Show that X is a sub-gaussian random vector with dependent sub-gaussian coordinates
$begingroup$
Let $X in R^n$ be a zero mean, random vector with sub-gaussian coordinates $X_i$.
prove that X is a sub-gaussian random vector no matter if coordinates are independent or dependent.
It is easy to prove the result in the case of independent coordinates.
When it comes to the case of dependent coordinates, I think of the definition of multivariate normal distribution but don't know if it works for sub-gaussian family.
Assume a random vector Z $in R^n$ has independent zero mean, unit variance, sub-gaussian coordinates and denote $Sigma_X$as the covariance matrix of X, we can find a Z such that $Sigma_X^{1/2} Z$ has the same distribution as X.
Because for $forall a in R^n$ $a^{T}Sigma_X^{1/2} in R^n$, given the case of independent coordinates, we can say for $forall a in R^n, a^{T}Sigma_X^{1/2}Z = a^{T}X$ is sub-gaussion distributed. So X is a sub-gaussian random vector.
I am not sure if the proof is right for the whole sub-gaussian family, because I am not sure we can find a Z that $Sigma_X^{1/2} Z$ is distributed same as X.
Any suggestions and ideas?
probability statistics normal-distribution
asked Jan 13 at 18:32
DylonDylon
63
$endgroup$
add a comment |
$begingroup$
Let $X in R^n$ be a zero mean, random vector with sub-gaussian coordinates $X_i$.
prove that X is a sub-gaussian random vector no matter if coordinates are independent or dependent.
It is easy to prove the result in the case of independent coordinates.
When it comes to the case of dependent coordinates, I think of the definition of multivariate normal distribution but don't know if it works for sub-gaussian family.
Assume a random vector Z $in R^n$ has independent zero mean, unit variance, sub-gaussian coordinates and denote $Sigma_X$as the covariance matrix of X, we can find a Z such that $Sigma_X^{1/2} Z$ has the same distribution as X.
Because for $forall a in R^n$ $a^{T}Sigma_X^{1/2} in R^n$, given the case of independent coordinates, we can say for $forall a in R^n, a^{T}Sigma_X^{1/2}Z = a^{T}X$ is sub-gaussion distributed. So X is a sub-gaussian random vector.
I am not sure if the proof is right for the whole sub-gaussian family, because I am not sure we can find a Z that $Sigma_X^{1/2} Z$ is distributed same as X.
Any suggestions and ideas?
probability statistics normal-distribution
asked Jan 13 at 18:32
DylonDylon
63
$endgroup$
add a comment |
$begingroup$
Let $X in R^n$ be a zero mean, random vector with sub-gaussian coordinates $X_i$.
prove that X is a sub-gaussian random vector no matter if coordinates are independent or dependent.
It is easy to prove the result in the case of independent coordinates.
When it comes to the case of dependent coordinates, I think of the definition of multivariate normal distribution but don't know if it works for sub-gaussian family.
Assume a random vector Z $in R^n$ has independent zero mean, unit variance, sub-gaussian coordinates and denote $Sigma_X$as the covariance matrix of X, we can find a Z such that $Sigma_X^{1/2} Z$ has the same distribution as X.
Because for $forall a in R^n$ $a^{T}Sigma_X^{1/2} in R^n$, given the case of independent coordinates, we can say for $forall a in R^n, a^{T}Sigma_X^{1/2}Z = a^{T}X$ is sub-gaussion distributed. So X is a sub-gaussian random vector.
I am not sure if the proof is right for the whole sub-gaussian family, because I am not sure we can find a Z that $Sigma_X^{1/2} Z$ is distributed same as X.
Any suggestions and ideas?
probability statistics normal-distribution
asked Jan 13 at 18:32
DylonDylon
63
$endgroup$
Let $X in R^n$ be a zero mean, random vector with sub-gaussian coordinates $X_i$.
prove that X is a sub-gaussian random vector no matter if coordinates are independent or dependent.
It is easy to prove the result in the case of independent coordinates.
