Weights of Observations

8/3/2019 Weights of Observations

1/20

Weights of Observations


2/20

Introduction

Weights can be assigned to observations

according to their relative quality

Example: Interior angles of a traverse aremeasured 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


3/20

Relation to Covariance Matrix

With correlated observations, weights are related to the

inverse of the covariance matrix,.

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.

2

0W

W ijijq !


4/20

Recall, Covariance Matrix

-

!7

2

2

2

11

2212

1211

nnn

n

n

xxxxx

xxxxx

xxxxx

WWW

WWW

WWW

.

/1//

.

.

For independent observations, the off-diagonal terms

are all zero.


5/20

Cofactor Matrix

7!2

0

1

WQ

We can also define a cofactor matrix which is related to the

covariance matrix.

The weight matrix is then:

12

0

1 7!! WQW


6/20

Weight Matrix forIndependent

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:

2

2

0

i

iwW

W!

If the weight, wi = 1, then

is the variance of an observation of unit weight (reference variance)

22

0 iWW !

2

0W


7/20

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

2

1

i

iwW

!


8/20

Simple Weighted Mean Example

3.1523

5.1525.1529.151!

!

y

A distance is measured three times, giving values of 151.9, 152.5,

and 152.5. Compute the mean.

Same answer by weighted mean. The value 152.5 appears twice

so it can be given a relative weight of 2.

3.1523

5.15229.151!

v!y


9/20

Weighted Mean Formula

!

!!

n

ii

n

i

ii

w

zw

z

1

1


10/20

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 1 = 0.010 m

D2 = 1097.241 m 2 = 0.005 m

Solution: First, compute the weights.

2

22

2

2

2

22

1

1

m000,40

)m005.0(

11

m000,10)m010.0(

11

!!!

!!!

W

W

w

w


11/20

Example - Continued

Now, compute the weighted mean.

m243.1097

40,000m10,000m

1097.241m40,000m1097.253m10,000m22

22

!

vv!

D

D

Notice that the value is much closer to the more

precise observation.


12/20

Standard Deviations Weighted Case

When computing a weighted mean, you want anindication 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


13/20

Standard Deviation Formulas

11

2

0

!

!

n

vw

S

n

i

ii

)1(1

2

!

!

nw

vw

Si

n

i

ii

i

Standard deviation

of unit weight

Standard deviation

of observation, i

Standard deviation

of the weighted

mean

!

!

!n

i

i

n

i

ii

M

wn

vw

S

1

1

2

)1(


14/20

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 toconsider entire lines of levels rather than

individual setups. Weights are:

Inversely proportional to line length

Inversely proportional to number of setups


15/20

Angle Example

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.


16/20

Differential Leveling Example

Four different routes were taken to determine the elevation

difference between two benchmarks (see table). Computed the

weighted mean elevation difference.


17/20

Example - ContinuedWeights: (note that weights are multiplied by 12 to produce

integers, but this is not necessary)

Compute weighted mean:

ft366.2524

78.608

24612

30.25238.25441.25635.2512

!!

vvvv!

M

M

What about significant figures?


18/20

Example - Continued

ft090.01

1

2

0 !

!

!

n

vw

S

n

i

ii

Compute residuals

Compute standard deviation of unit weight

Compute standard deviation of the mean

ft018.0

)1(1

1

2

!

!

!

!

n

i

i

n

i

ii

M

wn

vw

S


19/20

Example - Continued

Standard deviations of weighted observations:


20/20

Summary

Weighting allows us to consider differentprecisions of individual observations

So far, the examples have been with simplemeans

Soon, we will look at least squares adjustmentwith weights

In adjustments involving observations ofdifferent types (e.g. angles and distances) it isessential to use weights

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Weights of Observations