the expected norm of matrices with sub-gaussian entries
$begingroup$
It is the exercise 4.4.6 in the book High-dimension-probability, https://www.math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.pdf
Given the theorem that $$Prob{||A||>CK(sqrt m + sqrt n +t)} leq 2exp(-t^2)\for~t > 0$$
In results like this, C and c will always denote some positive absolute constants
Prove that $$E||A|| leq CK(sqrt m + sqrt n)$$
The book gives the hint that it can be solved directly by using the theorem above.
So I tried to calculate out the expection of operator norm of A:$$E||A||=int_0^{CK(sqrt m + sqrt n)}P(||A||>a)da + int_{CK(sqrt m + sqrt n)}^infty P(||A||>a)da\ leq int_0^{CK(sqrt m + sqrt n)}P(||A||>a)da + int_0^infty 2exp(-t^2)dt \ = int_0^{CK(sqrt m + sqrt n)}P(||A||>t)dt + sqrt pi$$
And I get stuck here. It seems not a right direction because I don't know the structure when t is smaller than 0.
Hint or helpful material are welcome.
probability statistics
asked Jan 7 at 6:38
DylonDylon
62
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add a comment |
$begingroup$
It is the exercise 4.4.6 in the book High-dimension-probability, https://www.math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.pdf
Given the theorem that $$Prob{||A||>CK(sqrt m + sqrt n +t)} leq 2exp(-t^2)\for~t > 0$$
In results like this, C and c will always denote some positive absolute constants
Prove that $$E||A|| leq CK(sqrt m + sqrt n)$$
The book gives the hint that it can be solved directly by using the theorem above.
So I tried to calculate out the expection of operator norm of A:$$E||A||=int_0^{CK(sqrt m + sqrt n)}P(||A||>a)da + int_{CK(sqrt m + sqrt n)}^infty P(||A||>a)da\ leq int_0^{CK(sqrt m + sqrt n)}P(||A||>a)da + int_0^infty 2exp(-t^2)dt \ = int_0^{CK(sqrt m + sqrt n)}P(||A||>t)dt + sqrt pi$$
And I get stuck here. It seems not a right direction because I don't know the structure when t is smaller than 0.
Hint or helpful material are welcome.
probability statistics
asked Jan 7 at 6:38
DylonDylon
62
$endgroup$
$begingroup$
Hi, and welcome to M.SE. Questions posed as:-"This is an exercise: solve it for me" are poorly received from other members. I you want to help other members help you, provide context for your question: for example, what did you tried to solve it? And where you get stuck?
$endgroup$
– Daniele Tampieri
Jan 7 at 6:57
$begingroup$
A complete qustion would include: 1. the book's mention that " In results like this, C and c will always denote some positive absolute constants", and 2. your own attempts. Where are they?
$endgroup$
– Did
Jan 7 at 7:10
$begingroup$
changes have been made. any suggestions?
$endgroup$
– Dylon
Jan 7 at 7:59
add a comment |
$begingroup$
It is the exercise 4.4.6 in the book High-dimension-probability, https://www.math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.pdf
Given the theorem that $$Prob{||A||>CK(sqrt m + sqrt n +t)} leq 2exp(-t^2)\for~t > 0$$
In results like this, C and c will always denote some positive absolute constants
Prove that $$E||A|| leq CK(sqrt m + sqrt n)$$
The book gives the hint that it can be solved directly by using the theorem above.
So I tried to calculate out the expection of operator norm of A:$$E||A||=int_0^{CK(sqrt m + sqrt n)}P(||A||>a)da + int_{CK(sqrt m + sqrt n)}^infty P(||A||>a)da\ leq int_0^{CK(sqrt m + sqrt n)}P(||A||>a)da + int_0^infty 2exp(-t^2)dt \ = int_0^{CK(sqrt m + sqrt n)}P(||A||>t)dt + sqrt pi$$
And I get stuck here. It seems not a right direction because I don't know the structure when t is smaller than 0.
