Date post: | 19-Dec-2015 |
Category: |
Documents |
View: | 216 times |
Download: | 2 times |
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, Empircal 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
Introduce the concept of prediction error and the norms that quantify it
Develop the Least Squares Solution
Develop the Minimum Length Solution
Determine the covariance of these solutions
Gmest = dprean estimate of the model parameters can be
used to predict the data
but the prediction may not match the observed data
(e.g. due to observational error)dpre ≠ dobs
e = dobs -dprethis mismatch leads us to define the
prediction error
e = 0when the model parameters exactly predict the data
B)A)
0 5 100
5
10
15
z
d
0 5 100
5
10
15
z
d
zi
diobsdipre ei
example of prediction errorfor line fit to data
0 1 2 3 4 5 6 7 8 9 10-1
0
1
z
e
0 1 2 3 4 5 6 7 8 9 10-1
0
1
z
|e|
0 1 2 3 4 5 6 7 8 9 10-1
0
1
z
|e|2
0 1 2 3 4 5 6 7 8 9 10-1
0
1
z
|e|10
higher norms give increaing weight to largest element of e
guiding principle for solving an inverse problem
find the mest
that minimizes E=||e||
withe = dobs –dpreand dpre = Gmest
B)A)
0 5 100
0.1
0.2
0.3
0.4
0.5
d
p(d)
0 5 100
0.1
0.2
0.3
0.4
0.5
d
p(d)
Answer is related to the distribution of the error. Are outliers common or rare?
long tailsoutliers common
outliers unimportantuse low norm
gives low weight to outliers
short tailsoutliers uncommonoutliers important
use high normgives high weight to outliers
L2 norm of error is its Euclidian length
so E is the square of the Euclidean lengthmimimize E
Principle of Least Squares
= eTe
ai = Σj δij bj = bi
Kronecker delta(elements of identity matrix) [I]ij = δij
a = Ib = bai = Σj δij bj = bi
i
Least Squares will fail
when more than one solution minimizes the error
the inverse problem is “underdetermined”
Method of Lagrange Multipliersminimize L with constraintsC1=0, C2=0, …
equivalent to
minimize Φ=L+λ1C1+λ2C2+…with no constraints λs called “Lagrange Multipliers”
Least Squares Solutionmest= [GTG ]-1GTdMinimum Length Solutionmest=GT [GGT ]-1dboth have the linear formm=Md
but ifm=Mdthen[cov m] = M [cov d] MT
when data are uncorrelated with uniform variance σd2
[cov d]=σd2I
so
Least Squares Solution[cov m] = [GTG ]-1GTσd2 G[GTG ]-1[cov m] = σd
2 [GTG ]-1
Minimum Length Solution[cov m] = GT [GGT ]-1 σd2 [GGT ]-1G[cov m] = σd
2 GT [GGT ]-2G
Least Squares Solution[cov m] = [GTG ]-1GTσd2 G[GTG ]-1[cov m] = σd
2 [GTG ]-1
Minimum Length Solution[cov m] = GT [GGT ]-1 σd2 [GGT ]-1G[cov m] = σd
2 GT [GGT ]-2G memorize
where to obtain the value of σd2 a priori value – based on knowledge of accuracy
of measurement technique
my ruler has 1 mm divisions, so σd≈ mm½
a posteriori value – based on prediction error
variance critically dependent on experiment design (structure of G)
1 12 123
1234
1 12 23 34… …
which is the better way to weigh a set of boxes ?
0 10 20 30 40 50 60 70 80 90 100
-2
0
2m
z
0 10 20 30 40 50 60 70 80 90 1000
0.5
1
sm
z
A)
B)
miest
σmi
i
i
0 1 2 3 4
0
1
2
3
4
500
1000
1500
2000
0 1 2 3 4
0
1
2
3
4
500
1000
1500
2000
2500
3000
-5 0 5-10
-5
0
5
10
-5 0 5-10
-5
0
5
10
-5 0 5-10
-5
0
5
10
-5 0 5-10
-5
0
5
10
z
m1
m20 40
4m1
0
4
m20 4E Ezd d
A) B)
C) D)
Relationship between[cov m] and Error Surface
for a linear problemcurvature is related to GTGE = (Gm-d)T(Gm-d) =
mT[GTG]m-dTGm-mTGTd+dTdso
∂2E/ ∂mi∂mj = [GTG] ij