Mixing
Bounding sub-Gaussian tail events by Gaussian tail events?
$begingroup$
Background
I have come across a problem where I want to derive a concentration inequality for sub-gaussian random variables. More precisely, I want to bound the spectrum of a certain empirical covariance matrix, where $(x_i)$ are sub-gaussian random vectors forming the rows of a matrix $X in mathbb{R}^{Ttimes n}$. Without specififying exactly what the dependence structure of the $x_i$ are, let us just say I want to bound the probability of an event of the form
$$
left|frac{1}{T}X^*X-Sigma right |_infty geq f(T,dots).
$$
My argument relies on a fact which holds (more or less) uniquely for Gaussian
random variables (I have a quotient of moment generating functions somewhere in my argument) and I have verified it to be true in the Gaussian case, but let us not dwell on this.
Question
Now, somehow the notion of sub-Gaussianity is meant to capture the fact the distribution is less wild than a Gaussian (in particular dispalyed by the moment generating definition so in the scalar case my question is trivial).
So, this got me thinking, is there a generic way to prove such an
inequality for the Gaussian case and then use some sort of
"extension/domination lemma" to give the result for the sub-Gaussian
case?
I can already see one problem with such an approach since the entries of $X^*X$ really are sub-exponential and not sub-Gaussian anymore (although one can imagine them being dominated by $chi$-squareds).
I would be very interested in any links, references, ideas, partial results that could help me better understand this problem. I am not interested in finding a different way to prove my bound, but really would like to explore this sort of "domination" argument!
probability probability-theory random-matrices concentration-of-measure
asked Jan 25 at 8:29
sortofamathematiciansortofamathematician
537
$endgroup$
add a comment |
$begingroup$
Background
I have come across a problem where I want to derive a concentration inequality for sub-gaussian random variables. More precisely, I want to bound the spectrum of a certain empirical covariance matrix, where $(x_i)$ are sub-gaussian random vectors forming the rows of a matrix $X in mathbb{R}^{Ttimes n}$. Without specififying exactly what the dependence structure of the $x_i$ are, let us just say I want to bound the probability of an event of the form
$$
left|frac{1}{T}X^*X-Sigma right |_infty geq f(T,dots).
$$
My argument relies on a fact which holds (more or less) uniquely for Gaussian
random variables (I have a quotient of moment generating functions somewhere in my argument) and I have verified it to be true in the Gaussian case, but let us not dwell on this.
Question
Now, somehow the notion of sub-Gaussianity is meant to capture the fact the distribution is less wild than a Gaussian (in particular dispalyed by the moment generating definition so in the scalar case my question is trivial).
So, this got me thinking, is there a generic way to prove such an
inequality for the Gaussian case and then use some sort of
"extension/domination lemma" to give the result for the sub-Gaussian
case?
I can already see one problem with such an approach since the entries of $X^*X$ really are sub-exponential and not sub-Gaussian anymore (although one can imagine them being dominated by $chi$-squareds).
I would be very interested in any links, references, ideas, partial results that could help me better understand this problem. I am not interested in finding a different way to prove my bound, but really would like to explore this sort of "domination" argument!
probability probability-theory random-matrices concentration-of-measure
asked Jan 25 at 8:29
sortofamathematiciansortofamathematician
537
$endgroup$
add a comment |
$begingroup$
Background
I have come across a problem where I want to derive a concentration inequality for sub-gaussian random variables. More precisely, I want to bound the spectrum of a certain empirical covariance matrix, where $(x_i)$ are sub-gaussian random vectors forming the rows of a matrix $X in mathbb{R}^{Ttimes n}$. Without specififying exactly what the dependence structure of the $x_i$ are, let us just say I want to bound the probability of an event of the form
$$
left|frac{1}{T}X^*X-Sigma right |_infty geq f(T,dots).
$$
My argument relies on a fact which holds (more or less) uniquely for Gaussian
random variables (I have a quotient of moment generating functions somewhere in my argument) and I have verified it to be true in the Gaussian case, but let us not dwell on this.
Question
Now, somehow the notion of sub-Gaussianity is meant to capture the fact the distribution is less wild than a Gaussian (in particular dispalyed by the moment generating definition so in the scalar case my question is trivial).
So, this got me thinking, is there a generic way to prove such an
inequality for the Gaussian case and then use some sort of
"extension/domination lemma" to give the result for the sub-Gaussian
case?