Hint or helpful material are welcome.
probability statistics
asked Jan 7 at 6:38
DylonDylon
62
$endgroup$
It is the exercise 4.4.6 in the book High-dimension-probability, https://www.math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.pdf
Given the theorem that $$Prob{||A||>CK(sqrt m + sqrt n +t)} leq 2exp(-t^2)\for~t > 0$$
In results like this, C and c will always denote some positive absolute constants
Prove that $$E||A|| leq CK(sqrt m + sqrt n)$$
The book gives the hint that it can be solved directly by using the theorem above.
So I tried to calculate out the expection of operator norm of A:$$E||A||=int_0^{CK(sqrt m + sqrt n)}P(||A||>a)da + int_{CK(sqrt m + sqrt n)}^infty P(||A||>a)da\ leq int_0^{CK(sqrt m + sqrt n)}P(||A||>a)da + int_0^infty 2exp(-t^2)dt \ = int_0^{CK(sqrt m + sqrt n)}P(||A||>t)dt + sqrt pi$$
And I get stuck here. It seems not a right direction because I don't know the structure when t is smaller than 0.
Hint or helpful material are welcome.
probability statistics
probability statistics
asked Jan 7 at 6:38
DylonDylon
62
asked Jan 7 at 6:38
DylonDylon
62
edited Jan 7 at 7:58
Dylon
asked Jan 7 at 6:38
DylonDylon
62
asked Jan 7 at 6:38
DylonDylon
62
asked Jan 7 at 6:38
DylonDylon
62
62
$begingroup$
Hi, and welcome to M.SE. Questions posed as:-"This is an exercise: solve it for me" are poorly received from other members. I you want to help other members help you, provide context for your question: for example, what did you tried to solve it? And where you get stuck?
$endgroup$
– Daniele Tampieri
Jan 7 at 6:57
$begingroup$
A complete qustion would include: 1. the book's mention that " In results like this, C and c will always denote some positive absolute constants", and 2. your own attempts. Where are they?
$endgroup$
– Did
Jan 7 at 7:10
$begingroup$
changes have been made. any suggestions?
$endgroup$
– Dylon
Jan 7 at 7:59
add a comment |
$begingroup$
Hi, and welcome to M.SE. Questions posed as:-"This is an exercise: solve it for me" are poorly received from other members. I you want to help other members help you, provide context for your question: for example, what did you tried to solve it? And where you get stuck?
$endgroup$
– Daniele Tampieri
Jan 7 at 6:57
$begingroup$
A complete qustion would include: 1. the book's mention that " In results like this, C and c will always denote some positive absolute constants", and 2. your own attempts. Where are they?
$endgroup$
– Did
Jan 7 at 7:10
$begingroup$
changes have been made. any suggestions?
$endgroup$
– Dylon
Jan 7 at 7:59
$begingroup$
Hi, and welcome to M.SE. Questions posed as:-"This is an exercise: solve it for me" are poorly received from other members. I you want to help other members help you, provide context for your question: for example, what did you tried to solve it? And where you get stuck?
$endgroup$
– Daniele Tampieri
Jan 7 at 6:57
$begingroup$
Hi, and welcome to M.SE. Questions posed as:-"This is an exercise: solve it for me" are poorly received from other members. I you want to help other members help you, provide context for your question: for example, what did you tried to solve it? And where you get stuck?
$endgroup$
– Daniele Tampieri
Jan 7 at 6:57
$begingroup$
A complete qustion would include: 1. the book's mention that " In results like this, C and c will always denote some positive absolute constants", and 2. your own attempts. Where are they?
$endgroup$
– Did
Jan 7 at 7:10
$begingroup$
A complete qustion would include: 1. the book's mention that " In results like this, C and c will always denote some positive absolute constants", and 2. your own attempts. Where are they?
$endgroup$
– Did
Jan 7 at 7:10
$begingroup$
changes have been made. any suggestions?
$endgroup$
– Dylon
Jan 7 at 7:59
$begingroup$
changes have been made. any suggestions?
