About consistency in an inverse problem formulation
I'm a beginner with inverse problems and I was reading about regularization techniques.
Consider the problem:
$$d=Kf_{text{true}}$$
$d$ is a data vector, $K$ is an linear operator, $d=hat{d}+eta$ where $eta$ is the noise and $delta=||eta||$ (noise level).
In regularization by filtering (linear problems), Vogel (Computational Methods for Inverse Problems) defines the error as the error due to truncation, plus the error due to noise:
$$e_{alpha} = e^{text{truc}}_{alpha} +e^{text{noise}}_{alpha}$$
and in a model for an inverse problem, is important to choose $alpha$ such that:
$$e^{text{truc}}_{alpha} to 0, e^{text{noise}}_{alpha} to 0 (star)$$
when $deltato 0$. This is a kind of consistency of the method.
A real problem does not satisfy the $(star)$ condition, but the model of the problem needs to satisfy it, at least in agreement with my interpretation.
Now, here is my question: why is important that a classical
regularization model of an inverse problem satisfies $(star)$?
For example, I've heard that in statistics crooss validation does not satisfy $(star)$, and some statisticians said that is not really important because a real problem does not satisfy $(star)$.
My intention is not to create a discussion, but rather have a proper justification of why consistency is necessary or not in the modeling of an inverse problem.
Thank you very much.
statistics regularization inverse-problems
edited 16 hours ago
amWhy
192k28224439
asked Feb 11 '15 at 18:35
Hiperion
91311028
add a comment |
I'm a beginner with inverse problems and I was reading about regularization techniques.
Consider the problem:
$$d=Kf_{text{true}}$$
$d$ is a data vector, $K$ is an linear operator, $d=hat{d}+eta$ where $eta$ is the noise and $delta=||eta||$ (noise level).
In regularization by filtering (linear problems), Vogel (Computational Methods for Inverse Problems) defines the error as the error due to truncation, plus the error due to noise:
$$e_{alpha} = e^{text{truc}}_{alpha} +e^{text{noise}}_{alpha}$$
and in a model for an inverse problem, is important to choose $alpha$ such that:
$$e^{text{truc}}_{alpha} to 0, e^{text{noise}}_{alpha} to 0 (star)$$
when $deltato 0$. This is a kind of consistency of the method.
A real problem does not satisfy the $(star)$ condition, but the model of the problem needs to satisfy it, at least in agreement with my interpretation.
Now, here is my question: why is important that a classical
regularization model of an inverse problem satisfies $(star)$?
For example, I've heard that in statistics crooss validation does not satisfy $(star)$, and some statisticians said that is not really important because a real problem does not satisfy $(star)$.
My intention is not to create a discussion, but rather have a proper justification of why consistency is necessary or not in the modeling of an inverse problem.
Thank you very much.
statistics regularization inverse-problems
edited 16 hours ago
amWhy
192k28224439
asked Feb 11 '15 at 18:35
Hiperion
91311028
add a comment |
I'm a beginner with inverse problems and I was reading about regularization techniques.
Consider the problem:
$$d=Kf_{text{true}}$$
$d$ is a data vector, $K$ is an linear operator, $d=hat{d}+eta$ where $eta$ is the noise and $delta=||eta||$ (noise level).
In regularization by filtering (linear problems), Vogel (Computational Methods for Inverse Problems) defines the error as the error due to truncation, plus the error due to noise:
$$e_{alpha} = e^{text{truc}}_{alpha} +e^{text{noise}}_{alpha}$$
and in a model for an inverse problem, is important to choose $alpha$ such that:
$$e^{text{truc}}_{alpha} to 0, e^{text{noise}}_{alpha} to 0 (star)$$
when $deltato 0$. This is a kind of consistency of the method.
A real problem does not satisfy the $(star)$ condition, but the model of the problem needs to satisfy it, at least in agreement with my interpretation.
Now, here is my question: why is important that a classical
regularization model of an inverse problem satisfies $(star)$?
For example, I've heard that in statistics crooss validation does not satisfy $(star)$, and some statisticians said that is not really important because a real problem does not satisfy $(star)$.
My intention is not to create a discussion, but rather have a proper justification of why consistency is necessary or not in the modeling of an inverse problem.
Thank you very much.
statistics regularization inverse-problems
edited 16 hours ago
amWhy
192k28224439
asked Feb 11 '15 at 18:35
Hiperion
91311028
I'm a beginner with inverse problems and I was reading about regularization techniques.
