Concentration of norm
Multi tool use
Let $(X_1,...,X_n)in mathbb{R}^n$ be a random vector with independent sub-gaussian coordinates $X_i$ that satisfy $mathbb{E}X_i^2=1$. Then
$$||||X||_2-sqrt{n}||_{psi_2}leq CK^2$$,
Where $K=max||X_i||_{psi_2}$.
In the book "High dimensional probability", it is claimed that we can assume $Kgeq 1$ and $C$ is a universal constant.
But it is not clear why we can do so. Because the above expression is not homogeneous. I tried to change variables but if I change $X_i$ then the property of unit variance no longer holds.
probability random-variables normal-distribution
asked 2 days ago
S_AlexS_Alex
1159
add a comment |
Let $(X_1,...,X_n)in mathbb{R}^n$ be a random vector with independent sub-gaussian coordinates $X_i$ that satisfy $mathbb{E}X_i^2=1$. Then
$$||||X||_2-sqrt{n}||_{psi_2}leq CK^2$$,
Where $K=max||X_i||_{psi_2}$.
In the book "High dimensional probability", it is claimed that we can assume $Kgeq 1$ and $C$ is a universal constant.
But it is not clear why we can do so. Because the above expression is not homogeneous. I tried to change variables but if I change $X_i$ then the property of unit variance no longer holds.
probability random-variables normal-distribution
asked 2 days ago
S_AlexS_Alex
1159
First, what is $psi_2$? An Orlicz norm? Second, doesn't $K geq 1$ give a weaker inequality, so it can be assumed trivially? It's not clear to me what the question is.
– snarski
2 days ago
It is sub-gaussian norm. if you prove for $Kgeq1$, it is not clear for me how you can deduce the result for the case that maximum of sub-gaussian norms is less than 1, which is not trivial.
– S_Alex
2 days ago
add a comment |
Let $(X_1,...,X_n)in mathbb{R}^n$ be a random vector with independent sub-gaussian coordinates $X_i$ that satisfy $mathbb{E}X_i^2=1$. Then
$$||||X||_2-sqrt{n}||_{psi_2}leq CK^2$$,
Where $K=max||X_i||_{psi_2}$.
In the book "High dimensional probability", it is claimed that we can assume $Kgeq 1$ and $C$ is a universal constant.
But it is not clear why we can do so. Because the above expression is not homogeneous. I tried to change variables but if I change $X_i$ then the property of unit variance no longer holds.
probability random-variables normal-distribution
asked 2 days ago
S_AlexS_Alex
1159
Let $(X_1,...,X_n)in mathbb{R}^n$ be a random vector with independent sub-gaussian coordinates $X_i$ that satisfy $mathbb{E}X_i^2=1$. Then
$$||||X||_2-sqrt{n}||_{psi_2}leq CK^2$$,
Where $K=max||X_i||_{psi_2}$.
In the book "High dimensional probability", it is claimed that we can assume $Kgeq 1$ and $C$ is a universal constant.
But it is not clear why we can do so. Because the above expression is not homogeneous. I tried to change variables but if I change $X_i$ then the property of unit variance no longer holds.
probability random-variables normal-distribution
probability random-variables normal-distribution
asked 2 days ago
S_AlexS_Alex
1159
asked 2 days ago
S_AlexS_Alex
1159
edited 2 days ago
S_Alex
asked 2 days ago
S_AlexS_Alex
1159
asked 2 days ago
S_AlexS_Alex
1159
asked 2 days ago
S_AlexS_Alex
1159
1159
First, what is $psi_2$? An Orlicz norm? Second, doesn't $K geq 1$ give a weaker inequality, so it can be assumed trivially? It's not clear to me what the question is.
– snarski
2 days ago
It is sub-gaussian norm. if you prove for $Kgeq1$, it is not clear for me how you can deduce the result for the case that maximum of sub-gaussian norms is less than 1, which is not trivial.
– S_Alex
2 days ago
add a comment |
First, what is $psi_2$? An Orlicz norm? Second, doesn't $K geq 1$ give a weaker inequality, so it can be assumed trivially? It's not clear to me what the question is.
– snarski
2 days ago
It is sub-gaussian norm. if you prove for $Kgeq1$, it is not clear for me how you can deduce the result for the case that maximum of sub-gaussian norms is less than 1, which is not trivial.
– S_Alex
2 days ago
First, what is $psi_2$? An Orlicz norm? Second, doesn't $K geq 1$ give a weaker inequality, so it can be assumed trivially? It's not clear to me what the question is.
– snarski
2 days ago
First, what is $psi_2$? An Orlicz norm? Second, doesn't $K geq 1$ give a weaker inequality, so it can be assumed trivially? It's not clear to me what the question is.
– snarski
2 days ago
It is sub-gaussian norm. if you prove for $Kgeq1$, it is not clear for me how you can deduce the result for the case that maximum of sub-gaussian norms is less than 1, which is not trivial.
– S_Alex
2 days ago
It is sub-gaussian norm. if you prove for $Kgeq1$, it is not clear for me how you can deduce the result for the case that maximum of sub-gaussian norms is less than 1, which is not trivial.
– S_Alex
2 days ago
add a comment |
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First, what is $psi_2$? An Orlicz norm? Second, doesn't $K geq 1$ give a weaker inequality, so it can be assumed trivially? It's not clear to me what the question is.
– snarski
2 days ago
It is sub-gaussian norm. if you prove for $Kgeq1$, it is not clear for me how you can deduce the result for the case that maximum of sub-gaussian norms is less than 1, which is not trivial.
– S_Alex
2 days ago