Mixing
Concentration of the measure of a general covariance-like matrix
$begingroup$
I consider a random matrix of the type : $M_n = frac{1}{n} X_n D_n X_n^intercal in mathbb{R}^{n times n}$, in which all matrices are square of size $n$. $D_n$ is a deterministic diagonal matrix with elements $D_{n,mu}$ which are bounded : $$sup_mu |D_{n,mu}| leq rho < +infty.$$ The matrix elements of the matrix $X_n$ are standard i.i.d. Gaussian variables (mean zero, variance $1$). I consider a function $f : mathbb{R} to mathbb{R}$ which is Lipschitz (I can basically make any other reasonable assumption on $f$), and I'm interested in the concentration, as $n$ grows, of the linear spectral statistic:
$$G_{n,f}(X_n) = frac{1}{n} mathrm{Tr} f(M_n)$$
There are some classic RMT results (for instance from the book of Anderson,Guionnet,Zeitouni : Link (Section 2.3 or 4.4), or in the previous paper of Guionnet&Zeitouni Link (these results can also be found in many other places).
For instance (Corollary 1.8.b of the Guionnet-Zeitouni), if one assumes that all the elements of $D_n$ are positive, and that $x mapsto f(x^2)$ is Lipschitz with Lipschitz constant $L > 0$, then one has:
$$mathbb{P}left(left|G_{n,f}(X_n) - mathbb{E}G_{n,f}(X_n)right| geq tright) leq expleft(- frac{1}{2 rho L}n^2 t^2right) $$
I have found other similar results in the litterature, but I couldn't find anything concerning the case of non-positive $D_n$. My main question is : does there exists similar concentration results concerning this case ? I don't know if the lack of study concerning this case is due to a real theoretical difficulty or if it comes from the fact that these matrices are less natural as they are not empirical covariance matrices...
Any help would be appreciated !
Thanks a lot :)
probability matrices probability-theory random-matrices concentration-of-measure
asked Jan 19 at 2:32
seampseamp
28712
$endgroup$
add a comment |
$begingroup$
I consider a random matrix of the type : $M_n = frac{1}{n} X_n D_n X_n^intercal in mathbb{R}^{n times n}$, in which all matrices are square of size $n$. $D_n$ is a deterministic diagonal matrix with elements $D_{n,mu}$ which are bounded : $$sup_mu |D_{n,mu}| leq rho < +infty.$$ The matrix elements of the matrix $X_n$ are standard i.i.d. Gaussian variables (mean zero, variance $1$). I consider a function $f : mathbb{R} to mathbb{R}$ which is Lipschitz (I can basically make any other reasonable assumption on $f$), and I'm interested in the concentration, as $n$ grows, of the linear spectral statistic:
$$G_{n,f}(X_n) = frac{1}{n} mathrm{Tr} f(M_n)$$
There are some classic RMT results (for instance from the book of Anderson,Guionnet,Zeitouni : Link (Section 2.3 or 4.4), or in the previous paper of Guionnet&Zeitouni Link (these results can also be found in many other places).
For instance (Corollary 1.8.b of the Guionnet-Zeitouni), if one assumes that all the elements of $D_n$ are positive, and that $x mapsto f(x^2)$ is Lipschitz with Lipschitz constant $L > 0$, then one has:
$$mathbb{P}left(left|G_{n,f}(X_n) - mathbb{E}G_{n,f}(X_n)right| geq tright) leq expleft(- frac{1}{2 rho L}n^2 t^2right) $$
I have found other similar results in the litterature, but I couldn't find anything concerning the case of non-positive $D_n$. My main question is : does there exists similar concentration results concerning this case ? I don't know if the lack of study concerning this case is due to a real theoretical difficulty or if it comes from the fact that these matrices are less natural as they are not empirical covariance matrices...
Any help would be appreciated !
