A Case of Identity Crisis · A Case of Identity Crisis Preserving Identifiability in Linear...
Transcript of A Case of Identity Crisis · A Case of Identity Crisis Preserving Identifiability in Linear...
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A Case of Identity CrisisPreserving Identifiability in Linear Compartment Models
Seth GerberdingDepartment of Mathematical Sciences
University of South Dakota
July 22, 2019
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Plan of Attack
Outline of my talk:Set Up:
1 Linear Compartment Model2 Input-Output Equation3 Identifiability
Results1 Removing the Leak2 Moving the Ouput3 New Models
Fin/ Nemo ModelsWing/ Tweety Bird Models
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It’s a Set Up: Part 1Linear Compartment ModelsComponents of a mode:
CompartmentsInputOutputEdges (Parameters, kijs)Leaks (Optional) (Special kind of parameter, k0j)
k32
kn,n−1
k21
k1n
k43
1
2 3
n
out
in
k01
Cycle(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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It’s a Set Up: Part 1Linear Compartment ModelsComponents of a mode:
CompartmentsInputOutputEdges (Parameters, kijs)Leaks (Optional) (Special kind of parameter, k0j)
k32
kn,n−1
k21
k1n
k43
1
2 3
n
out
in
k01
Cycle(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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It’s a Set Up: Part 1Linear Compartment ModelsComponents of a mode:
CompartmentsInputOutputEdges (Parameters, kijs)Leaks (Optional) (Special kind of parameter, k0j)
k32
kn,n−1
k21
k1n
k43
1
2 3
n
out
in
k01
Cycle(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
![Page 6: A Case of Identity Crisis · A Case of Identity Crisis Preserving Identifiability in Linear Compartment Models Seth Gerberding Department of Mathematical Sciences University of South](https://reader033.fdocuments.us/reader033/viewer/2022060316/5f0c1d337e708231d433cf03/html5/thumbnails/6.jpg)
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It’s a Set Up: Part 1Linear Compartment ModelsComponents of a mode:
CompartmentsInputOutputEdges (Parameters, kijs)Leaks (Optional) (Special kind of parameter, k0j)
k32
kn,n−1
k21
k1n
k43
1
2 3
n
out
in
k01
Cycle(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
![Page 7: A Case of Identity Crisis · A Case of Identity Crisis Preserving Identifiability in Linear Compartment Models Seth Gerberding Department of Mathematical Sciences University of South](https://reader033.fdocuments.us/reader033/viewer/2022060316/5f0c1d337e708231d433cf03/html5/thumbnails/7.jpg)
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It’s a Set Up: Part 1Linear Compartment ModelsComponents of a mode:
CompartmentsInputOutputEdges (Parameters, kijs)Leaks (Optional) (Special kind of parameter, k0j)
k32
kn,n−1
k21
k1n
k43
1
2 3
n
out
in
k01
Cycle(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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It’s still a Set-Up
Input-Output EquationEquation which holds along any solution any solution of the ODE,involving only input and output variablesGeneral Equation:det(∂I −A)yi =
∑j∈In(−1)i+j det(∂I −Aji)uj
Trick lies in computing the determinantsGives us coefficientsDerive a coefficient map
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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It’s still a Set-Up
Input-Output EquationEquation which holds along any solution any solution of the ODE,involving only input and output variablesGeneral Equation:det(∂I −A)yi =
∑j∈In(−1)i+j det(∂I −Aji)uj
Trick lies in computing the determinantsGives us coefficientsDerive a coefficient map
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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It’s still a Set-Up
Input-Output EquationEquation which holds along any solution any solution of the ODE,involving only input and output variablesGeneral Equation:det(∂I −A)yi =
∑j∈In(−1)i+j det(∂I −Aji)uj
Trick lies in computing the determinantsGives us coefficientsDerive a coefficient map
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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It’s still a Set-Up
Input-Output EquationEquation which holds along any solution any solution of the ODE,involving only input and output variablesGeneral Equation:det(∂I −A)yi =
∑j∈In(−1)i+j det(∂I −Aji)uj
Trick lies in computing the determinantsGives us coefficientsDerive a coefficient map
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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It’s totally a Set-Up
Coefficient MapGoes from the parameters to coefficients of input-output equationExample:
c : R5 7→ R5
(k01, k21, k12, k23, k32) 7→(k21 + k32, k32k01, k12k32 + k01k21, k12 + k21, k01k12k21k23, k32) (1)
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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It’s totally a Set-Up
Coefficient MapGoes from the parameters to coefficients of input-output equationExample:
c : R5 7→ R5
(k01, k21, k12, k23, k32) 7→(k21 + k32, k32k01, k12k32 + k01k21, k12 + k21, k01k12k21k23, k32) (1)
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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OMG we get it-it’s a set-up!
