sample covariance matrix
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Suppose two covariance function estimators, with the same formula except for a coefficient. Then make two sample covariance matrix(SCM) from each of the functions. Why should these matrices differ in negativeness or positiveness feature? To be clear $R=[R_{i-j}]$ should be non-negative definite(positive semi-definite) but $tilde{R} =[tilde R_{i-j}]$ is not necessarily non-negative definite.
$$y(t)=phi(t)theta$$
$$R_{k}=frac{1}{N}sum_{t=1}^{N-k}y(t)y^T(t+k)quad R_{-k}=R_k^Tquad kge 0$$
$$tilde{R}_{k}=frac{1}{N-k}sum_{t=1}^{N-k}y(t)y^T(t+k)quad tilde R_{-k}=tilde R_k^Tquad kge 0$$
The SCM is built by the following formulla.
$$R=begin{bmatrix}R_0&R_1&cdots&R_{N-k}\
R_1^T&R_0&cdots& R_{N-k-1} \
vdots\
R_{N-k}^T&cdots&&R_0 end{bmatrix}$$
They are used in system identification.
main problem
matrices covariance system-identification
asked Nov 17 at 16:36
V.Ajall
74
|
show 3 more comments
up vote
-1
down vote
favorite
Suppose two covariance function estimators, with the same formula except for a coefficient. Then make two sample covariance matrix(SCM) from each of the functions. Why should these matrices differ in negativeness or positiveness feature? To be clear $R=[R_{i-j}]$ should be non-negative definite(positive semi-definite) but $tilde{R} =[tilde R_{i-j}]$ is not necessarily non-negative definite.
$$y(t)=phi(t)theta$$
$$R_{k}=frac{1}{N}sum_{t=1}^{N-k}y(t)y^T(t+k)quad R_{-k}=R_k^Tquad kge 0$$
$$tilde{R}_{k}=frac{1}{N-k}sum_{t=1}^{N-k}y(t)y^T(t+k)quad tilde R_{-k}=tilde R_k^Tquad kge 0$$
The SCM is built by the following formulla.
$$R=begin{bmatrix}R_0&R_1&cdots&R_{N-k}\
R_1^T&R_0&cdots& R_{N-k-1} \
vdots\
R_{N-k}^T&cdots&&R_0 end{bmatrix}$$
They are used in system identification.
main problem
matrices covariance system-identification
asked Nov 17 at 16:36
V.Ajall
74
What is the matrix here to be clear ? Please make your question non ambiguous, thanks!
– Ezy
Nov 17 at 17:07
@Ezy Clear enough?
– V.Ajall
Nov 18 at 7:14
Much better! Now i understand the question :)
– Ezy
Nov 18 at 12:15
And R is dimension NxN ? In other words you let k run up to N-1 ?
– Ezy
Nov 18 at 12:19
y is $ntimes 1$ so R is $nNtimes nN$ and yes we let k run up to $N-1$
– V.Ajall
Nov 18 at 12:45
|
show 3 more comments
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Suppose two covariance function estimators, with the same formula except for a coefficient. Then make two sample covariance matrix(SCM) from each of the functions. Why should these matrices differ in negativeness or positiveness feature? To be clear $R=[R_{i-j}]$ should be non-negative definite(positive semi-definite) but $tilde{R} =[tilde R_{i-j}]$ is not necessarily non-negative definite.
$$y(t)=phi(t)theta$$
$$R_{k}=frac{1}{N}sum_{t=1}^{N-k}y(t)y^T(t+k)quad R_{-k}=R_k^Tquad kge 0$$
$$tilde{R}_{k}=frac{1}{N-k}sum_{t=1}^{N-k}y(t)y^T(t+k)quad tilde R_{-k}=tilde R_k^Tquad kge 0$$
The SCM is built by the following formulla.
$$R=begin{bmatrix}R_0&R_1&cdots&R_{N-k}\
R_1^T&R_0&cdots& R_{N-k-1} \
vdots\
R_{N-k}^T&cdots&&R_0 end{bmatrix}$$
They are used in system identification.
main problem
matrices covariance system-identification
asked Nov 17 at 16:36
V.Ajall
74
Suppose two covariance function estimators, with the same formula except for a coefficient. Then make two sample covariance matrix(SCM) from each of the functions. Why should these matrices differ in negativeness or positiveness feature? To be clear $R=[R_{i-j}]$ should be non-negative definite(positive semi-definite) but $tilde{R} =[tilde R_{i-j}]$ is not necessarily non-negative definite.
$$y(t)=phi(t)theta$$
$$R_{k}=frac{1}{N}sum_{t=1}^{N-k}y(t)y^T(t+k)quad R_{-k}=R_k^Tquad kge 0$$
$$tilde{R}_{k}=frac{1}{N-k}sum_{t=1}^{N-k}y(t)y^T(t+k)quad tilde R_{-k}=tilde R_k^Tquad kge 0$$
The SCM is built by the following formulla.