Thanks a lot :)
probability matrices probability-theory random-matrices concentration-of-measure
asked Jan 19 at 2:32
seampseamp
28712
$endgroup$
add a comment |
$begingroup$
I consider a random matrix of the type : $M_n = frac{1}{n} X_n D_n X_n^intercal in mathbb{R}^{n times n}$, in which all matrices are square of size $n$. $D_n$ is a deterministic diagonal matrix with elements $D_{n,mu}$ which are bounded : $$sup_mu |D_{n,mu}| leq rho < +infty.$$ The matrix elements of the matrix $X_n$ are standard i.i.d. Gaussian variables (mean zero, variance $1$). I consider a function $f : mathbb{R} to mathbb{R}$ which is Lipschitz (I can basically make any other reasonable assumption on $f$), and I'm interested in the concentration, as $n$ grows, of the linear spectral statistic:
$$G_{n,f}(X_n) = frac{1}{n} mathrm{Tr} f(M_n)$$
There are some classic RMT results (for instance from the book of Anderson,Guionnet,Zeitouni : Link (Section 2.3 or 4.4), or in the previous paper of Guionnet&Zeitouni Link (these results can also be found in many other places).
For instance (Corollary 1.8.b of the Guionnet-Zeitouni), if one assumes that all the elements of $D_n$ are positive, and that $x mapsto f(x^2)$ is Lipschitz with Lipschitz constant $L > 0$, then one has:
$$mathbb{P}left(left|G_{n,f}(X_n) - mathbb{E}G_{n,f}(X_n)right| geq tright) leq expleft(- frac{1}{2 rho L}n^2 t^2right) $$
I have found other similar results in the litterature, but I couldn't find anything concerning the case of non-positive $D_n$. My main question is : does there exists similar concentration results concerning this case ? I don't know if the lack of study concerning this case is due to a real theoretical difficulty or if it comes from the fact that these matrices are less natural as they are not empirical covariance matrices...
Any help would be appreciated !
Thanks a lot :)
probability matrices probability-theory random-matrices concentration-of-measure
asked Jan 19 at 2:32
seampseamp
28712
$endgroup$
I consider a random matrix of the type : $M_n = frac{1}{n} X_n D_n X_n^intercal in mathbb{R}^{n times n}$, in which all matrices are square of size $n$. $D_n$ is a deterministic diagonal matrix with elements $D_{n,mu}$ which are bounded : $$sup_mu |D_{n,mu}| leq rho < +infty.$$ The matrix elements of the matrix $X_n$ are standard i.i.d. Gaussian variables (mean zero, variance $1$). I consider a function $f : mathbb{R} to mathbb{R}$ which is Lipschitz (I can basically make any other reasonable assumption on $f$), and I'm interested in the concentration, as $n$ grows, of the linear spectral statistic:
$$G_{n,f}(X_n) = frac{1}{n} mathrm{Tr} f(M_n)$$
There are some classic RMT results (for instance from the book of Anderson,Guionnet,Zeitouni : Link (Section 2.3 or 4.4), or in the previous paper of Guionnet&Zeitouni Link (these results can also be found in many other places).
For instance (Corollary 1.8.b of the Guionnet-Zeitouni), if one assumes that all the elements of $D_n$ are positive, and that $x mapsto f(x^2)$ is Lipschitz with Lipschitz constant $L > 0$, then one has:
$$mathbb{P}left(left|G_{n,f}(X_n) - mathbb{E}G_{n,f}(X_n)right| geq tright) leq expleft(- frac{1}{2 rho L}n^2 t^2right) $$
I have found other similar results in the litterature, but I couldn't find anything concerning the case of non-positive $D_n$. My main question is : does there exists similar concentration results concerning this case ? I don't know if the lack of study concerning this case is due to a real theoretical difficulty or if it comes from the fact that these matrices are less natural as they are not empirical covariance matrices...
Any help would be appreciated !
Thanks a lot :)
probability matrices probability-theory random-matrices concentration-of-measure
probability matrices probability-theory random-matrices concentration-of-measure
asked Jan 19 at 2:32
seampseamp
28712
asked Jan 19 at 2:32
seampseamp
28712
asked Jan 19 at 2:32
seampseamp
28712
asked Jan 19 at 2:32
seampseamp
28712
asked Jan 19 at 2:32
seampseamp
28712
28712
add a comment |
add a comment |
1 Answer
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There are similar concentration results if you wish to examine singular values instead of eigenvalues. Joel Tropp explains this trick here.