IdentifiabilityOnly know 2 things: input and output dataGiven the coefficient values, can we figure out what the parametersare?For example: Given(k21 + k32, k32k01, k12k32 + k01k21, k12 + k21, k01k12k21k23, k32)Can we figure out (k01, k21, k12, k23, k32)?
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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OMG we get it-it’s a set-up!
IdentifiabilityOnly know 2 things: input and output dataGiven the coefficient values, can we figure out what the parametersare?For example: Given(k21 + k32, k32k01, k12k32 + k01k21, k12 + k21, k01k12k21k23, k32)Can we figure out (k01, k21, k12, k23, k32)?
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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OMG we get it-it’s a set-up!
IdentifiabilityOnly know 2 things: input and output dataGiven the coefficient values, can we figure out what the parametersare?For example: Given(k21 + k32, k32k01, k12k32 + k01k21, k12 + k21, k01k12k21k23, k32)Can we figure out (k01, k21, k12, k23, k32)?
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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At this point it’s obviously a set-up
Key tools for identifiability:A model is identifiable iff the Jacobian Matrix of the coefficient maphas full rankFull rank means:
1 Given n parameters...2 There exists a n× n submatrix of the Jacobian matrix...3 such that the determinant of the submatrix is nonzero.
We need only one to be nonzero.
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At this point it’s obviously a set-up
Key tools for identifiability:A model is identifiable iff the Jacobian Matrix of the coefficient maphas full rankFull rank means:
1 Given n parameters...2 There exists a n× n submatrix of the Jacobian matrix...3 such that the determinant of the submatrix is nonzero.
We need only one to be nonzero.
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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At this point it’s obviously a set-up
Key tools for identifiability:A model is identifiable iff the Jacobian Matrix of the coefficient maphas full rankFull rank means:
1 Given n parameters...2 There exists a n× n submatrix of the Jacobian matrix...3 such that the determinant of the submatrix is nonzero.
We need only one to be nonzero.
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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At this point it’s obviously a set-up
Key tools for identifiability:A model is identifiable iff the Jacobian Matrix of the coefficient maphas full rankFull rank means:
1 Given n parameters...2 There exists a n× n submatrix of the Jacobian matrix...3 such that the determinant of the submatrix is nonzero.
We need only one to be nonzero.
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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Example Part 1
X1
X2
X3
X4
k14
k43
k21
k32
in
out
X ′1(t) = −k21x1 + k14x4
X ′2(t) = −k32x2 + k21x1
X ′3(t) = −k43x3 + k32x2
X ′4(t) = −k14x4 + k43x3
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Example Part 1
X1
X2
X3
X4
k14
k43
k21
k32
in
out
X ′1(t) = −k21x1 + k14x4
X ′2(t) = −k32x2 + k21x1
X ′3(t) = −k43x3 + k32x2
X ′4(t) = −k14x4 + k43x3
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Example Part 2
X ′1(t) =− k21x1 + k14x4
X ′2(t) =− k32x2 + k21x1
X ′3(t) =− k43x3 + k32x2
X ′4(t) =− k14x4 + k43x3
A matrix can be built from those ODEs:
A =
−k21 0 0 k14k21 −k32 0 00 k32 −k43 00 0 k43 −k14
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Example Part 2
X ′1(t) =− k21x1 + k14x4
X ′2(t) =− k32x2 + k21x1
X ′3(t) =− k43x3 + k32x2
X ′4(t) =− k14x4 + k43x3
A matrix can be built from those ODEs:
A =
−k21 0 0 k14k21 −k32 0 00 k32 −k43 00 0 k43 −k14
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Example Part 3
The input-out equation for this model is:det(∂I −A)y1 = det(∂I −A)11u1
The (∂I −A) matrix is:
(∂I −A) =
ddt + k21 0 0 −k14−k21
ddt + k32 0 0
0 −k32ddt + k43 0
0 0 −k43ddt + k14
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Example Part 4
1 From the input-output equation, derive a coefficient map
(k21, ..., k1n) 7−→ (c1, c2, ...)
