Physics and Engineering of Radiation Detection

9.3. Probability 541

likelihood solution will become

μ∗ =

∑N

∑i=1xi/σ

N

i=1(1/σ)

=

1

N

∑N

i=1

xi, (9.3.43)

which is nothing but the expression for calculating simple mean. Hence we have
found that calculating mean by this method require that the variable is distributed
normally and that each measurement has the same error associated with it. Measur-
ing activity of a radioactive source falls into this category provided all the conditions
including the state of the detector does not change with time.
Let us now try to calculate the error in the calculation of the solution we just
obtained. Note that we are interested in finding out the spread ofμaboutμ∗and not
the errors in individual measurements. To do this we make use of the argument that
for large number of measurements (N→∞),L(μ) approaches a normal distribution.
Hence we can write

L(μ)=

1

σt

√

2 π

e−(μ−μ

∗) (^2) / 2 σ (^2) t
(9.3.44)
.
Here the subscripttinσtis meant to differentiate the standard deviation ofμfrom
that ofx. Again the condition 9.3.23 can be used to obtain the maximum likelihood
solution for this distribution. We first take the natural logarithm of both sides of
the above equation to obtain
ln(L)=ln

[(

e−(μ−μ

∗) (^2) / 2 σ (^2) t)
(σt)−^1 (2π)−^1 /^2

]

= −

(μ−μ∗)^2

2 σt^2

−ln(σt)−

ln(2π)

2

. (9.3.45)

Differentiating this twice with respect toμ∗gives

∂ln(L)

∂μ∗

=

μ−μ∗

σ^2 t

⇒

∂^2 ln(L)

∂μ∗^2

= −

1

σ^2 t

. (9.3.46)

Hence the error in calculation ofμ,isgivenby

μ=σt=

[

−

∂^2 ln(L)

∂μ∗^2

]− 1 / 2

. (9.3.47)

This general expression for computing errors is of central importance in likelihood
method and is extensively used in data analysis. Let us now use this expression to
derive the expression for the total error in measurements when each measurement
is characterized by its own errorσi. This can be done by differentiating equation