Are Linear Maps resistant to Noise?
$begingroup$
Let's assume I have a $m times m$ matrix $M$ with Frobenius norm $1$ and a unit vector $x in S^{m-1}$. I also have a second $m times m$ matrix $M^*$ which is obtained from the first one plus some injected noise $eta$, where every entry $eta_{i,j}$ comes from a Gaussian distribution with mean $0$, such that $||eta||_F = 1/10$.
What can we say about the following expected value $$mathbb{E}[| Mx - M^*x|]?$$
EDIT: How it has been suggested in an answer, $mathbb{E}|eta(x)| <1/10$ since $|eta(x)| < 1/10$ because of the Frobenius norm of the noise, so the equality only holds when vector $x$
is aligned with the highest eigenvector of the noise matrix $eta$
, but this happens with low probability, and this probability should (intuitively) change with respect to $m$
: in a higher dimensional space the probability of two random vectors having a similar direction is lower than in a 2-dimensional one.
Is there a way to improve the bound on this expected value?
linear-algebra probability matrices linear-transformations
edited Jan 14 at 15:45
Omnomnomnom
129k792185
asked Jan 13 at 20:17
AlfredAlfred
407
$endgroup$
|
show 3 more comments
$begingroup$
Let's assume I have a $m times m$ matrix $M$ with Frobenius norm $1$ and a unit vector $x in S^{m-1}$. I also have a second $m times m$ matrix $M^*$ which is obtained from the first one plus some injected noise $eta$, where every entry $eta_{i,j}$ comes from a Gaussian distribution with mean $0$, such that $||eta||_F = 1/10$.
What can we say about the following expected value $$mathbb{E}[| Mx - M^*x|]?$$
EDIT: How it has been suggested in an answer, $mathbb{E}|eta(x)| <1/10$ since $|eta(x)| < 1/10$ because of the Frobenius norm of the noise, so the equality only holds when vector $x$
is aligned with the highest eigenvector of the noise matrix $eta$
, but this happens with low probability, and this probability should (intuitively) change with respect to $m$
: in a higher dimensional space the probability of two random vectors having a similar direction is lower than in a 2-dimensional one.
Is there a way to improve the bound on this expected value?
linear-algebra probability matrices linear-transformations
edited Jan 14 at 15:45
Omnomnomnom
129k792185
asked Jan 13 at 20:17
AlfredAlfred
407
$endgroup$
$begingroup$
What exactly is $|x|$? Is it the Euclidean norm?
$endgroup$
– Omnomnomnom
Jan 13 at 21:36
$begingroup$
And what is $|eta|$ in this context?
$endgroup$
– Omnomnomnom
Jan 13 at 21:39
$begingroup$
Sorry, $||eta||$ is the Frobenius norm, while $||x||$ is the euclidean norm. I'll edit. Thanks for the correction
$endgroup$
– Alfred
Jan 13 at 22:55
1
$begingroup$
You’re computing $Bbb E |eta x|$ for what that’s worth
$endgroup$
– Omnomnomnom
Jan 13 at 23:57
1
$begingroup$
We could say $$Bbb E|eta(x)| leq Bbb E |eta|_2 leq Bbb E | eta|_F = 1/10$$ perhaps that helps
$endgroup$
– Omnomnomnom
Jan 14 at 0:00
|
show 3 more comments
$begingroup$
Let's assume I have a $m times m$ matrix $M$ with Frobenius norm $1$ and a unit vector $x in S^{m-1}$. I also have a second $m times m$ matrix $M^*$ which is obtained from the first one plus some injected noise $eta$, where every entry $eta_{i,j}$ comes from a Gaussian distribution with mean $0$, such that $||eta||_F = 1/10$.
What can we say about the following expected value $$mathbb{E}[| Mx - M^*x|]?$$
EDIT: How it has been suggested in an answer, $mathbb{E}|eta(x)| <1/10$ since $|eta(x)| < 1/10$ because of the Frobenius norm of the noise, so the equality only holds when vector $x$
is aligned with the highest eigenvector of the noise matrix $eta$
, but this happens with low probability, and this probability should (intuitively) change with respect to $m$
: in a higher dimensional space the probability of two random vectors having a similar direction is lower than in a 2-dimensional one.
Is there a way to improve the bound on this expected value?
linear-algebra probability matrices linear-transformations
edited Jan 14 at 15:45
Omnomnomnom
129k792185
asked Jan 13 at 20:17
AlfredAlfred
407
$endgroup$
Let's assume I have a $m times m$ matrix $M$ with Frobenius norm $1$ and a unit vector $x in S^{m-1}$. I also have a second $m times m$ matrix $M^*$ which is obtained from the first one plus some injected noise $eta$, where every entry $eta_{i,j}$ comes from a Gaussian distribution with mean $0$, such that $||eta||_F = 1/10$.
What can we say about the following expected value $$mathbb{E}[| Mx - M^*x|]?$$
EDIT: How it has been suggested in an answer, $mathbb{E}|eta(x)| <1/10$ since $|eta(x)| < 1/10$ because of the Frobenius norm of the noise, so the equality only holds when vector $x$
is aligned with the highest eigenvector of the noise matrix $eta$
, but this happens with low probability, and this probability should (intuitively) change with respect to $m$
: in a higher dimensional space the probability of two random vectors having a similar direction is lower than in a 2-dimensional one.