2 Take the Jacobian matrix of the coefficient map3 If the Jacobian matrix is full rank, then the model is identifiable4 The example I showed was identifiable
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Example Part 4
1 From the input-output equation, derive a coefficient map
(k21, ..., k1n) 7−→ (c1, c2, ...)
2 Take the Jacobian matrix of the coefficient map3 If the Jacobian matrix is full rank, then the model is identifiable4 The example I showed was identifiable
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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Example Part 4
1 From the input-output equation, derive a coefficient map
(k21, ..., k1n) 7−→ (c1, c2, ...)
2 Take the Jacobian matrix of the coefficient map3 If the Jacobian matrix is full rank, then the model is identifiable4 The example I showed was identifiable
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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Pause...
Any questions about the Set-Up?
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Results: Finding identity
Three main results:Removing the LeakMoving the output in cycle modelsNew Models
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A Brief Digression
A note on elementary symmetric polynomialsEssential for proving the resultsGiven a set X = {x1, ..., xn}The mth elementary symmetric polynomial is:
em =∑
j1<j2<···<jm
xj1 ...xjm
Easier with an example
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A Brief Digression
A note on elementary symmetric polynomialsEssential for proving the resultsGiven a set X = {x1, ..., xn}The mth elementary symmetric polynomial is:
em =∑
j1<j2<···<jm
xj1 ...xjm
Easier with an example
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A Brief Digression
A note on elementary symmetric polynomialsEssential for proving the resultsGiven a set X = {x1, ..., xn}The mth elementary symmetric polynomial is:
em =∑
j1<j2<···<jm
xj1 ...xjm
Easier with an example
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It’s elementary, my dear Watson
Say X = {x1, x2, x3}. The Elementary symmetric polynomials on X are:e0 = 1
e1 = x1 + x2 + x3
e2 = x1x2 + x1x3 + x2x3
e3 = x1x2x3
IMPORTANT PROPERTY
∂em∂xi
=∑
j2<...<jm
xj2 ...xjm =: em−1{xi}
Elementary symmetric polynomials are linearly independent
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It’s elementary, my dear Watson
Say X = {x1, x2, x3}. The Elementary symmetric polynomials on X are:e0 = 1
e1 = x1 + x2 + x3
e2 = x1x2 + x1x3 + x2x3
e3 = x1x2x3
IMPORTANT PROPERTY
∂em∂xi
=∑
j2<...<jm
xj2 ...xjm =: em−1{xi}
Elementary symmetric polynomials are linearly independent
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It’s elementary, my dear Watson
Say X = {x1, x2, x3}. The Elementary symmetric polynomials on X are:e0 = 1
e1 = x1 + x2 + x3
e2 = x1x2 + x1x3 + x2x3
e3 = x1x2x3
IMPORTANT PROPERTY
∂em∂xi
=∑
j2<...<jm
xj2 ...xjm =: em−1{xi}
Elementary symmetric polynomials are linearly independent
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It’s elementary, my dear Watson
Say X = {x1, x2, x3}. The Elementary symmetric polynomials on X are:e0 = 1
e1 = x1 + x2 + x3
e2 = x1x2 + x1x3 + x2x3
e3 = x1x2x3
IMPORTANT PROPERTY
∂em∂xi
=∑
j2<...<jm
xj2 ...xjm =: em−1{xi}
Elementary symmetric polynomials are linearly independent
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It’s elementary, my dear Watson
Say X = {x1, x2, x3}. The Elementary symmetric polynomials on X are:e0 = 1
e1 = x1 + x2 + x3
e2 = x1x2 + x1x3 + x2x3
e3 = x1x2x3
IMPORTANT PROPERTY
∂em∂xi
=∑
j2<...<jm
xj2 ...xjm =: em−1{xi}
Elementary symmetric polynomials are linearly independent
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Removing the Leak
Given an identifiable model with a leak, does removing the leak preserveidentifiability?YES! for certain models (cycle, catenary, mammillary)
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Removing the Leak
Given an identifiable model with a leak, does removing the leak preserveidentifiability?YES! for certain models (cycle, catenary, mammillary)
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Basic Proof
TheoremLet M be a catenary, cycle, or mammillary model that has at least oneinput and exactly one leak. If M is generically locally identifiable from thecoefficient map, then so is the model M obtained from M by removing theleak.