When it comes to the case of dependent coordinates, I think of the definition of multivariate normal distribution but don't know if it works for sub-gaussian family.
Assume a random vector Z $in R^n$ has independent zero mean, unit variance, sub-gaussian coordinates and denote $Sigma_X$as the covariance matrix of X, we can find a Z such that $Sigma_X^{1/2} Z$ has the same distribution as X.
Because for $forall a in R^n$ $a^{T}Sigma_X^{1/2} in R^n$, given the case of independent coordinates, we can say for $forall a in R^n, a^{T}Sigma_X^{1/2}Z = a^{T}X$ is sub-gaussion distributed. So X is a sub-gaussian random vector.
I am not sure if the proof is right for the whole sub-gaussian family, because I am not sure we can find a Z that $Sigma_X^{1/2} Z$ is distributed same as X.
Any suggestions and ideas?
probability statistics normal-distribution
probability statistics normal-distribution
asked Jan 13 at 18:32
DylonDylon
63
asked Jan 13 at 18:32
DylonDylon
63
edited Jan 15 at 1:36
Dylon
asked Jan 13 at 18:32
DylonDylon
63
asked Jan 13 at 18:32
DylonDylon
63
asked Jan 13 at 18:32
DylonDylon
63
63
add a comment |
add a comment |
1 Answer
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$begingroup$
Let $X=(X_1,...X_n) in R^n$ be a random vector with sub-gaussians coordinates $X_i$. By definition it is sub-gaussian iff the random variable $langle X,xrangle$ is sub-gaussian for any $x in mathbb{R}^n$.
Consider $x in mathbb{R}^n$ We will show that $Y:= langle X,xrangle$ satisfies $$lVert YrVert_p leq K_2 sqrt{p} enspace forall pgeq 1$$
(one of the equivalent definitions for being sub-gaussian).
Indeed,
begin{align*}
lVert YrVert_p= lVert{displaystyle sum_{i=1}^nx_iX_i}rVert_p leqsum_{i=1}^n|x_i|lVert X_i rVert_p\
leq (sum_{i=1}^n|x_i|K_{2i})sqrt{p} && (lVert X_i rVert_p leq K_{2i}sqrt{p}, enspace text{since $X_i$ are sub-gaussian}) \
implieslVert YrVert_p leq Ksqrt{p} enspace forall pgeq 1 && (text{ where $K=sum_{i=1}^n|x_i|K_{2i}$})
end{align*}
I used prop 2.5.2 and definition 3.4.1 from the following book of Roman Vershynin High-Dimensional Probability
answered Apr 14 at 22:44
sakassakas
12
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Check out our Code of Conduct.
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1 Answer
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$begingroup$
Let $X=(X_1,...X_n) in R^n$ be a random vector with sub-gaussians coordinates $X_i$. By definition it is sub-gaussian iff the random variable $langle X,xrangle$ is sub-gaussian for any $x in mathbb{R}^n$.
Consider $x in mathbb{R}^n$ We will show that $Y:= langle X,xrangle$ satisfies $$lVert YrVert_p leq K_2 sqrt{p} enspace forall pgeq 1$$
(one of the equivalent definitions for being sub-gaussian).
Indeed,
begin{align*}
lVert YrVert_p= lVert{displaystyle sum_{i=1}^nx_iX_i}rVert_p leqsum_{i=1}^n|x_i|lVert X_i rVert_p\
leq (sum_{i=1}^n|x_i|K_{2i})sqrt{p} && (lVert X_i rVert_p leq K_{2i}sqrt{p}, enspace text{since $X_i$ are sub-gaussian}) \
implieslVert YrVert_p leq Ksqrt{p} enspace forall pgeq 1 && (text{ where $K=sum_{i=1}^n|x_i|K_{2i}$})
end{align*}
I used prop 2.5.2 and definition 3.4.1 from the following book of Roman Vershynin High-Dimensional Probability
answered Apr 14 at 22:44
sakassakas
12
New contributor
sakas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Let $X=(X_1,...X_n) in R^n$ be a random vector with sub-gaussians coordinates $X_i$. By definition it is sub-gaussian iff the random variable $langle X,xrangle$ is sub-gaussian for any $x in mathbb{R}^n$.
