Global Positioning System Reference
In-Depth Information
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Since a linear transformation of a random variable with multivariate normal distri-
bution results in another multivariate normal distribution according to Theorem 1, it
follows that
z
is distributed as
N
n
,
r
I
r
F
T
Ax
o
−
O
2
0
z
∼
σ
(4.258)
O
r
I
n
−
r
n
−
The random variables
z
1
and
z
2
are stochastically independent, as are the individ-
ual components according to Theorem 3. From Theorem 2 it follows that
2
0
I
)
z
2
∼
σ
N
n
−
r
(
o
,
(4.259)
Thus
n
0
,
0
[13
2
z
2
i
∼
σ
(4.260)
z
2
i
σ
0
∼
n(
0
,
1
)
(4.261)
Lin
—
-
——
Lon
*PgE
are unit variate normal distributed. As listed in Appendix A5, the square of a stan-
dardized normal distributed variable has a chi-square distribution with one degree of
freedom. In addition, the sum of chi-square distributed variables is also a chi-square
distribution with a degree of freedom equal to the sum of the individual degrees of
freedom. Using these functions of random variables, it follows that
v
T
Pv
n
−
r
[13
R
σ
z
2
z
2
i
σ
z
2
n
0
=
=
0
∼ χ
(4.262)
−
r
0
σ
i
=
1
ha
s a chi-square distribution with
n
−
r
degrees of freedom.
4.9.3 Testing
v
T
Pv
and
v
T
Pv
∆
Co
mbining the result of (4.262) with the expression for the a posteriori variance of
un
it weight of Table 4.1, we obtain the formulation for a fundamental statistical test
in
least-squares estimation:
2
0
v
T
Pv
σ
σ
n
=
(n
−
r)
∼ χ
(4.263)
−
r
0
0
σ
No
te that
n
r
is the degree of freedom of the adjustment. If there is no rank deficiency
in
the design matrix, the degree of freedom is
n
−
−
u
. Based on the statistics (4.263), the
tes
t can be performed to find out whether the adjustment is distorted. The formulation
of
the hypothesis is as follows:
2
0
2
H
0
:
σ
0
=σ
(4.264)
