownμandσ,itisinfact

possible to obtainexactconfidence intervals on the meanμ, for a sample of any

sizeN, using Student’st-distribution. This is discussed in subsection 31.7.5.

31.3.6 Estimation of several quantities simultaneously

Suppose one uses a samplex 1 ,x 2 ,...,xNto calculate the values of several es-

timatorsaˆ 1 ,aˆ 2 ,...,aˆM(collectively denoted byaˆ) of the quantitiesa 1 ,a 2 ,...,aM

(collectively denoted bya) that describe the population from which the sample was

drawn. The joint sampling distribution of these estimators is anM-dimensional

PDFP(aˆ|a) given by

P(aˆ|a)dMˆa=P(x|a)dNx.

Sample valuesx 1 ,x 2 ,...,xNare drawn independently from a Gaussian distribution with

meanμand standard deviationσ. Suppose we choose the sample mean ̄xand sample stan-

dard deviationsrespectively as estimatorsμˆandσˆ. Find the joint sampling distribution of

these estimators.

Since each data valuexiin the sample is assumed to be independent of the others, the
joint probability distribution of sample values is given by

P(x|μ, σ)=(2πσ^2 )−N/^2 exp

[

−

∑

i(xi−μ)

2

2 σ^2

]

.

We may rewrite the sum in the exponent as follows:
∑

i

(xi−μ)^2 =

∑

i

(xi− ̄x+x ̄−μ)^2

=

∑

i

(xi− ̄x)^2 +2( ̄x−μ)

∑

i

(xi− ̄x)+

∑

i

( ̄x−μ)^2

=Ns^2 +N( ̄x−μ)^2 ,