Information Technology Reference
In-Depth Information
N
ξ
=
1
ξ
j
p
j
=
,
0
(1.7)
j
=
implying that the average value is the best representation of the data. Gauss determined
that the variance defined by (
1.3
) in terms of the error (
1.6
) takes the form
N
2
2
σ
≡
1
ξ
j
p
j
(1.8)
j
=
and can be used to measure how well the average characterizes the ensemble of mea-
surements. Note that it is not necessary to introduce
p
j
for the following argument and,
although its introduction would not change the presentation in any substantial way, the
discussion is somewhat simpler without it.
Gauss used the statistical independence of the measured quantities to prove that the
average value gave their best representation and that, with a couple of physically rea-
sonable assumptions, the associated statistical distribution was normal, an unfortunate
name that had not been introduced at that time. We present a modified version of his
arguments here to lay bare the requirements of normalcy. The probability
I
of obtaining
a value in the interval (
Q
,
Q
+
Q
)
in any measurement is given by
I
=
P
(
Q
)
Q
(1.9)
and in a sequence of
N
measurements the data are replaced with the deviations from the
average, that is, by the errors
ξ
1
,...,ξ
N
, allowing us to segment the range of values
into
N
intervals,
P
(ξ
j
)ξ
j
≡
probability of observing the deviation
ξ
j
.
(1.10)
In (
1.10
) the probability of making the
N
independent measurements in the ensemble
together with the property that the probability of the occurrence of any two independent
events is given by the product of their individual probabilities, and assuming
ξ
j
=
ξ
for all
j
,is
N
N
I
=
P
(ξ
j
)ξ
j
=
P
(ξ
1
)
P
(ξ
2
)...
P
(ξ
N
)ξ
.
(1.11)
j
=
1
According to Gauss the estima
tio
n of the value for
Q
appears
plausible
if the ensemble
of measurements resulting in
Q
is the most
probabl
e
.
Thus,
Q
is determined in such
a way that the probability
I
is a maximum for
Q
=
Q
. To determine this form of the
probability density we impose the condition
d
ln
I
dQ
=
0
(1.12)
and use (
1.11
) to obtain
ln
P
ξ
j
∂ξ
j
,
N
N
ξ
j
∂
d
l
n
I
dQ
=
d
dQ
∂
ln
I
∂ξ
j
=−
(1.13)
j
=
1
j
=
1
