Transcript Weights of Observations

Weights of Observations
Introduction
• Weights can be assigned to observations
according to their relative quality
• Example: Interior angles of a traverse are
measured – half of them by an inexperienced
operator and the other half by the best
instrument person. Relative weight should be
applied.
• Weight is inversely proportional to variance
Relation to Covariance Matrix
With correlated observations, weights are related to the
inverse of the covariance matrix, Σ.
For convenience, we introduce the concept of a
cofactor. The cofactor is related to its associated
covariance element by a scale factor which is the
inverse of the reference variance.
 ij
qij  2
0
Recall, Covariance Matrix
  x21

 x2 x1


 

 xn x1
x x
 x2
1 2
2

x x
1 n
  x1xn 

  x2 xn 

 

  x2n 
For independent observations, the off-diagonal terms
are all zero.
Cofactor Matrix
We can also define a cofactor matrix which is related to the
covariance matrix.
Q
1

2
0

The weight matrix is then:
W  Q 1   02 1
Weight Matrix for Independent
Observations
• Covariance matrix is diagonal
• Inverse is also diagonal, where each diagonal term is
the reciprocal of the corresponding variance element
• Therefore, the weight for observation i is:
 02
wi  2
i
If the weight, wi = 1, then  02   i2
 02 is the variance of an observation of unit weight (reference variance)
Reference Variance
• It is an arbitrary scale factor (a priori)
• A convenient value is 1 (one)
• In that case the weight of an independent observation
is the reciprocal of its variance
wi 
1
 i2
Simple Weighted Mean Example
A distance is measured three times, giving values of 151.9, 152.5,
and 152.5. Compute the mean.
y
151.9  152.5  152.5
 152.3
3
Same answer by weighted mean. The value 152.5 appears twice
so it can be given a relative weight of 2.
y
151.9  2 152.5
 152.3
3
Weighted Mean Formula
n
z
w z
i 1
n
i i
w
i 1
i
Weighted Mean – Example 2
A line was measured twice, using two different total stations. The
distance observations are listed below along with the computed
standard deviations based on the instrument specifications. Compute
the weighted mean.
D1 = 1097.253 m
D2 = 1097.241 m
σ1 = 0.010 m
σ2 = 0.005 m
Solution: First, compute the weights.
w1 
1
 12
1

1
2

10
,
000
m
(0.010m) 2
1
2
w2  2 

40
,
000
m
 2 (0.005m) 2
Example - Continued
Now, compute the weighted mean.
10,000m  2 1097.253m  40,000m  2 1097.241m
D
10,000m  2  40,000m  2
D  1097.243m
Notice that the value is much closer to the more
precise observation.
Standard Deviations – Weighted Case
• When computing a weighted mean, you want an
indication of standard deviation of observations.
• Since there are different weights, there will be
different standard deviations
• A single representative value is the standard
deviation of an observation of unit weight
• We can also compute standard deviation for a
particular observation
• And compute the standard deviation of the
weighted mean
Standard Deviation Formulas
n
Standard deviation
of unit weight
Standard deviation
of observation, i
S0 
2
w
v
 ii
i 1
n 1
n
Si 
2
w
v
 ii
i 1
wi (n  1)
n
Standard deviation
of the weighted
mean
SM 
2
w
v
 ii
i 1
n
(n  1) wi
i 1
Weights for Angles and Leveling
• If all other conditions are equal, angle weights
are directly proportional to the number of turns
• For differential leveling it is conventional to
consider entire lines of levels rather than
individual setups. Weights are:
– Inversely proportional to line length
– Inversely proportional to number of setups
Angle Example 9.2
This example asks for an “adjustment” and uses the
concept of a correction factor which has not been
described at this point. We will skip this type of
problem until we get to the topic of least squares
adjustment.
Differential Leveling Example
Four different routes were taken to determine the elevation
difference between two benchmarks (see table). Computed the
weighted mean elevation difference.
Example - Continued
Weights: (note that weights are multiplied by 12 to produce
integers, but this is not necessary)
Compute weighted mean:
12  25.35  6  25.41  4  25.38  2  25.30
M
12  6  4  2
608.78
M
 25.366ft
24
What about significant figures?
Example - Continued
Compute residuals
n
Compute standard deviation of unit weight
S0 
2
w
v
 ii
i 1
n 1
 0.090ft
n
Compute standard deviation of the mean
SM 
2
w
v
 ii
i 1
n
(n  1) wi
i 1
 0.018ft
Example - Continued
Standard deviations of weighted observations:
Summary
• Weighting allows us to consider different
precisions of individual observations
• So far, the examples have been with simple
means
• Soon, we will look at least squares adjustment
with weights
• In adjustments involving observations of
different types (e.g. angles and distances) it is
essential to use weights