Proof.Proposition 4.7 from Gross, Harrington, Meshkat, and Shiu (2019) statescatenary, cycle, and mammillary models, with no leaks, are locallyidentifiable from the coefficient map. Then, by Theorem 4.3 from Grosset. al., adding a leak preserves identifiability. Thus, both M and M aregenerically locally identifiable.
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Basic Proof
TheoremLet M be a catenary, cycle, or mammillary model that has at least oneinput and exactly one leak. If M is generically locally identifiable from thecoefficient map, then so is the model M obtained from M by removing theleak.
Proof.Proposition 4.7 from Gross, Harrington, Meshkat, and Shiu (2019) statescatenary, cycle, and mammillary models, with no leaks, are locallyidentifiable from the coefficient map. Then, by Theorem 4.3 from Grosset. al., adding a leak preserves identifiability. Thus, both M and M aregenerically locally identifiable.
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Moving the Output
The Question:In a cycle model, does removing the output preserve identifiability?
For example:Output is located in compartment 1...
k32
kn,n−1
k21
k1n
k43
1
2 3
n
out
in
Cycle
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Moving the Output
The Question:In a cycle model, does removing the output preserve identifiability?
... moved to compartment 3
k32
kn,n−1
k21
k1n
k43
1
2 3
n
out
in
Cycle
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Solution to output distribution
Key takeaways: Let p be the output compartment.Input-ouput equation:(
ddt
n(e0) +
ddt
n−1e1 + · · ·+ d
dten−1
)yn =(∏n+1
i=p+1 ki,i−1
(ddt
p−2e∗0 +
ddt
p−3e∗1 + ...+ e∗p−2)
))u1
Coefficient Map:
c : Rn → Rn+p−2
where
(k21, ..., k1n) 7−→ (e1, ..., en−1,
n+1∏i=p+1
ki,i−1e∗0, ...,
n+1∏i=p+1
ki,i−1e∗p−2)
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Solution to output distribution
Key takeaways: Let p be the output compartment.Input-ouput equation:(
ddt
n(e0) +
ddt
n−1e1 + · · ·+ d
dten−1
)yn =(∏n+1
i=p+1 ki,i−1
(ddt
p−2e∗0 +
ddt
p−3e∗1 + ...+ e∗p−2)
))u1
Coefficient Map:
c : Rn → Rn+p−2
where
(k21, ..., k1n) 7−→ (e1, ..., en−1,
n+1∏i=p+1
ki,i−1e∗0, ...,
n+1∏i=p+1
ki,i−1e∗p−2)
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Solution to output distribution pt. 2
Selected submatrix of the Jacobian:
J =
1 1 . . . 1 . . . 1
e1{k21} e1{k32} . . . e1{kp+1,p} . . . e1{k1n}...
. . ....
en−2{k21} en−2{k32} . . . en−2{kp+1,p} . . . en−2k1n}0 0 . . . en−p+1{kp+1,p} . . . en−p+1{k1n}
Determinant is nonzeroSince determinant is nonzero, model is generically locally identifiable!
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Solution to output distribution pt. 2
Selected submatrix of the Jacobian:
J =
1 1 . . . 1 . . . 1
e1{k21} e1{k32} . . . e1{kp+1,p} . . . e1{k1n}...
