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Science Forum Index » Math - Numerical Analysis Forum » Weighted, constrained least square fitting...
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| rizwindu... |
 Posted: Mon May 19, 2008 1:12 am |
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Guest
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Hi all,
I'm not an expert in linear algebra, so be easy on me. Most of what
I've got so far is from books, but need a bit of help with something.
Attempting to solve least squares problem
Y = Hx
where x is unknowns, Y are observations. Subject to the constraints
Ax = t
where t are constrained values (from Y) and A are the corresponding
rows of H.
I'm also trying to do this weighted, allowing different weights for
different observations in Y. I've got a solution for the constrained
problem and a solution to the weighted problem, but not sure of the
joint solution.
My book has a term Z in the initial problem (Y = Hx + Z), described as
an error term with E(Z) = 0 and E(ZZ') = V, and the term V appears in
the solution. Initially I'm using the identity matrix as V, and it
gives the correct constrained solution. I believe V may allow me to
weight the problem by setting elements on the leading diagional to
different values (i.e. 0 for observations to ignore, 1 for ones to use
and 0.5 for ones to sort of use), but this doesn't give me the results
I'm expecting.
Any ideas ? |
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| Allen McIntosh... |
| Posted: Mon May 19, 2008 6:33 am |
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Guest
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rizwindu wrote:
Quote: Attempting to solve least squares problem
Y = Hx
Ax = t
where t are constrained values (from Y) and A are the corresponding
rows of H.
My book has a term Z in the initial problem (Y = Hx + Z), described as
an error term with E(Z) = 0 and E(ZZ') = V, and the term V appears in
the solution. Initially I'm using the identity matrix as V, and it
gives the correct constrained solution. I believe V may allow me to
weight the problem by setting elements on the leading diagonal to
different values (i.e. 0 for observations to ignore, 1 for ones to use
and 0.5 for ones to sort of use), but this doesn't give me the results
I'm expecting.
Any ideas ?
If you read your book carefully (and if it is any good) it should tell
you to make the variances inversely proportional to the weights. |
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| aruzinsky... |
| Posted: Mon May 19, 2008 7:00 am |
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Guest
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On May 19, 5:12 am, rizwindu <bgeo... at (no spam) gmail.com> wrote:
Quote: Hi all,
I'm not an expert in linear algebra, so be easy on me. Most of what
I've got so far is from books, but need a bit of help with something.
Attempting to solve least squares problem
Y = Hx
where x is unknowns, Y are observations. Subject to the constraints
Ax = t
where t are constrained values (from Y) and A are the corresponding
rows of H.
I'm also trying to do this weighted, allowing different weights for
different observations in Y. I've got a solution for the constrained
problem and a solution to the weighted problem, but not sure of the
joint solution.
My book has a term Z in the initial problem (Y = Hx + Z), described as
an error term with E(Z) = 0 and E(ZZ') = V, and the term V appears in
the solution. Initially I'm using the identity matrix as V, and it
gives the correct constrained solution. I believe V may allow me to
weight the problem by setting elements on the leading diagional to
different values (i.e. 0 for observations to ignore, 1 for ones to use
and 0.5 for ones to sort of use), but this doesn't give me the results
I'm expecting.
Any ideas ?
Expectations "E" have no place in numerical analysis.
You can practically solve the problem with only LS methods because the
constrained portion corresponds to infinite weights. For an
approximate solution, the weights for the constrained part can be made
very large; how large depends on the numerical stability of the
particular LS algorithm used.
solve
D*[ HT AT ]T x = D*[ yT tT ]T
where
D = diag(W^0.5 c*I)
c is as a large constant as possible while maintaining numerical
stability.
W is a diagonal matrix of weights
Of course, this is just multiplying rows by Wii^0.5 and c. |
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| Carl Barron... |
| Posted: Tue May 20, 2008 12:14 am |
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Guest
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rizwindu <bgeorge at (no spam) gmail.com> wrote:
Quote: Hi all,
I'm not an expert in linear algebra, so be easy on me. Most of what
I've got so far is from books, but need a bit of help with something.
Attempting to solve least squares problem
Y = Hx
where x is unknowns, Y are observations. Subject to the constraints
Ax = t
where t are constrained values (from Y) and A are the corresponding
rows of H.
I'm also trying to do this weighted, allowing different weights for
different observations in Y. I've got a solution for the constrained
problem and a solution to the weighted problem, but not sure of the
joint solution.
My book has a term Z in the initial problem (Y = Hx + Z), described as
an error term with E(Z) = 0 and E(ZZ') = V, and the term V appears in
the solution. Initially I'm using the identity matrix as V, and it
gives the correct constrained solution. I believe V may allow me to
weight the problem by setting elements on the leading diagional to
different values (i.e. 0 for observations to ignore, 1 for ones to use
and 0.5 for ones to sort of use), but this doesn't give me the results
I'm expecting.
Any ideas ?
Different approach:
let A^T = Q |R| where Q^TQ = I and R is square upper triangular.
|0|
Q^Tx - |y| dim(y) = dim(R)
|z|
solve Ry = t
MQ = |A1 A2| where A1y + A2z makes sense
we want min||Mx-v|| suject to Ax=t
solve min|| A2z - (b-A1y)|| for z
x = Q|y|
|z| |
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