$endgroup$
– Dylon
Jan 7 at 7:59
add a comment |
1 Answer
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The question asks to show that, if $Xgeqslant0$ almost surely and $$P(Xgeqslant c+x)leqslant2e^{-x^2}$$ for some nonnegative $c$, for every nonnegative $x$, then $E(X)leqslant c$ (simply use $X=|A|/CK$ and $c=sqrt m+sqrt n$). Without further properties of $X$, this does not hold. But one has $$E(X)=int_0^infty P(Xgeqslant x)dxleqslantint_0^cdx+int_0^infty P(Xgeqslant c+x)dx$$ which leads to $$E(X)leqslant c+int_0^infty 2e^{-x^2}dx=c+sqrtpi$$ One can refine this slightly, replacing the bound $2e^{-x^2}$ by $$min{1,2e^{-x^2}}$$ but this will not yield $E(X)leqslant c$, only $E(X)leqslant c+$some other absolute constant.
However, if $m$ and $n$ are positive integers, then $$sqrt m+sqrt ngeqslant2>sqrtpi$$ hence, in the original notations, one gets $$E(|A|)leqslant C'K(sqrt m+sqrt n)$$ with $$C'=2C$$ In view of the book's mention that "In results like this, C and c will always denote some positive absolute constants" (whose exact values may vary from line to line, as one usually adds), this last inequality might be what the authors had in mind.
answered Jan 7 at 13:55
DidDid
247k23222458
$endgroup$
$begingroup$
Thanks. It seems clear now when constant C is clarified.
$endgroup$
– Dylon
Jan 10 at 5:58
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
The question asks to show that, if $Xgeqslant0$ almost surely and $$P(Xgeqslant c+x)leqslant2e^{-x^2}$$ for some nonnegative $c$, for every nonnegative $x$, then $E(X)leqslant c$ (simply use $X=|A|/CK$ and $c=sqrt m+sqrt n$). Without further properties of $X$, this does not hold. But one has $$E(X)=int_0^infty P(Xgeqslant x)dxleqslantint_0^cdx+int_0^infty P(Xgeqslant c+x)dx$$ which leads to $$E(X)leqslant c+int_0^infty 2e^{-x^2}dx=c+sqrtpi$$ One can refine this slightly, replacing the bound $2e^{-x^2}$ by $$min{1,2e^{-x^2}}$$ but this will not yield $E(X)leqslant c$, only $E(X)leqslant c+$some other absolute constant.
However, if $m$ and $n$ are positive integers, then $$sqrt m+sqrt ngeqslant2>sqrtpi$$ hence, in the original notations, one gets $$E(|A|)leqslant C'K(sqrt m+sqrt n)$$ with $$C'=2C$$ In view of the book's mention that "In results like this, C and c will always denote some positive absolute constants" (whose exact values may vary from line to line, as one usually adds), this last inequality might be what the authors had in mind.
answered Jan 7 at 13:55
DidDid
247k23222458
$endgroup$
$begingroup$
Thanks. It seems clear now when constant C is clarified.
$endgroup$
– Dylon
Jan 10 at 5:58
add a comment |
$begingroup$
The question asks to show that, if $Xgeqslant0$ almost surely and $$P(Xgeqslant c+x)leqslant2e^{-x^2}$$ for some nonnegative $c$, for every nonnegative $x$, then $E(X)leqslant c$ (simply use $X=|A|/CK$ and $c=sqrt m+sqrt n$). Without further properties of $X$, this does not hold. But one has $$E(X)=int_0^infty P(Xgeqslant x)dxleqslantint_0^cdx+int_0^infty P(Xgeqslant c+x)dx$$ which leads to $$E(X)leqslant c+int_0^infty 2e^{-x^2}dx=c+sqrtpi$$ One can refine this slightly, replacing the bound $2e^{-x^2}$ by $$min{1,2e^{-x^2}}$$ but this will not yield $E(X)leqslant c$, only $E(X)leqslant c+$some other absolute constant.
However, if $m$ and $n$ are positive integers, then $$sqrt m+sqrt ngeqslant2>sqrtpi$$ hence, in the original notations, one gets $$E(|A|)leqslant C'K(sqrt m+sqrt n)$$ with $$C'=2C$$ In view of the book's mention that "In results like this, C and c will always denote some positive absolute constants" (whose exact values may vary from line to line, as one usually adds), this last inequality might be what the authors had in mind.
answered Jan 7 at 13:55
DidDid
247k23222458
$endgroup$
$begingroup$
Thanks. It seems clear now when constant C is clarified.