. . ....
en−2{k21} en−2{k32} . . . en−2{kp+1,p} . . . en−2k1n}0 0 . . . en−p+1{kp+1,p} . . . en−p+1{k1n}
Determinant is nonzeroSince determinant is nonzero, model is generically locally identifiable!
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Solution to output distribution pt. 2
Selected submatrix of the Jacobian:
J =
1 1 . . . 1 . . . 1
e1{k21} e1{k32} . . . e1{kp+1,p} . . . e1{k1n}...
. . ....
en−2{k21} en−2{k32} . . . en−2{kp+1,p} . . . en−2k1n}0 0 . . . en−p+1{kp+1,p} . . . en−p+1{k1n}
Determinant is nonzeroSince determinant is nonzero, model is generically locally identifiable!
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Introducing, New Models!
A new ”family” of modelsFin ModelNemo ModelWing ModelTweety Bird Model
Nemo models derived from Fin modelsTweety Bird models derived from Wing Models
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Introducing, New Models!
A new ”family” of modelsFin ModelNemo ModelWing ModelTweety Bird Model
Nemo models derived from Fin modelsTweety Bird models derived from Wing Models
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Introducing, New Models!
A new ”family” of modelsFin ModelNemo ModelWing ModelTweety Bird Model
Nemo models derived from Fin modelsTweety Bird models derived from Wing Models
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Fin and Nemo Models
k32
kn,n−1
k21
k1n
k43
k14
k13k12
1
2 3
n
4
out
in
Fin
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Fin and Nemo Models
k32
kn,n−1
k21
k1n
k43
k141
2 3
n
4
out
in
Nemo
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Finding Nemo: an existential crisis
How to find Nemo:By showing Fin models are identifiable, we can show Nemo models areidentifiable.
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Finding Nemo: The Lucky Fin
For a Fin Model:Input-Output Equation... Too long for this slide...Coefficient Map
c : R2n−2 → R2n−1
such that
(k12, ..., k1n, k21, ..., kn,n−1) 7−→
(e′1, ..., e′n−1, e
∗1, e
∗2+
2∑i=2
Piei2−i, ..., e
∗j+
j∑i=2
Pieij−i, ..., e
∗n+
n∑i=2
Piein−i)
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Finding Nemo: The Lucky Fin
For a Fin Model:Input-Output Equation... Too long for this slide...Coefficient Map
c : R2n−2 → R2n−1
such that
(k12, ..., k1n, k21, ..., kn,n−1) 7−→
(e′1, ..., e′n−1, e
∗1, e
∗2+
2∑i=2
Piei2−i, ..., e
∗j+
j∑i=2
Pieij−i, ..., e
∗n+
n∑i=2
Piein−i)
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Finding Nemo: The Lucky Fin
For a Fin Model:Input-Output Equation... Too long for this slide...Coefficient Map
c : R2n−2 → R2n−1
such that
(k12, ..., k1n, k21, ..., kn,n−1) 7−→
(e′1, ..., e′n−1, e
∗1, e
∗2+
2∑i=2
Piei2−i, ..., e
∗j+
j∑i=2
Pieij−i, ..., e
∗n+
n∑i=2
Piein−i)
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Finding Nemo: The Lucky Fin, Chapter 2
Selected Submatrix of the coefficient map Jacobian:New proof approachBreak into “block” matrix
J(c) =
[W XY Z
]
Only need to show W and Z have nonzero determinants
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Finding Nemo: The Lucky Fin, Chapter 2
Selected Submatrix of the coefficient map Jacobian:New proof approachBreak into “block” matrix
J(c) =
[W XY Z
]
Only need to show W and Z have nonzero determinants
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Finding Nemo: The Lucky Fin, Chapter 2, pt. 2
W =
e∗n−1{k21}+ βn
2 e∗n−1{k32}+ βn3 . . . e∗n−1 + αn−3
2 + βnn−1
0 e′1{k32} . . . e′1{kn,n−1}...