$$R=begin{bmatrix}R_0&R_1&cdots&R_{N-k}\
R_1^T&R_0&cdots& R_{N-k-1} \
vdots\
R_{N-k}^T&cdots&&R_0 end{bmatrix}$$
They are used in system identification.
main problem
matrices covariance system-identification
matrices covariance system-identification
asked Nov 17 at 16:36
V.Ajall
74
asked Nov 17 at 16:36
V.Ajall
74
edited Nov 18 at 15:18
asked Nov 17 at 16:36
V.Ajall
74
asked Nov 17 at 16:36
V.Ajall
74
asked Nov 17 at 16:36
V.Ajall
74
74
What is the matrix here to be clear ? Please make your question non ambiguous, thanks!
– Ezy
Nov 17 at 17:07
@Ezy Clear enough?
– V.Ajall
Nov 18 at 7:14
Much better! Now i understand the question :)
– Ezy
Nov 18 at 12:15
And R is dimension NxN ? In other words you let k run up to N-1 ?
– Ezy
Nov 18 at 12:19
y is $ntimes 1$ so R is $nNtimes nN$ and yes we let k run up to $N-1$
– V.Ajall
Nov 18 at 12:45
|
show 3 more comments
What is the matrix here to be clear ? Please make your question non ambiguous, thanks!
– Ezy
Nov 17 at 17:07
@Ezy Clear enough?
– V.Ajall
Nov 18 at 7:14
Much better! Now i understand the question :)
– Ezy
Nov 18 at 12:15
And R is dimension NxN ? In other words you let k run up to N-1 ?
– Ezy
Nov 18 at 12:19
y is $ntimes 1$ so R is $nNtimes nN$ and yes we let k run up to $N-1$
– V.Ajall
Nov 18 at 12:45
What is the matrix here to be clear ? Please make your question non ambiguous, thanks!
– Ezy
Nov 17 at 17:07
What is the matrix here to be clear ? Please make your question non ambiguous, thanks!
– Ezy
Nov 17 at 17:07
@Ezy Clear enough?
– V.Ajall
Nov 18 at 7:14
@Ezy Clear enough?
– V.Ajall
Nov 18 at 7:14
Much better! Now i understand the question :)
– Ezy
Nov 18 at 12:15
Much better! Now i understand the question :)
– Ezy
Nov 18 at 12:15
And R is dimension NxN ? In other words you let k run up to N-1 ?
– Ezy
Nov 18 at 12:19
And R is dimension NxN ? In other words you let k run up to N-1 ?
– Ezy
Nov 18 at 12:19
y is $ntimes 1$ so R is $nNtimes nN$ and yes we let k run up to $N-1$
– V.Ajall
Nov 18 at 12:45
y is $ntimes 1$ so R is $nNtimes nN$ and yes we let k run up to $N-1$
– V.Ajall
Nov 18 at 12:45
|
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
0
down vote
For $N=2$ and $y(0)=(1,-1)/sqrt{2}$ and $y(1)=(-1,-1)/sqrt{2}$ you see that the first matrix has determinant 0 while the second one has determinant $>0$ so they have different signature.
answered Nov 18 at 12:23
Ezy
54429
please check the changes I have made in the question.
– V.Ajall
Nov 18 at 12:46
I updated my answer. Going forward you need to make questions non ambiguous
– Ezy
Nov 18 at 13:01
Thank you but I am looking for the proof of first one not an example only! :)
– V.Ajall
Nov 18 at 13:04
There is no proof because the claim is wrong. I gave counter example
– Ezy
Nov 18 at 13:05
Seems that your counter example is wrong since $R_0$ has determinant 0.5 and $R_1$ has determinant 0 and matrix $R$ is nonnegative which is not contradict what has mentioned in the stated problem.
– V.Ajall
Nov 18 at 13:34
|
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
For $N=2$ and $y(0)=(1,-1)/sqrt{2}$ and $y(1)=(-1,-1)/sqrt{2}$ you see that the first matrix has determinant 0 while the second one has determinant $>0$ so they have different signature.
answered Nov 18 at 12:23
Ezy
54429
please check the changes I have made in the question.
– V.Ajall
Nov 18 at 12:46
I updated my answer. Going forward you need to make questions non ambiguous
– Ezy
Nov 18 at 13:01
Thank you but I am looking for the proof of first one not an example only! :)
– V.Ajall
Nov 18 at 13:04
There is no proof because the claim is wrong. I gave counter example
– Ezy
Nov 18 at 13:05
Seems that your counter example is wrong since $R_0$ has determinant 0.5 and $R_1$ has determinant 0 and matrix $R$ is nonnegative which is not contradict what has mentioned in the stated problem.
