Mathematical Methods for Physics and Engineering : A Comprehensive Guide

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31.5 MAXIMUM-LIKELIHOOD METHOD


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246810 12 14


L(x;τ)

τ

Figure 31.8 The likelihood functionL(x;τ) (normalised to unit area) for the
sample values given in the worked example in subsection 31.5.4 and indicated
here by short vertical lines.

68.3% Bayesian central confidence interval. Even whenL(x;a) is not Gaussian,


however, (31.85) is often used as a measure of the standard error.


For the sample data given in subsection 31.5.4, use the likelihood function to estimate the
standard errorˆσˆτin the ML estimatorˆτand obtain the Bayesian68%central confidence
interval onτ.

We showed in (31.67) that the likelihood function in this case is given by


L(x;τ)=

1


τN

exp

[



1


τ

(x 1 +x 2 +···+xN)

]


.


wherexi,i=1, 2 ,...,N, denotes the sample value andN= 10. This likelihood function is
plotted in figure 31.8, after normalising (numerically) to unit area. The short vertical lines
in the figure indicate the sample values. We see that the likelihood function peaks at the
ML estimateτˆ=3.77 that we found in subsection 31.5.4. Also, from (31.77), we have


∂^2 lnL
∂τ^2

=


N


τ

2

(


1 −


2


τN

∑N


i=1

xi

)


.


Remembering thatτˆ=



ixi/N, our estimate of the standard error inˆτis

σˆˆτ=

(



∂^2 lnL
∂τ^2


∣∣



τ=ˆτ

)− 1 / 2


=


ˆτ

N

=1. 19 ,


which is precisely the estimate of the standard error we obtained in subsection 31.5.4.
It should be noted, however, that in general we would not expect the two estimates of
standard error made by the different methods to be identical.
In order to calculate the Bayesian 68% central confidence interval, we must determine
the valuesa−anda+that satisfy (31.84) withα=β=0.16. In this case, the calculation
can be performed analytically but is somewhat tedious. It is trivial, however, to determine
a−anda+numerically and we find the confidence interval to be [3. 16 , 6 .20]. Thus we can
quote our result with 68% central confidence limits as


τ=3. 77 +2− 0 ..^4361.
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