Mixing
Centered and bounded implies subgaussian
$begingroup$
There's a result that any $B$-bounded centered random variable $X$ (i.e., $mathbb{E}(X)=0$ and $|X|<B$) is sub-Gaussian with parameter $B sqrt{2 pi}$.
Does it still true in $n$-dimensions? If yes can we calculate its parameter?
probability probability-theory statistics random-variables cryptography
asked Jan 18 at 12:42
C.S.C.S.
235
$endgroup$
add a comment |
$begingroup$
There's a result that any $B$-bounded centered random variable $X$ (i.e., $mathbb{E}(X)=0$ and $|X|<B$) is sub-Gaussian with parameter $B sqrt{2 pi}$.
Does it still true in $n$-dimensions? If yes can we calculate its parameter?
probability probability-theory statistics random-variables cryptography
asked Jan 18 at 12:42
C.S.C.S.
235
$endgroup$
$begingroup$
what is your definition of sub-gaussianity for higher dimensions ? Something like there exists $C$, $v$ such that $mathbb P[lVert X rVert>t]leq C e^{-v t^2}$. Or maybe being sub-gaussian component wise ?
$endgroup$
– P. Quinton
Jan 18 at 12:47
$begingroup$
@P.Quinton . In higher dimension, $X$ is sub-Gaussian if $langle X, mathbf{u} rangle$ is sub-Gaussian, for all unit vectors $mathbf{u}$ .
$endgroup$
– C.S.
Jan 18 at 12:50
$begingroup$
what would be the parameter then ?
$endgroup$
– P. Quinton
Jan 18 at 13:22
add a comment |
$begingroup$
There's a result that any $B$-bounded centered random variable $X$ (i.e., $mathbb{E}(X)=0$ and $|X|<B$) is sub-Gaussian with parameter $B sqrt{2 pi}$.
Does it still true in $n$-dimensions? If yes can we calculate its parameter?
probability probability-theory statistics random-variables cryptography
asked Jan 18 at 12:42
C.S.C.S.
235
$endgroup$
There's a result that any $B$-bounded centered random variable $X$ (i.e., $mathbb{E}(X)=0$ and $|X|<B$) is sub-Gaussian with parameter $B sqrt{2 pi}$.
Does it still true in $n$-dimensions? If yes can we calculate its parameter?
probability probability-theory statistics random-variables cryptography
probability probability-theory statistics random-variables cryptography
asked Jan 18 at 12:42
C.S.C.S.
235
asked Jan 18 at 12:42
C.S.C.S.
235
asked Jan 18 at 12:42
C.S.C.S.
235
asked Jan 18 at 12:42
C.S.C.S.
235
asked Jan 18 at 12:42
C.S.C.S.
235
235
$begingroup$
what is your definition of sub-gaussianity for higher dimensions ? Something like there exists $C$, $v$ such that $mathbb P[lVert X rVert>t]leq C e^{-v t^2}$. Or maybe being sub-gaussian component wise ?
$endgroup$
– P. Quinton
Jan 18 at 12:47
$begingroup$
@P.Quinton . In higher dimension, $X$ is sub-Gaussian if $langle X, mathbf{u} rangle$ is sub-Gaussian, for all unit vectors $mathbf{u}$ .
$endgroup$
– C.S.
Jan 18 at 12:50
$begingroup$
what would be the parameter then ?
$endgroup$
– P. Quinton
Jan 18 at 13:22
add a comment |
$begingroup$
what is your definition of sub-gaussianity for higher dimensions ? Something like there exists $C$, $v$ such that $mathbb P[lVert X rVert>t]leq C e^{-v t^2}$. Or maybe being sub-gaussian component wise ?
$endgroup$
– P. Quinton
Jan 18 at 12:47
$begingroup$
@P.Quinton . In higher dimension, $X$ is sub-Gaussian if $langle X, mathbf{u} rangle$ is sub-Gaussian, for all unit vectors $mathbf{u}$ .
$endgroup$
– C.S.
Jan 18 at 12:50
$begingroup$
what would be the parameter then ?
$endgroup$
– P. Quinton
Jan 18 at 13:22
$begingroup$
what is your definition of sub-gaussianity for higher dimensions ? Something like there exists $C$, $v$ such that $mathbb P[lVert X rVert>t]leq C e^{-v t^2}$. Or maybe being sub-gaussian component wise ?
$endgroup$
– P. Quinton
Jan 18 at 12:47
$begingroup$
what is your definition of sub-gaussianity for higher dimensions ? Something like there exists $C$, $v$ such that $mathbb P[lVert X rVert>t]leq C e^{-v t^2}$. Or maybe being sub-gaussian component wise ?
$endgroup$
– P. Quinton
Jan 18 at 12:47
$begingroup$
@P.Quinton . In higher dimension, $X$ is sub-Gaussian if $langle X, mathbf{u} rangle$ is sub-Gaussian, for all unit vectors $mathbf{u}$ .
$endgroup$
– C.S.
Jan 18 at 12:50
$begingroup$
@P.Quinton . In higher dimension, $X$ is sub-Gaussian if $langle X, mathbf{u} rangle$ is sub-Gaussian, for all unit vectors $mathbf{u}$ .
$endgroup$
– C.S.
Jan 18 at 12:50
$begingroup$
what would be the parameter then ?
$endgroup$
– P. Quinton
Jan 18 at 13:22
$begingroup$
what would be the parameter then ?
$endgroup$
– P. Quinton
Jan 18 at 13:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose that $mathbf X=(X_1,dots,X_n)^T$ is bounded in the sense that for any $i$, $|X_i|leq B_i$ for some $mathbf B=(B_1,dots,B_n)^T$, suppose also that $mathbb E[mathbf X]=mathbf 0$.
