Mathematical Methods for Physics and Engineering : A Comprehensive Guide

STATISTICS

Differentiating with respect toμand setting the result equal to zero gives

∂lnL

∂μ

= 1 TV−^1 (x−μ 1 )=0.

Thus, the ML estimator is given by

μˆ=

1 TV−^1 x

1 TV−^11

=

∑

i,j(V

− (^1) )ijxj
∑
i,j(V−^1 )ij

.

In the case of uncorrelated errors in measurement, (V−^1 )ij=δij/σ^2 i and our estimator
reduces to that given in (31.72).

In all the examples considered so far, the likelihood function has been effectively

one-dimensional, either instrinsically or under the assumption that the values of

all but one of the parameters are known in advance. As the following example

involving two parameters shows, the application of the ML method to the

estimation of several parameters simultaneously is straightforward.

In an experimentNmeasurementsxiof some quantity are made. Suppose the random error

on each sample value is drawn independently from a Gaussian distribution of mean zero but

unknown standard deviationσ(which is the same for each measurement). Calculate the ML

estimates of the true valueμof the quantity being measured and the standard deviationσ

of the random errors.

In this case the log-likelihood function is given by

lnL(x;μ, σ)=−

1

2

∑N

i=1

[

ln(2πσ^2 )+

(xi−μ)^2

σ^2

]

.

Taking partial derivatives of lnLwith respect toμandσand setting the results equal to
zero at the joint estimateμ,ˆσˆ,weobtain

∑N

i=1

xi−μˆ

σˆ^2

=0, (31.73)

∑N

i=1

(xi−μˆ)^2

σˆ^3

−

∑N

i=1

1

σˆ

=0. (31.74)

In principle, one should solve these two equations simultaneously forμˆandσˆ, but in this
case we notice that the first is solved immediately by

μˆ=

1

N

∑N

i=1

xi= ̄x,

where ̄xis the sample mean. Substituting this result into the second equation, we find

σˆ=

√√

√

√^1

N

∑N

i=1

(xi−x ̄)^2 =s,

wheresis the sample standard deviation. As shown in subsection 31.4.3,sis a biased
estimator ofσ. The reason why the ML method may produce a biased estimator is
discussed in the next subsection.