As for the matrix being indefinite, if you assume $D_n$ is fixed (as you do above), it may be possible to follow a similar proof as that above. However, it seems as though you are dealing with a random indefinite symmetric matrix, which is a difficult object to understand. There are many results on random Hermitian matrices, but typically they involve bounding the largest singular value (which corresponds to the eigenvalue of largest magnitude). Indeed, there are fewer results in general for indefinite matrices due to the fact that the covariance matrix is always positive semidefinite.
I would say that I'm willing to help, but I think this would lead me down a tangent that I wouldn't be able to dig myself out of. In general, you may just want to consider $M_n M_n^{top}$.
answered Jan 24 at 23:22
OldGodzillaOldGodzilla
58227
$endgroup$
$begingroup$
What's true is that actually the $f$ I am interested in is a function of the absolute values of the eigenvalues of $M_n$, so of its singular values. I will look more precisely into the reference. My matrix $D_n$ is deterministic, so I will try to see if I can make something with the proof technique of the reference you gave.
$endgroup$
– seamp
Jan 25 at 0:43
$begingroup$
Something I don't really get is that all these calculations seem to bound the largest eigenvalue of $M_n$ (or $M_n M_n^intercal$), but that's not what I'm interested in here...
$endgroup$
– seamp
Jan 25 at 0:47
add a comment |
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$begingroup$
There are similar concentration results if you wish to examine singular values instead of eigenvalues. Joel Tropp explains this trick here.
As for the matrix being indefinite, if you assume $D_n$ is fixed (as you do above), it may be possible to follow a similar proof as that above. However, it seems as though you are dealing with a random indefinite symmetric matrix, which is a difficult object to understand. There are many results on random Hermitian matrices, but typically they involve bounding the largest singular value (which corresponds to the eigenvalue of largest magnitude). Indeed, there are fewer results in general for indefinite matrices due to the fact that the covariance matrix is always positive semidefinite.
I would say that I'm willing to help, but I think this would lead me down a tangent that I wouldn't be able to dig myself out of. In general, you may just want to consider $M_n M_n^{top}$.
answered Jan 24 at 23:22
OldGodzillaOldGodzilla
58227
$endgroup$
$begingroup$
What's true is that actually the $f$ I am interested in is a function of the absolute values of the eigenvalues of $M_n$, so of its singular values. I will look more precisely into the reference. My matrix $D_n$ is deterministic, so I will try to see if I can make something with the proof technique of the reference you gave.
$endgroup$
– seamp
Jan 25 at 0:43
$begingroup$
Something I don't really get is that all these calculations seem to bound the largest eigenvalue of $M_n$ (or $M_n M_n^intercal$), but that's not what I'm interested in here...
$endgroup$
– seamp
Jan 25 at 0:47
add a comment |
$begingroup$
There are similar concentration results if you wish to examine singular values instead of eigenvalues. Joel Tropp explains this trick here.
As for the matrix being indefinite, if you assume $D_n$ is fixed (as you do above), it may be possible to follow a similar proof as that above. However, it seems as though you are dealing with a random indefinite symmetric matrix, which is a difficult object to understand. There are many results on random Hermitian matrices, but typically they involve bounding the largest singular value (which corresponds to the eigenvalue of largest magnitude). Indeed, there are fewer results in general for indefinite matrices due to the fact that the covariance matrix is always positive semidefinite.
I would say that I'm willing to help, but I think this would lead me down a tangent that I wouldn't be able to dig myself out of. In general, you may just want to consider $M_n M_n^{top}$.
answered Jan 24 at 23:22
OldGodzillaOldGodzilla
58227
$endgroup$
$begingroup$
What's true is that actually the $f$ I am interested in is a function of the absolute values of the eigenvalues of $M_n$, so of its singular values. I will look more precisely into the reference. My matrix $D_n$ is deterministic, so I will try to see if I can make something with the proof technique of the reference you gave.
$endgroup$
– seamp
Jan 25 at 0:43
$begingroup$
Something I don't really get is that all these calculations seem to bound the largest eigenvalue of $M_n$ (or $M_n M_n^intercal$), but that's not what I'm interested in here...
$endgroup$
– seamp
Jan 25 at 0:47
add a comment |
$begingroup$
There are similar concentration results if you wish to examine singular values instead of eigenvalues. Joel Tropp explains this trick here.