Consider the problem:
$$d=Kf_{text{true}}$$
$d$ is a data vector, $K$ is an linear operator, $d=hat{d}+eta$ where $eta$ is the noise and $delta=||eta||$ (noise level).
In regularization by filtering (linear problems), Vogel (Computational Methods for Inverse Problems) defines the error as the error due to truncation, plus the error due to noise:
$$e_{alpha} = e^{text{truc}}_{alpha} +e^{text{noise}}_{alpha}$$
and in a model for an inverse problem, is important to choose $alpha$ such that:
$$e^{text{truc}}_{alpha} to 0, e^{text{noise}}_{alpha} to 0 (star)$$
when $deltato 0$. This is a kind of consistency of the method.
A real problem does not satisfy the $(star)$ condition, but the model of the problem needs to satisfy it, at least in agreement with my interpretation.
Now, here is my question: why is important that a classical
regularization model of an inverse problem satisfies $(star)$?
For example, I've heard that in statistics crooss validation does not satisfy $(star)$, and some statisticians said that is not really important because a real problem does not satisfy $(star)$.
My intention is not to create a discussion, but rather have a proper justification of why consistency is necessary or not in the modeling of an inverse problem.
Thank you very much.
statistics regularization inverse-problems
statistics regularization inverse-problems
edited 16 hours ago
amWhy
192k28224439
asked Feb 11 '15 at 18:35
Hiperion
91311028
edited 16 hours ago
amWhy
192k28224439
asked Feb 11 '15 at 18:35
Hiperion
91311028
edited 16 hours ago
amWhy
192k28224439
edited 16 hours ago
amWhy
192k28224439
edited 16 hours ago
amWhy
192k28224439
192k28224439
asked Feb 11 '15 at 18:35
Hiperion
91311028
asked Feb 11 '15 at 18:35
Hiperion
91311028
asked Feb 11 '15 at 18:35
Hiperion
91311028
91311028
add a comment |
add a comment |
1 Answer
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One proves as much as is possible. The ideal thing to prove is a theorem which gives an explicit upper bound on the error in terms of the noise level in some norm. The condition you mention is a weaker version of this; it states that the error vanishes as the noise does, but does not given an explicit bound.
The condition gives some faith in the method being reasonable. If one has other mathematical or practical justifications for the method being reasonable, maybe one can ignore this one.
answered 17 hours ago
Tommi Brander
956922
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
votes
One proves as much as is possible. The ideal thing to prove is a theorem which gives an explicit upper bound on the error in terms of the noise level in some norm. The condition you mention is a weaker version of this; it states that the error vanishes as the noise does, but does not given an explicit bound.
The condition gives some faith in the method being reasonable. If one has other mathematical or practical justifications for the method being reasonable, maybe one can ignore this one.
answered 17 hours ago
Tommi Brander
956922
add a comment |
One proves as much as is possible. The ideal thing to prove is a theorem which gives an explicit upper bound on the error in terms of the noise level in some norm. The condition you mention is a weaker version of this; it states that the error vanishes as the noise does, but does not given an explicit bound.
The condition gives some faith in the method being reasonable. If one has other mathematical or practical justifications for the method being reasonable, maybe one can ignore this one.
answered 17 hours ago
Tommi Brander
956922
add a comment |
One proves as much as is possible. The ideal thing to prove is a theorem which gives an explicit upper bound on the error in terms of the noise level in some norm. The condition you mention is a weaker version of this; it states that the error vanishes as the noise does, but does not given an explicit bound.
The condition gives some faith in the method being reasonable. If one has other mathematical or practical justifications for the method being reasonable, maybe one can ignore this one.
answered 17 hours ago
Tommi Brander
956922
One proves as much as is possible. The ideal thing to prove is a theorem which gives an explicit upper bound on the error in terms of the noise level in some norm. The condition you mention is a weaker version of this; it states that the error vanishes as the noise does, but does not given an explicit bound.
The condition gives some faith in the method being reasonable. If one has other mathematical or practical justifications for the method being reasonable, maybe one can ignore this one.
answered 17 hours ago
Tommi Brander
956922
answered 17 hours ago
Tommi Brander
956922
answered 17 hours ago
Tommi Brander
956922
answered 17 hours ago
Tommi Brander
956922
956922
add a comment |
add a comment |
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