Then for any unit vector $mathbf u$, $X_{mathbf u}=langle mathbf X,mathbf urangle$ is such that $mathbb E[X_{mathbf u}]=0$ by linearity and using triangular inequality
begin{align*}
|X_{mathbf u}|&leq sum_{i} |u_i| |X_i|\
&leq sum_i |u_i| B_i\
&:= B_{mathbf u}
end{align*}
Hence $X_{mathbf u}$ is sub-Gaussian with parameter $B_{mathbf u}sqrt{2pi}$.
Now I'm not sure about what the parameter is in your definition but I'm pretty sure it can be found easily at this point.
answered Jan 18 at 13:22
P. QuintonP. Quinton
1,8511213
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Suppose that $mathbf X=(X_1,dots,X_n)^T$ is bounded in the sense that for any $i$, $|X_i|leq B_i$ for some $mathbf B=(B_1,dots,B_n)^T$, suppose also that $mathbb E[mathbf X]=mathbf 0$.
Then for any unit vector $mathbf u$, $X_{mathbf u}=langle mathbf X,mathbf urangle$ is such that $mathbb E[X_{mathbf u}]=0$ by linearity and using triangular inequality
begin{align*}
|X_{mathbf u}|&leq sum_{i} |u_i| |X_i|\
&leq sum_i |u_i| B_i\
&:= B_{mathbf u}
end{align*}
Hence $X_{mathbf u}$ is sub-Gaussian with parameter $B_{mathbf u}sqrt{2pi}$.
Now I'm not sure about what the parameter is in your definition but I'm pretty sure it can be found easily at this point.
answered Jan 18 at 13:22
P. QuintonP. Quinton
1,8511213
$endgroup$
add a comment |
$begingroup$
Suppose that $mathbf X=(X_1,dots,X_n)^T$ is bounded in the sense that for any $i$, $|X_i|leq B_i$ for some $mathbf B=(B_1,dots,B_n)^T$, suppose also that $mathbb E[mathbf X]=mathbf 0$.
Then for any unit vector $mathbf u$, $X_{mathbf u}=langle mathbf X,mathbf urangle$ is such that $mathbb E[X_{mathbf u}]=0$ by linearity and using triangular inequality
begin{align*}
|X_{mathbf u}|&leq sum_{i} |u_i| |X_i|\
&leq sum_i |u_i| B_i\
&:= B_{mathbf u}
end{align*}
Hence $X_{mathbf u}$ is sub-Gaussian with parameter $B_{mathbf u}sqrt{2pi}$.
Now I'm not sure about what the parameter is in your definition but I'm pretty sure it can be found easily at this point.
answered Jan 18 at 13:22
P. QuintonP. Quinton
1,8511213
$endgroup$
add a comment |
$begingroup$
Suppose that $mathbf X=(X_1,dots,X_n)^T$ is bounded in the sense that for any $i$, $|X_i|leq B_i$ for some $mathbf B=(B_1,dots,B_n)^T$, suppose also that $mathbb E[mathbf X]=mathbf 0$.
Then for any unit vector $mathbf u$, $X_{mathbf u}=langle mathbf X,mathbf urangle$ is such that $mathbb E[X_{mathbf u}]=0$ by linearity and using triangular inequality
begin{align*}
|X_{mathbf u}|&leq sum_{i} |u_i| |X_i|\
&leq sum_i |u_i| B_i\
&:= B_{mathbf u}
end{align*}
Hence $X_{mathbf u}$ is sub-Gaussian with parameter $B_{mathbf u}sqrt{2pi}$.
Now I'm not sure about what the parameter is in your definition but I'm pretty sure it can be found easily at this point.
answered Jan 18 at 13:22
P. QuintonP. Quinton
1,8511213
$endgroup$
Suppose that $mathbf X=(X_1,dots,X_n)^T$ is bounded in the sense that for any $i$, $|X_i|leq B_i$ for some $mathbf B=(B_1,dots,B_n)^T$, suppose also that $mathbb E[mathbf X]=mathbf 0$.
Then for any unit vector $mathbf u$, $X_{mathbf u}=langle mathbf X,mathbf urangle$ is such that $mathbb E[X_{mathbf u}]=0$ by linearity and using triangular inequality
begin{align*}
|X_{mathbf u}|&leq sum_{i} |u_i| |X_i|\
&leq sum_i |u_i| B_i\
&:= B_{mathbf u}
end{align*}
Hence $X_{mathbf u}$ is sub-Gaussian with parameter $B_{mathbf u}sqrt{2pi}$.
Now I'm not sure about what the parameter is in your definition but I'm pretty sure it can be found easily at this point.
answered Jan 18 at 13:22
P. QuintonP. Quinton
1,8511213
answered Jan 18 at 13:22
P. QuintonP. Quinton
1,8511213
answered Jan 18 at 13:22
P. QuintonP. Quinton
1,8511213
answered Jan 18 at 13:22
P. QuintonP. Quinton
1,8511213
1,8511213
add a comment |
add a comment |
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$begingroup$
what is your definition of sub-gaussianity for higher dimensions ? Something like there exists $C$, $v$ such that $mathbb P[lVert X rVert>t]leq C e^{-v t^2}$. Or maybe being sub-gaussian component wise ?
$endgroup$
– P. Quinton
Jan 18 at 12:47
$begingroup$
@P.Quinton . In higher dimension, $X$ is sub-Gaussian if $langle X, mathbf{u} rangle$ is sub-Gaussian, for all unit vectors $mathbf{u}$ .
$endgroup$
– C.S.
Jan 18 at 12:50
$begingroup$
what would be the parameter then ?
$endgroup$
– P. Quinton
Jan 18 at 13:22