$endgroup$
– Dylon
Jan 10 at 5:58
add a comment |
$begingroup$
The question asks to show that, if $Xgeqslant0$ almost surely and $$P(Xgeqslant c+x)leqslant2e^{-x^2}$$ for some nonnegative $c$, for every nonnegative $x$, then $E(X)leqslant c$ (simply use $X=|A|/CK$ and $c=sqrt m+sqrt n$). Without further properties of $X$, this does not hold. But one has $$E(X)=int_0^infty P(Xgeqslant x)dxleqslantint_0^cdx+int_0^infty P(Xgeqslant c+x)dx$$ which leads to $$E(X)leqslant c+int_0^infty 2e^{-x^2}dx=c+sqrtpi$$ One can refine this slightly, replacing the bound $2e^{-x^2}$ by $$min{1,2e^{-x^2}}$$ but this will not yield $E(X)leqslant c$, only $E(X)leqslant c+$some other absolute constant.
However, if $m$ and $n$ are positive integers, then $$sqrt m+sqrt ngeqslant2>sqrtpi$$ hence, in the original notations, one gets $$E(|A|)leqslant C'K(sqrt m+sqrt n)$$ with $$C'=2C$$ In view of the book's mention that "In results like this, C and c will always denote some positive absolute constants" (whose exact values may vary from line to line, as one usually adds), this last inequality might be what the authors had in mind.
answered Jan 7 at 13:55
DidDid
247k23222458
$endgroup$
The question asks to show that, if $Xgeqslant0$ almost surely and $$P(Xgeqslant c+x)leqslant2e^{-x^2}$$ for some nonnegative $c$, for every nonnegative $x$, then $E(X)leqslant c$ (simply use $X=|A|/CK$ and $c=sqrt m+sqrt n$). Without further properties of $X$, this does not hold. But one has $$E(X)=int_0^infty P(Xgeqslant x)dxleqslantint_0^cdx+int_0^infty P(Xgeqslant c+x)dx$$ which leads to $$E(X)leqslant c+int_0^infty 2e^{-x^2}dx=c+sqrtpi$$ One can refine this slightly, replacing the bound $2e^{-x^2}$ by $$min{1,2e^{-x^2}}$$ but this will not yield $E(X)leqslant c$, only $E(X)leqslant c+$some other absolute constant.
However, if $m$ and $n$ are positive integers, then $$sqrt m+sqrt ngeqslant2>sqrtpi$$ hence, in the original notations, one gets $$E(|A|)leqslant C'K(sqrt m+sqrt n)$$ with $$C'=2C$$ In view of the book's mention that "In results like this, C and c will always denote some positive absolute constants" (whose exact values may vary from line to line, as one usually adds), this last inequality might be what the authors had in mind.
answered Jan 7 at 13:55
DidDid
247k23222458
answered Jan 7 at 13:55
DidDid
247k23222458
answered Jan 7 at 13:55
DidDid
247k23222458
answered Jan 7 at 13:55
DidDid
247k23222458
247k23222458
$begingroup$
Thanks. It seems clear now when constant C is clarified.
$endgroup$
– Dylon
Jan 10 at 5:58
add a comment |
$begingroup$
Thanks. It seems clear now when constant C is clarified.
$endgroup$
– Dylon
Jan 10 at 5:58
$begingroup$
Thanks. It seems clear now when constant C is clarified.
$endgroup$
– Dylon
Jan 10 at 5:58
$begingroup$
Thanks. It seems clear now when constant C is clarified.
$endgroup$
– Dylon
Jan 10 at 5:58
add a comment |
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$begingroup$
Hi, and welcome to M.SE. Questions posed as:-"This is an exercise: solve it for me" are poorly received from other members. I you want to help other members help you, provide context for your question: for example, what did you tried to solve it? And where you get stuck?
$endgroup$
– Daniele Tampieri
Jan 7 at 6:57
$begingroup$
A complete qustion would include: 1. the book's mention that " In results like this, C and c will always denote some positive absolute constants", and 2. your own attempts. Where are they?
$endgroup$
– Did
Jan 7 at 7:10
$begingroup$
changes have been made. any suggestions?
$endgroup$
– Dylon
Jan 7 at 7:59