Concentration of Gaussian random matrices
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I know that if $X$ be a random matrix distributed as $N(0,I_n)$, then $frac{1}{n}X^TX$ is concentrated around the identity matrix. Does any body know the probability of $P(||frac{1}{n}X^TX-I||<delta)$?
probability-theory normal-distribution concentration-of-measure
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add a comment |
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I know that if $X$ be a random matrix distributed as $N(0,I_n)$, then $frac{1}{n}X^TX$ is concentrated around the identity matrix. Does any body know the probability of $P(||frac{1}{n}X^TX-I||<delta)$?
probability-theory normal-distribution concentration-of-measure
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1
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A relatively recent paper by Vershynin (arxiv/abs/1004.3484) shows how to answer this question. He's mainly interested in asymptotics, but working through the proof of Proposition 2.1 would give you the result you want.
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– cdipaolo
Nov 27 '18 at 4:05
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Another approach that would be looser yet would be probably a bit easier to compute is to apply Tropp's matrix concentration inequalities and bound the norm of a Gaussian vector with high probability. See Section 1.6.3 of arxiv/abs/1501.01571 for this.
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– cdipaolo
Nov 27 '18 at 4:18
add a comment |
$begingroup$
I know that if $X$ be a random matrix distributed as $N(0,I_n)$, then $frac{1}{n}X^TX$ is concentrated around the identity matrix. Does any body know the probability of $P(||frac{1}{n}X^TX-I||<delta)$?
probability-theory normal-distribution concentration-of-measure
$endgroup$
I know that if $X$ be a random matrix distributed as $N(0,I_n)$, then $frac{1}{n}X^TX$ is concentrated around the identity matrix. Does any body know the probability of $P(||frac{1}{n}X^TX-I||<delta)$?
probability-theory normal-distribution concentration-of-measure
probability-theory normal-distribution concentration-of-measure
asked Nov 27 '18 at 3:44
S_AlexS_Alex
1409
1409
1
$begingroup$
A relatively recent paper by Vershynin (arxiv/abs/1004.3484) shows how to answer this question. He's mainly interested in asymptotics, but working through the proof of Proposition 2.1 would give you the result you want.
$endgroup$
– cdipaolo
Nov 27 '18 at 4:05
$begingroup$
Another approach that would be looser yet would be probably a bit easier to compute is to apply Tropp's matrix concentration inequalities and bound the norm of a Gaussian vector with high probability. See Section 1.6.3 of arxiv/abs/1501.01571 for this.
$endgroup$
– cdipaolo
Nov 27 '18 at 4:18
add a comment |
1
$begingroup$
A relatively recent paper by Vershynin (arxiv/abs/1004.3484) shows how to answer this question. He's mainly interested in asymptotics, but working through the proof of Proposition 2.1 would give you the result you want.
$endgroup$
– cdipaolo
Nov 27 '18 at 4:05
$begingroup$
Another approach that would be looser yet would be probably a bit easier to compute is to apply Tropp's matrix concentration inequalities and bound the norm of a Gaussian vector with high probability. See Section 1.6.3 of arxiv/abs/1501.01571 for this.
$endgroup$
– cdipaolo
Nov 27 '18 at 4:18
1
1
$begingroup$
A relatively recent paper by Vershynin (arxiv/abs/1004.3484) shows how to answer this question. He's mainly interested in asymptotics, but working through the proof of Proposition 2.1 would give you the result you want.
$endgroup$
– cdipaolo
Nov 27 '18 at 4:05
$begingroup$
A relatively recent paper by Vershynin (arxiv/abs/1004.3484) shows how to answer this question. He's mainly interested in asymptotics, but working through the proof of Proposition 2.1 would give you the result you want.
$endgroup$
– cdipaolo
Nov 27 '18 at 4:05
$begingroup$
Another approach that would be looser yet would be probably a bit easier to compute is to apply Tropp's matrix concentration inequalities and bound the norm of a Gaussian vector with high probability. See Section 1.6.3 of arxiv/abs/1501.01571 for this.
$endgroup$
– cdipaolo
Nov 27 '18 at 4:18
$begingroup$
Another approach that would be looser yet would be probably a bit easier to compute is to apply Tropp's matrix concentration inequalities and bound the norm of a Gaussian vector with high probability. See Section 1.6.3 of arxiv/abs/1501.01571 for this.
$endgroup$
– cdipaolo
Nov 27 '18 at 4:18
add a comment |
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$begingroup$
A relatively recent paper by Vershynin (arxiv/abs/1004.3484) shows how to answer this question. He's mainly interested in asymptotics, but working through the proof of Proposition 2.1 would give you the result you want.
$endgroup$
– cdipaolo
Nov 27 '18 at 4:05
$begingroup$
Another approach that would be looser yet would be probably a bit easier to compute is to apply Tropp's matrix concentration inequalities and bound the norm of a Gaussian vector with high probability. See Section 1.6.3 of arxiv/abs/1501.01571 for this.
$endgroup$
– cdipaolo
Nov 27 '18 at 4:18