I can already see one problem with such an approach since the entries of $X^*X$ really are sub-exponential and not sub-Gaussian anymore (although one can imagine them being dominated by $chi$-squareds).
I would be very interested in any links, references, ideas, partial results that could help me better understand this problem. I am not interested in finding a different way to prove my bound, but really would like to explore this sort of "domination" argument!
probability probability-theory random-matrices concentration-of-measure
asked Jan 25 at 8:29
sortofamathematiciansortofamathematician
537
$endgroup$
Background
I have come across a problem where I want to derive a concentration inequality for sub-gaussian random variables. More precisely, I want to bound the spectrum of a certain empirical covariance matrix, where $(x_i)$ are sub-gaussian random vectors forming the rows of a matrix $X in mathbb{R}^{Ttimes n}$. Without specififying exactly what the dependence structure of the $x_i$ are, let us just say I want to bound the probability of an event of the form
$$
left|frac{1}{T}X^*X-Sigma right |_infty geq f(T,dots).
$$
My argument relies on a fact which holds (more or less) uniquely for Gaussian
random variables (I have a quotient of moment generating functions somewhere in my argument) and I have verified it to be true in the Gaussian case, but let us not dwell on this.
Question
Now, somehow the notion of sub-Gaussianity is meant to capture the fact the distribution is less wild than a Gaussian (in particular dispalyed by the moment generating definition so in the scalar case my question is trivial).
So, this got me thinking, is there a generic way to prove such an
inequality for the Gaussian case and then use some sort of
"extension/domination lemma" to give the result for the sub-Gaussian
case?
I can already see one problem with such an approach since the entries of $X^*X$ really are sub-exponential and not sub-Gaussian anymore (although one can imagine them being dominated by $chi$-squareds).
I would be very interested in any links, references, ideas, partial results that could help me better understand this problem. I am not interested in finding a different way to prove my bound, but really would like to explore this sort of "domination" argument!
probability probability-theory random-matrices concentration-of-measure
probability probability-theory random-matrices concentration-of-measure
asked Jan 25 at 8:29
sortofamathematiciansortofamathematician
537
asked Jan 25 at 8:29
sortofamathematiciansortofamathematician
537
edited Jan 25 at 8:40
sortofamathematician
asked Jan 25 at 8:29
sortofamathematiciansortofamathematician
537
asked Jan 25 at 8:29
sortofamathematiciansortofamathematician
537
asked Jan 25 at 8:29
sortofamathematiciansortofamathematician
537
537
add a comment |
add a comment |
1 Answer
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$begingroup$
You can use covering number arguments to get similar results for sub-gaussian distributions sometimes. The result that you are interested in is Theorem 4.6.1 (see Eq 4.22) in the book[1]. The book is available online and the proof uses covering number argument.
Also, Lemma 6.2.3 in [1] shows an example where the moment generating function of sub-gaussian random vector was bounded by mgf of a gaussian vector.
[1]: High-Dimensional Probability, Roman Vershynin. https://www.math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.html
answered Feb 16 at 3:36
AnkitpAnkitp
21029
$endgroup$
$begingroup$
@Anktip Thank you. I'm aware of the reference and the theorem you cite is precisely the one I am playing around with (without the independence assumption). I have at this point kind of realized that the question might be a bit too imprecise for this exchange so I will accept your answer as sufficient.
$endgroup$
– sortofamathematician
Feb 16 at 17:20
$begingroup$
It will be useful for others, including me, if you could share other examples that you may have found where "domination arguments" was used.
$endgroup$
– Ankitp
Feb 16 at 19:38
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
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oldest
votes
active
oldest
votes
$begingroup$
You can use covering number arguments to get similar results for sub-gaussian distributions sometimes. The result that you are interested in is Theorem 4.6.1 (see Eq 4.22) in the book[1]. The book is available online and the proof uses covering number argument.
Also, Lemma 6.2.3 in [1] shows an example where the moment generating function of sub-gaussian random vector was bounded by mgf of a gaussian vector.
[1]: High-Dimensional Probability, Roman Vershynin. https://www.math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.html
answered Feb 16 at 3:36
AnkitpAnkitp
21029
$endgroup$
$begingroup$
@Anktip Thank you. I'm aware of the reference and the theorem you cite is precisely the one I am playing around with (without the independence assumption). I have at this point kind of realized that the question might be a bit too imprecise for this exchange so I will accept your answer as sufficient.
$endgroup$
– sortofamathematician
Feb 16 at 17:20
$begingroup$
It will be useful for others, including me, if you could share other examples that you may have found where "domination arguments" was used.