0 e′n−2{k32} . . . e′n−2{kn,n−1}
det(W ) = 0
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The Lucky Fin, Chapter 2, pt. 2, scene 2
Z =
1 0 0 . . . 0
e∗1{k1n} γ22 0 . . . 0σ3 γ23 γ33 . . . 0...
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σj . . . γjj . . . 0...
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σn−1 γ2n−1 . . . γjn−1 . . . γn−1n−1
det(Z) = 0
Thus, det(J)= 0, and the model is generically locally identifiable
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Finding Nemo: Just keep swimming
Technique:When you remove those edges...Coefficient map slightly changesThe W submatrix is essentially unchangedOnly Z has to be really addressed
Remove the column corresponding to the removed parameterRemove the row corresponding to the same parameter
While Z is smaller, it still has a nonzero determinant
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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Finding Nemo: Just keep swimming
Technique:When you remove those edges...Coefficient map slightly changesThe W submatrix is essentially unchangedOnly Z has to be really addressed
Remove the column corresponding to the removed parameterRemove the row corresponding to the same parameter
While Z is smaller, it still has a nonzero determinant
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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Finding Nemo: Just keep swimming
Technique:When you remove those edges...Coefficient map slightly changesThe W submatrix is essentially unchangedOnly Z has to be really addressed
Remove the column corresponding to the removed parameterRemove the row corresponding to the same parameter
While Z is smaller, it still has a nonzero determinant
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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Finding Nemo: Just keep swimming
Technique:When you remove those edges...Coefficient map slightly changesThe W submatrix is essentially unchangedOnly Z has to be really addressed
Remove the column corresponding to the removed parameterRemove the row corresponding to the same parameter
While Z is smaller, it still has a nonzero determinant
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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Finding Nemo: Just keep swimming
Technique:When you remove those edges...Coefficient map slightly changesThe W submatrix is essentially unchangedOnly Z has to be really addressed
Remove the column corresponding to the removed parameterRemove the row corresponding to the same parameter
While Z is smaller, it still has a nonzero determinant
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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Finding Nemo: Just keep swimming
Technique:When you remove those edges...Coefficient map slightly changesThe W submatrix is essentially unchangedOnly Z has to be really addressed
Remove the column corresponding to the removed parameterRemove the row corresponding to the same parameter
While Z is smaller, it still has a nonzero determinant
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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Nemo Found!
A little more explanation
Z =
1 0 0 . . . 0
e∗1{k1n} γ22 0 . . . 0σ3 γ23 γ33 . . . 0...
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σj . . . γjj . . . 0...
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σn−1 γ2n−1 . . . γjn−1 . . . γn−1n−1
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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Wing and Tweety-Bird models
k32
kn,n−1
k21
k1n
k43
k41
k31
kn1
1
2 3
n
4
out
in
Wing
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Wing and Tweety Bird Models
k32
kn,n−1
k21
k1n
k43
kn1
1
2 3
n
4
out
in
Tweety Bird
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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Finding Tweety Bird
Technique:Same idea as Fin/Nemo modelCoefficients are different...But the proof is similar enoughBoth models are identifiable
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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Finding Tweety Bird
Technique:Same idea as Fin/Nemo modelCoefficients are different...But the proof is similar enoughBoth models are identifiable
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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Finding Tweety Bird
Technique:Same idea as Fin/Nemo modelCoefficients are different...But the proof is similar enoughBoth models are identifiable
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
![Page 75: A Case of Identity Crisis · A Case of Identity Crisis Preserving Identifiability in Linear Compartment Models Seth Gerberding Department of Mathematical Sciences University of South](https://reader033.fdocuments.us/reader033/viewer/2022060316/5f0c1d337e708231d433cf03/html5/thumbnails/75.jpg)
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Finding Tweety Bird
Technique:Same idea as Fin/Nemo modelCoefficients are different...But the proof is similar enoughBoth models are identifiable
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019
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Acknowledgments
Acknowledgments
Sincere thanks to Dr. Anne Shiu, Nida Obatake, Thomas Yahl, and TexasA&M. It’s been a pleasure.
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The End
(S. Gerberding, USD) A Case of Identity Crisis July 22, 2019