Consider $x in mathbb{R}^n$ We will show that $Y:= langle X,xrangle$ satisfies $$lVert YrVert_p leq K_2 sqrt{p} enspace forall pgeq 1$$
(one of the equivalent definitions for being sub-gaussian).
Indeed,
begin{align*}
lVert YrVert_p= lVert{displaystyle sum_{i=1}^nx_iX_i}rVert_p leqsum_{i=1}^n|x_i|lVert X_i rVert_p\
leq (sum_{i=1}^n|x_i|K_{2i})sqrt{p} && (lVert X_i rVert_p leq K_{2i}sqrt{p}, enspace text{since $X_i$ are sub-gaussian}) \
implieslVert YrVert_p leq Ksqrt{p} enspace forall pgeq 1 && (text{ where $K=sum_{i=1}^n|x_i|K_{2i}$})
end{align*}
I used prop 2.5.2 and definition 3.4.1 from the following book of Roman Vershynin High-Dimensional Probability
answered Apr 14 at 22:44
sakassakas
12
New contributor
sakas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Let $X=(X_1,...X_n) in R^n$ be a random vector with sub-gaussians coordinates $X_i$. By definition it is sub-gaussian iff the random variable $langle X,xrangle$ is sub-gaussian for any $x in mathbb{R}^n$.
Consider $x in mathbb{R}^n$ We will show that $Y:= langle X,xrangle$ satisfies $$lVert YrVert_p leq K_2 sqrt{p} enspace forall pgeq 1$$
(one of the equivalent definitions for being sub-gaussian).
Indeed,
begin{align*}
lVert YrVert_p= lVert{displaystyle sum_{i=1}^nx_iX_i}rVert_p leqsum_{i=1}^n|x_i|lVert X_i rVert_p\
leq (sum_{i=1}^n|x_i|K_{2i})sqrt{p} && (lVert X_i rVert_p leq K_{2i}sqrt{p}, enspace text{since $X_i$ are sub-gaussian}) \
implieslVert YrVert_p leq Ksqrt{p} enspace forall pgeq 1 && (text{ where $K=sum_{i=1}^n|x_i|K_{2i}$})
end{align*}
I used prop 2.5.2 and definition 3.4.1 from the following book of Roman Vershynin High-Dimensional Probability
answered Apr 14 at 22:44
sakassakas
12
New contributor
sakas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let $X=(X_1,...X_n) in R^n$ be a random vector with sub-gaussians coordinates $X_i$. By definition it is sub-gaussian iff the random variable $langle X,xrangle$ is sub-gaussian for any $x in mathbb{R}^n$.
Consider $x in mathbb{R}^n$ We will show that $Y:= langle X,xrangle$ satisfies $$lVert YrVert_p leq K_2 sqrt{p} enspace forall pgeq 1$$
(one of the equivalent definitions for being sub-gaussian).
Indeed,
begin{align*}
lVert YrVert_p= lVert{displaystyle sum_{i=1}^nx_iX_i}rVert_p leqsum_{i=1}^n|x_i|lVert X_i rVert_p\
leq (sum_{i=1}^n|x_i|K_{2i})sqrt{p} && (lVert X_i rVert_p leq K_{2i}sqrt{p}, enspace text{since $X_i$ are sub-gaussian}) \
implieslVert YrVert_p leq Ksqrt{p} enspace forall pgeq 1 && (text{ where $K=sum_{i=1}^n|x_i|K_{2i}$})
end{align*}
I used prop 2.5.2 and definition 3.4.1 from the following book of Roman Vershynin High-Dimensional Probability
answered Apr 14 at 22:44
sakassakas
12
New contributor
sakas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 14 at 22:51
answered Apr 14 at 22:44
sakassakas
12
New contributor
sakas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 14 at 22:44
sakassakas
12
answered Apr 14 at 22:44
sakassakas
12
12
New contributor
sakas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
sakas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
sakas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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