Is there a way to improve the bound on this expected value?
linear-algebra probability matrices linear-transformations
linear-algebra probability matrices linear-transformations
edited Jan 14 at 15:45
Omnomnomnom
129k792185
asked Jan 13 at 20:17
AlfredAlfred
407
edited Jan 14 at 15:45
Omnomnomnom
129k792185
asked Jan 13 at 20:17
AlfredAlfred
407
edited Jan 14 at 15:45
Omnomnomnom
129k792185
edited Jan 14 at 15:45
Omnomnomnom
129k792185
edited Jan 14 at 15:45
Omnomnomnom
129k792185
129k792185
asked Jan 13 at 20:17
AlfredAlfred
407
asked Jan 13 at 20:17
AlfredAlfred
407
asked Jan 13 at 20:17
AlfredAlfred
407
407
$begingroup$
What exactly is $|x|$? Is it the Euclidean norm?
$endgroup$
– Omnomnomnom
Jan 13 at 21:36
$begingroup$
And what is $|eta|$ in this context?
$endgroup$
– Omnomnomnom
Jan 13 at 21:39
$begingroup$
Sorry, $||eta||$ is the Frobenius norm, while $||x||$ is the euclidean norm. I'll edit. Thanks for the correction
$endgroup$
– Alfred
Jan 13 at 22:55
1
$begingroup$
You’re computing $Bbb E |eta x|$ for what that’s worth
$endgroup$
– Omnomnomnom
Jan 13 at 23:57
1
$begingroup$
We could say $$Bbb E|eta(x)| leq Bbb E |eta|_2 leq Bbb E | eta|_F = 1/10$$ perhaps that helps
$endgroup$
– Omnomnomnom
Jan 14 at 0:00
|
show 3 more comments
$begingroup$
What exactly is $|x|$? Is it the Euclidean norm?
$endgroup$
– Omnomnomnom
Jan 13 at 21:36
$begingroup$
And what is $|eta|$ in this context?
$endgroup$
– Omnomnomnom
Jan 13 at 21:39
$begingroup$
Sorry, $||eta||$ is the Frobenius norm, while $||x||$ is the euclidean norm. I'll edit. Thanks for the correction
$endgroup$
– Alfred
Jan 13 at 22:55
1
$begingroup$
You’re computing $Bbb E |eta x|$ for what that’s worth
$endgroup$
– Omnomnomnom
Jan 13 at 23:57
1
$begingroup$
We could say $$Bbb E|eta(x)| leq Bbb E |eta|_2 leq Bbb E | eta|_F = 1/10$$ perhaps that helps
$endgroup$
– Omnomnomnom
Jan 14 at 0:00
$begingroup$
What exactly is $|x|$? Is it the Euclidean norm?
$endgroup$
– Omnomnomnom
Jan 13 at 21:36
$begingroup$
What exactly is $|x|$? Is it the Euclidean norm?
$endgroup$
– Omnomnomnom
Jan 13 at 21:36
$begingroup$
And what is $|eta|$ in this context?
$endgroup$
– Omnomnomnom
Jan 13 at 21:39
$begingroup$
And what is $|eta|$ in this context?
$endgroup$
– Omnomnomnom
Jan 13 at 21:39
$begingroup$
Sorry, $||eta||$ is the Frobenius norm, while $||x||$ is the euclidean norm. I'll edit. Thanks for the correction
$endgroup$
– Alfred
Jan 13 at 22:55
$begingroup$
Sorry, $||eta||$ is the Frobenius norm, while $||x||$ is the euclidean norm. I'll edit. Thanks for the correction
$endgroup$
– Alfred
Jan 13 at 22:55
1
1
$begingroup$
You’re computing $Bbb E |eta x|$ for what that’s worth
$endgroup$
– Omnomnomnom
Jan 13 at 23:57
$begingroup$
You’re computing $Bbb E |eta x|$ for what that’s worth
$endgroup$
– Omnomnomnom
Jan 13 at 23:57
1
1
$begingroup$
We could say $$Bbb E|eta(x)| leq Bbb E |eta|_2 leq Bbb E | eta|_F = 1/10$$ perhaps that helps
$endgroup$
– Omnomnomnom
Jan 14 at 0:00
$begingroup$
We could say $$Bbb E|eta(x)| leq Bbb E |eta|_2 leq Bbb E | eta|_F = 1/10$$ perhaps that helps
$endgroup$
– Omnomnomnom
Jan 14 at 0:00
|
show 3 more comments
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$begingroup$
What exactly is $|x|$? Is it the Euclidean norm?
$endgroup$
– Omnomnomnom
Jan 13 at 21:36
$begingroup$
And what is $|eta|$ in this context?
$endgroup$
– Omnomnomnom
Jan 13 at 21:39
$begingroup$
Sorry, $||eta||$ is the Frobenius norm, while $||x||$ is the euclidean norm. I'll edit. Thanks for the correction
$endgroup$
– Alfred
Jan 13 at 22:55
1
$begingroup$
You’re computing $Bbb E |eta x|$ for what that’s worth
$endgroup$
– Omnomnomnom
Jan 13 at 23:57
1
$begingroup$
We could say $$Bbb E|eta(x)| leq Bbb E |eta|_2 leq Bbb E | eta|_F = 1/10$$ perhaps that helps
$endgroup$
– Omnomnomnom
Jan 14 at 0:00