– V.Ajall
Nov 18 at 13:34
|
show 4 more comments
up vote
0
down vote
For $N=2$ and $y(0)=(1,-1)/sqrt{2}$ and $y(1)=(-1,-1)/sqrt{2}$ you see that the first matrix has determinant 0 while the second one has determinant $>0$ so they have different signature.
answered Nov 18 at 12:23
Ezy
54429
please check the changes I have made in the question.
– V.Ajall
Nov 18 at 12:46
I updated my answer. Going forward you need to make questions non ambiguous
– Ezy
Nov 18 at 13:01
Thank you but I am looking for the proof of first one not an example only! :)
– V.Ajall
Nov 18 at 13:04
There is no proof because the claim is wrong. I gave counter example
– Ezy
Nov 18 at 13:05
Seems that your counter example is wrong since $R_0$ has determinant 0.5 and $R_1$ has determinant 0 and matrix $R$ is nonnegative which is not contradict what has mentioned in the stated problem.
– V.Ajall
Nov 18 at 13:34
|
show 4 more comments
up vote
0
down vote
up vote
0
down vote
For $N=2$ and $y(0)=(1,-1)/sqrt{2}$ and $y(1)=(-1,-1)/sqrt{2}$ you see that the first matrix has determinant 0 while the second one has determinant $>0$ so they have different signature.
answered Nov 18 at 12:23
Ezy
54429
For $N=2$ and $y(0)=(1,-1)/sqrt{2}$ and $y(1)=(-1,-1)/sqrt{2}$ you see that the first matrix has determinant 0 while the second one has determinant $>0$ so they have different signature.
answered Nov 18 at 12:23
Ezy
54429
edited Nov 18 at 13:01
answered Nov 18 at 12:23
Ezy
54429
answered Nov 18 at 12:23
Ezy
54429
answered Nov 18 at 12:23
Ezy
54429
54429
please check the changes I have made in the question.
– V.Ajall
Nov 18 at 12:46
I updated my answer. Going forward you need to make questions non ambiguous
– Ezy
Nov 18 at 13:01
Thank you but I am looking for the proof of first one not an example only! :)
– V.Ajall
Nov 18 at 13:04
There is no proof because the claim is wrong. I gave counter example
– Ezy
Nov 18 at 13:05
Seems that your counter example is wrong since $R_0$ has determinant 0.5 and $R_1$ has determinant 0 and matrix $R$ is nonnegative which is not contradict what has mentioned in the stated problem.
– V.Ajall
Nov 18 at 13:34
|
show 4 more comments
please check the changes I have made in the question.
– V.Ajall
Nov 18 at 12:46
I updated my answer. Going forward you need to make questions non ambiguous
– Ezy
Nov 18 at 13:01
Thank you but I am looking for the proof of first one not an example only! :)
– V.Ajall
Nov 18 at 13:04
There is no proof because the claim is wrong. I gave counter example
– Ezy
Nov 18 at 13:05
Seems that your counter example is wrong since $R_0$ has determinant 0.5 and $R_1$ has determinant 0 and matrix $R$ is nonnegative which is not contradict what has mentioned in the stated problem.
– V.Ajall
Nov 18 at 13:34
please check the changes I have made in the question.
– V.Ajall
Nov 18 at 12:46
please check the changes I have made in the question.
– V.Ajall
Nov 18 at 12:46
I updated my answer. Going forward you need to make questions non ambiguous
– Ezy
Nov 18 at 13:01
I updated my answer. Going forward you need to make questions non ambiguous
– Ezy
Nov 18 at 13:01
Thank you but I am looking for the proof of first one not an example only! :)
– V.Ajall
Nov 18 at 13:04
Thank you but I am looking for the proof of first one not an example only! :)
– V.Ajall
Nov 18 at 13:04
There is no proof because the claim is wrong. I gave counter example
– Ezy
Nov 18 at 13:05
There is no proof because the claim is wrong. I gave counter example
– Ezy
Nov 18 at 13:05
Seems that your counter example is wrong since $R_0$ has determinant 0.5 and $R_1$ has determinant 0 and matrix $R$ is nonnegative which is not contradict what has mentioned in the stated problem.
– V.Ajall
Nov 18 at 13:34
Seems that your counter example is wrong since $R_0$ has determinant 0.5 and $R_1$ has determinant 0 and matrix $R$ is nonnegative which is not contradict what has mentioned in the stated problem.
– V.Ajall
Nov 18 at 13:34
|
show 4 more comments
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What is the matrix here to be clear ? Please make your question non ambiguous, thanks!
– Ezy
Nov 17 at 17:07
@Ezy Clear enough?
– V.Ajall
Nov 18 at 7:14
Much better! Now i understand the question :)
– Ezy
Nov 18 at 12:15
And R is dimension NxN ? In other words you let k run up to N-1 ?
– Ezy
Nov 18 at 12:19
y is $ntimes 1$ so R is $nNtimes nN$ and yes we let k run up to $N-1$
– V.Ajall
Nov 18 at 12:45