As for the matrix being indefinite, if you assume $D_n$ is fixed (as you do above), it may be possible to follow a similar proof as that above. However, it seems as though you are dealing with a random indefinite symmetric matrix, which is a difficult object to understand. There are many results on random Hermitian matrices, but typically they involve bounding the largest singular value (which corresponds to the eigenvalue of largest magnitude). Indeed, there are fewer results in general for indefinite matrices due to the fact that the covariance matrix is always positive semidefinite.
I would say that I'm willing to help, but I think this would lead me down a tangent that I wouldn't be able to dig myself out of. In general, you may just want to consider $M_n M_n^{top}$.
answered Jan 24 at 23:22
OldGodzillaOldGodzilla
58227
$endgroup$
There are similar concentration results if you wish to examine singular values instead of eigenvalues. Joel Tropp explains this trick here.
As for the matrix being indefinite, if you assume $D_n$ is fixed (as you do above), it may be possible to follow a similar proof as that above. However, it seems as though you are dealing with a random indefinite symmetric matrix, which is a difficult object to understand. There are many results on random Hermitian matrices, but typically they involve bounding the largest singular value (which corresponds to the eigenvalue of largest magnitude). Indeed, there are fewer results in general for indefinite matrices due to the fact that the covariance matrix is always positive semidefinite.
I would say that I'm willing to help, but I think this would lead me down a tangent that I wouldn't be able to dig myself out of. In general, you may just want to consider $M_n M_n^{top}$.
answered Jan 24 at 23:22
OldGodzillaOldGodzilla
58227
answered Jan 24 at 23:22
OldGodzillaOldGodzilla
58227
answered Jan 24 at 23:22
OldGodzillaOldGodzilla
58227
answered Jan 24 at 23:22
OldGodzillaOldGodzilla
58227
58227
$begingroup$
What's true is that actually the $f$ I am interested in is a function of the absolute values of the eigenvalues of $M_n$, so of its singular values. I will look more precisely into the reference. My matrix $D_n$ is deterministic, so I will try to see if I can make something with the proof technique of the reference you gave.
$endgroup$
– seamp
Jan 25 at 0:43
$begingroup$
Something I don't really get is that all these calculations seem to bound the largest eigenvalue of $M_n$ (or $M_n M_n^intercal$), but that's not what I'm interested in here...
$endgroup$
– seamp
Jan 25 at 0:47
add a comment |
$begingroup$
What's true is that actually the $f$ I am interested in is a function of the absolute values of the eigenvalues of $M_n$, so of its singular values. I will look more precisely into the reference. My matrix $D_n$ is deterministic, so I will try to see if I can make something with the proof technique of the reference you gave.
$endgroup$
– seamp
Jan 25 at 0:43
$begingroup$
Something I don't really get is that all these calculations seem to bound the largest eigenvalue of $M_n$ (or $M_n M_n^intercal$), but that's not what I'm interested in here...
$endgroup$
– seamp
Jan 25 at 0:47
$begingroup$
What's true is that actually the $f$ I am interested in is a function of the absolute values of the eigenvalues of $M_n$, so of its singular values. I will look more precisely into the reference. My matrix $D_n$ is deterministic, so I will try to see if I can make something with the proof technique of the reference you gave.
$endgroup$
– seamp
Jan 25 at 0:43
$begingroup$
What's true is that actually the $f$ I am interested in is a function of the absolute values of the eigenvalues of $M_n$, so of its singular values. I will look more precisely into the reference. My matrix $D_n$ is deterministic, so I will try to see if I can make something with the proof technique of the reference you gave.
$endgroup$
– seamp
Jan 25 at 0:43
$begingroup$
Something I don't really get is that all these calculations seem to bound the largest eigenvalue of $M_n$ (or $M_n M_n^intercal$), but that's not what I'm interested in here...
$endgroup$
– seamp
Jan 25 at 0:47
$begingroup$
Something I don't really get is that all these calculations seem to bound the largest eigenvalue of $M_n$ (or $M_n M_n^intercal$), but that's not what I'm interested in here...
$endgroup$
– seamp
Jan 25 at 0:47
add a comment |
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