Lecture 10

Nonuniquenessand

Localized Averages

SyllabusLecture 01 Describing Inverse ProblemsLecture 02 Probability and Measurement Error, Part 1Lecture 03 Probability and Measurement Error, Part 2 Lecture 04 The L2 Norm and Simple Least SquaresLecture 05 A Priori Information and Weighted Least SquaredLecture 06 Resolution and Generalized InversesLecture 07 Backus-Gilbert Inverse and the Trade Off of Resolution and VarianceLecture 08 The Principle of Maximum LikelihoodLecture 09 Inexact TheoriesLecture 10 Nonuniqueness and Localized AveragesLecture 11 Vector Spaces and Singular Value DecompositionLecture 12 Equality and Inequality ConstraintsLecture 13 L1 , L∞ Norm Problems and Linear ProgrammingLecture 14 Nonlinear Problems: Grid and Monte Carlo Searches Lecture 15 Nonlinear Problems: Newton’s Method Lecture 16 Nonlinear Problems: Simulated Annealing and Bootstrap Confidence Intervals Lecture 17 Factor AnalysisLecture 18 Varimax Factors, Empirical Orthogonal FunctionsLecture 19 Backus-Gilbert Theory for Continuous Problems; Radon’s ProblemLecture 20 Linear Operators and Their AdjointsLecture 21 Fréchet DerivativesLecture 22 Exemplary Inverse Problems, incl. Filter DesignLecture 23 Exemplary Inverse Problems, incl. Earthquake LocationLecture 24 Exemplary Inverse Problems, incl. Vibrational Problems

Purpose of the Lecture

Show that null vectors are the source of nonuniqueness

Show why some localized averages of model parameters are unique while others aren’t

Show how nonunique averages can be bounded using prior information on the bounds of the underlying model parameters

Introduce the Linear Programming Problem

Part 1

null vectors as the source of

nonuniqueness

in linear inverse problems

suppose two different solutions exactly satisfy the same data

since there are twothe solution is nonunique

then the difference between the solutions satisfies

the quantity

mnull = m(1) – m(2)

is called a null vector

it satisfies

G mnull = 0

an inverse problem can have more than one null vector

mnull(1) mnull(2) mnull(3)...

any linear combination of null vectors is a null vector

αmnull(1) + βmnull(2) +γmnull(3)is a null vector for any α, β, γ

suppose that a particular choice of model parameters

mparsatisfies

G mpar=dobs with error E

then

has the same error Efor any choice of αi

since

e = dobs-Gmgen = dobs-Gmpar + Σi αi 0

since

since αi is arbitrarythe solution is nonunique

hence

an inverse problem isnonunique

if it has null vectors

Gmexample

consider the inverse problem

a solution with zero error ismpar=[d1, d1, d1, d1]T

the null vectors are easy to work out

note that times any

of these vectors is zero

the general solution to the inverse problem is

Part 2

Why some localized averages are

unique

while others aren’t

let’s denote a weighted average of the model parameters as

<m> = aT mwhere a is the vector of weights

let’s denote a weighted average of the model parameters as

<m> = aT mwhere a is the vector of weights

a may or may not be “localized”

a = [0.25, 0.25, 0.25, 0.25]T

a = [0. 90, 0.07, 0.02, 0.01]T

not localized

localized near m1

examples

now compute the average of the general solution

now compute the average of the general solution

if this term is zero for all i,then <m> does not depend on αi,

so average is unique

an average <m>=aTm is unique

if the average of all the null vectors

is zero

if we just pick an averageout of the hat

because we like it ... its nicely localized

chances are that it will not zero all the null vectors

so the average will not be unique

relationship to model resolution R

relationship to model resolution R

aT is a linear combination of the rows of the data kernel G

if we just pick an averageout of the hat

because we like it ... its nicely localized

its not likely that it can be built out of the rows of G

so it will not be unique

suppose we pick aaverage that is not unique

is it of any use?

Part 3

bounding localized averages

even though they are nonunique

we will now show

if we can put weak bounds on mthey may translate into stronger

bounds on <m>

example

with

so

example

with

so

nonunique

but suppose mi is bounded0 > mi > 2d1

smallest α3= -d1largest α3= +d1

(2/3) d1 > <m> > (4/3)d1

smallest α3= -d1largest α3= +d1

(2/3) d1 > <m> > (4/3)d1

smallest α3= -d1largest α3= +d1

bounds on <m> tighter than bounds on mi

the question is how to do this in more complicated cases

Part 4

The Linear Programming Problem

the Linear Programming problem

the Linear Programming problemflipping sign

switches minimization to maximization

flipping signs of A and bswitches to ≥

in Business

unit profitquantity of each product profit

maximizes

no negative productionphysical limitations of factory

government regulationsetc

care about both profit z and product quantities x

in our case

am <m>

bounds on mnot needed Gm=d

first minimizethen

maximize

care only about <m>, not m

In MatLab

Example 1

simple data kernelone datum

sum of mi is zero

bounds|mi| ≤ 1average

unweighted average of K model parameters

2 4 6 8 10 12 14 16 18 200

0.5

1

1.5

width

boun

d on

|<m

>|

Kbounds on

absolute

value

of weigh

ted avera

ge

2 4 6 8 10 12 14 16 18 200

0.5

1

1.5

width

boun

d on

|<m

>|

Kbounds on

absolute

value

of weigh

ted avera

ge

if you know that the sum of 20 things is zeroand

if you know that the things are bounded by ±1then you know

the sum of 19 of the things is bounded by about ±0.1

2 4 6 8 10 12 14 16 18 200

0.5

1

1.5

width

boun

d on

|<m

>|

Kbounds on

absolute

value

of weigh

ted avera

ge

for K>10<m> has tigher boundsthan mi

Example 2

more complicated data kerneldk weighted average of first 5k/2 m’s

bounds0 ≤ mi ≤ 1average

localized average of 5 neighboringmodel parameters

02

46

810

-0.5 0

0.5 1

1.5

z

mG mtrue mi (zi)

depth, ziw

idth, w

(A)(B)

= =

=

≈

dobs j

i

j

i

02

46

810

-0.5 0

0.5 1

1.5

z

mG mtrue mi (zi)

depth, ziw

idth, w

(A)(B)

= =

=

≈

dobs j

i

j

i

complicated Gbut

reminiscent of Laplace Transform kernel

02

46

810

-0.5 0

0.5 1

1.5

z

mG mtrue mi (zi)

depth, ziw

idth, w

(A)(B)

= =

=

≈

dobs j

i

j

i

true mi increased with depth zi

02

46

810

-0.5 0

0.5 1

1.5

z

mG mtrue mi (zi)

depth, ziw

idth, w

(A)(B)

= =

=

≈

dobs j

i

j

i

minimum length solution

02

46

810

-0.5 0

0.5 1

1.5

z

mG mtrue mi (zi)

depth, ziw

idth, w

(A)(B)

= =

=

≈

dobs j

i

j

i

lower boundon solution

upperbound

on solution

02

46

810

-0.5 0

0.5 1

1.5

z

mG mtrue mi (zi)

depth, ziw

idth, w

(A)(B)

= =

=

≈

dobs j

i

j

i

lower boundon average

upperbound

onaverage