$endgroup$
– Ankitp
Feb 16 at 19:38
add a comment |
$begingroup$
You can use covering number arguments to get similar results for sub-gaussian distributions sometimes. The result that you are interested in is Theorem 4.6.1 (see Eq 4.22) in the book[1]. The book is available online and the proof uses covering number argument.
Also, Lemma 6.2.3 in [1] shows an example where the moment generating function of sub-gaussian random vector was bounded by mgf of a gaussian vector.
[1]: High-Dimensional Probability, Roman Vershynin. https://www.math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.html
answered Feb 16 at 3:36
AnkitpAnkitp
21029
$endgroup$
$begingroup$
@Anktip Thank you. I'm aware of the reference and the theorem you cite is precisely the one I am playing around with (without the independence assumption). I have at this point kind of realized that the question might be a bit too imprecise for this exchange so I will accept your answer as sufficient.
$endgroup$
– sortofamathematician
Feb 16 at 17:20
$begingroup$
It will be useful for others, including me, if you could share other examples that you may have found where "domination arguments" was used.
$endgroup$
– Ankitp
Feb 16 at 19:38
add a comment |
$begingroup$
You can use covering number arguments to get similar results for sub-gaussian distributions sometimes. The result that you are interested in is Theorem 4.6.1 (see Eq 4.22) in the book[1]. The book is available online and the proof uses covering number argument.
Also, Lemma 6.2.3 in [1] shows an example where the moment generating function of sub-gaussian random vector was bounded by mgf of a gaussian vector.
[1]: High-Dimensional Probability, Roman Vershynin. https://www.math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.html
answered Feb 16 at 3:36
AnkitpAnkitp
21029
$endgroup$
You can use covering number arguments to get similar results for sub-gaussian distributions sometimes. The result that you are interested in is Theorem 4.6.1 (see Eq 4.22) in the book[1]. The book is available online and the proof uses covering number argument.
Also, Lemma 6.2.3 in [1] shows an example where the moment generating function of sub-gaussian random vector was bounded by mgf of a gaussian vector.
[1]: High-Dimensional Probability, Roman Vershynin. https://www.math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.html
answered Feb 16 at 3:36
AnkitpAnkitp
21029
answered Feb 16 at 3:36
AnkitpAnkitp
21029
answered Feb 16 at 3:36
AnkitpAnkitp
21029
answered Feb 16 at 3:36
AnkitpAnkitp
21029
21029
$begingroup$
@Anktip Thank you. I'm aware of the reference and the theorem you cite is precisely the one I am playing around with (without the independence assumption). I have at this point kind of realized that the question might be a bit too imprecise for this exchange so I will accept your answer as sufficient.
$endgroup$
– sortofamathematician
Feb 16 at 17:20
$begingroup$
It will be useful for others, including me, if you could share other examples that you may have found where "domination arguments" was used.
$endgroup$
– Ankitp
Feb 16 at 19:38
add a comment |
$begingroup$
@Anktip Thank you. I'm aware of the reference and the theorem you cite is precisely the one I am playing around with (without the independence assumption). I have at this point kind of realized that the question might be a bit too imprecise for this exchange so I will accept your answer as sufficient.
$endgroup$
– sortofamathematician
Feb 16 at 17:20
$begingroup$
It will be useful for others, including me, if you could share other examples that you may have found where "domination arguments" was used.
$endgroup$
– Ankitp
Feb 16 at 19:38
$begingroup$
@Anktip Thank you. I'm aware of the reference and the theorem you cite is precisely the one I am playing around with (without the independence assumption). I have at this point kind of realized that the question might be a bit too imprecise for this exchange so I will accept your answer as sufficient.
$endgroup$
– sortofamathematician
Feb 16 at 17:20
$begingroup$
@Anktip Thank you. I'm aware of the reference and the theorem you cite is precisely the one I am playing around with (without the independence assumption). I have at this point kind of realized that the question might be a bit too imprecise for this exchange so I will accept your answer as sufficient.
$endgroup$
– sortofamathematician
Feb 16 at 17:20
$begingroup$
It will be useful for others, including me, if you could share other examples that you may have found where "domination arguments" was used.
$endgroup$
– Ankitp
Feb 16 at 19:38
$begingroup$
It will be useful for others, including me, if you could share other examples that you may have found where "domination arguments" was used.
$endgroup$
– Ankitp
Feb 16 at 19:38
add a comment |
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