Measuring “concentration” in an expansion
$begingroup$
Suppose we have a vector $v in mathbb{R}^{n}$ which we expand in some orthonormal basis ${g_{m}}$ of $mathbb{R}^{n}$:
begin{align*}
v = sum_{m=1}^{n} a_{m}g_{m}
end{align*}
I want to measure how "concentrated" $v$ is in the basis ${g_{m}}$. To wit, I would like some measurement that is maximized when $a_{m} = 0$ for all but one $m$, and minimized when all the $a_{m}$ have the same magnitude. Is there a nice way to measure this "concentration" (preferably easily computed), and is there standard terminology for it?
vectors linear-transformations
asked Dec 4 '18 at 20:49
user14717user14717
3,8281020
$endgroup$
|
show 1 more comment
$begingroup$
Suppose we have a vector $v in mathbb{R}^{n}$ which we expand in some orthonormal basis ${g_{m}}$ of $mathbb{R}^{n}$:
begin{align*}
v = sum_{m=1}^{n} a_{m}g_{m}
end{align*}
I want to measure how "concentrated" $v$ is in the basis ${g_{m}}$. To wit, I would like some measurement that is maximized when $a_{m} = 0$ for all but one $m$, and minimized when all the $a_{m}$ have the same magnitude. Is there a nice way to measure this "concentration" (preferably easily computed), and is there standard terminology for it?
vectors linear-transformations
asked Dec 4 '18 at 20:49
user14717user14717
3,8281020
$endgroup$
$begingroup$
Fix $p<q$. Then $|a|_q/|a|_p$ has the extrema that you want
$endgroup$
– Federico
Dec 4 '18 at 20:57
$begingroup$
Alternatively, you could consider $p_i=a_i^2/|a|_2^2$ and take the Shannon entropy of the $p_i$'s
$endgroup$
– Federico
Dec 4 '18 at 20:59
$begingroup$
After asking this question, I found this paper: ieeexplore.ieee.org/document/5238742
$endgroup$
– user14717
Dec 5 '18 at 0:27
$begingroup$
@Federico: Your answer is basically summarized in the linked paper, so it's good. Do you wish to move your comments to an answer?
$endgroup$
– user14717
Dec 5 '18 at 0:44
$begingroup$
Unfortunately I don't have access to the paper. I don't mind if you post your own answer and accept it. By the way, I'm curious to know why the Shannon entropy is an inferior choice, based on their "Robin Hood, Scaling, Rising Tide, Cloning, Bill Gates, and Babies" properties
$endgroup$
– Federico
Dec 5 '18 at 14:28
|
show 1 more comment
$begingroup$
Suppose we have a vector $v in mathbb{R}^{n}$ which we expand in some orthonormal basis ${g_{m}}$ of $mathbb{R}^{n}$:
begin{align*}
v = sum_{m=1}^{n} a_{m}g_{m}
end{align*}
I want to measure how "concentrated" $v$ is in the basis ${g_{m}}$. To wit, I would like some measurement that is maximized when $a_{m} = 0$ for all but one $m$, and minimized when all the $a_{m}$ have the same magnitude. Is there a nice way to measure this "concentration" (preferably easily computed), and is there standard terminology for it?
vectors linear-transformations
asked Dec 4 '18 at 20:49
user14717user14717
3,8281020
$endgroup$
Suppose we have a vector $v in mathbb{R}^{n}$ which we expand in some orthonormal basis ${g_{m}}$ of $mathbb{R}^{n}$:
begin{align*}
v = sum_{m=1}^{n} a_{m}g_{m}
end{align*}
I want to measure how "concentrated" $v$ is in the basis ${g_{m}}$. To wit, I would like some measurement that is maximized when $a_{m} = 0$ for all but one $m$, and minimized when all the $a_{m}$ have the same magnitude. Is there a nice way to measure this "concentration" (preferably easily computed), and is there standard terminology for it?
vectors linear-transformations
vectors linear-transformations
asked Dec 4 '18 at 20:49
user14717user14717
3,8281020
asked Dec 4 '18 at 20:49
user14717user14717
3,8281020
edited Dec 4 '18 at 20:56
user14717
asked Dec 4 '18 at 20:49
user14717user14717
3,8281020
asked Dec 4 '18 at 20:49
user14717user14717
3,8281020
asked Dec 4 '18 at 20:49
user14717user14717
3,8281020
3,8281020
$begingroup$
Fix $p<q$. Then $|a|_q/|a|_p$ has the extrema that you want
$endgroup$
– Federico
Dec 4 '18 at 20:57
$begingroup$
Alternatively, you could consider $p_i=a_i^2/|a|_2^2$ and take the Shannon entropy of the $p_i$'s
$endgroup$
– Federico
Dec 4 '18 at 20:59
$begingroup$
After asking this question, I found this paper: ieeexplore.ieee.org/document/5238742
$endgroup$
– user14717
Dec 5 '18 at 0:27
$begingroup$
@Federico: Your answer is basically summarized in the linked paper, so it's good. Do you wish to move your comments to an answer?
$endgroup$
– user14717
Dec 5 '18 at 0:44
$begingroup$
Unfortunately I don't have access to the paper. I don't mind if you post your own answer and accept it. By the way, I'm curious to know why the Shannon entropy is an inferior choice, based on their "Robin Hood, Scaling, Rising Tide, Cloning, Bill Gates, and Babies" properties
$endgroup$
– Federico
Dec 5 '18 at 14:28
|
show 1 more comment
$begingroup$
Fix $p<q$. Then $|a|_q/|a|_p$ has the extrema that you want
$endgroup$
– Federico
Dec 4 '18 at 20:57
$begingroup$
Alternatively, you could consider $p_i=a_i^2/|a|_2^2$ and take the Shannon entropy of the $p_i$'s
$endgroup$
– Federico
Dec 4 '18 at 20:59
$begingroup$
After asking this question, I found this paper: ieeexplore.ieee.org/document/5238742
$endgroup$
– user14717
Dec 5 '18 at 0:27
$begingroup$
@Federico: Your answer is basically summarized in the linked paper, so it's good. Do you wish to move your comments to an answer?
$endgroup$
– user14717
Dec 5 '18 at 0:44
$begingroup$
Unfortunately I don't have access to the paper. I don't mind if you post your own answer and accept it. By the way, I'm curious to know why the Shannon entropy is an inferior choice, based on their "Robin Hood, Scaling, Rising Tide, Cloning, Bill Gates, and Babies" properties
$endgroup$
– Federico
Dec 5 '18 at 14:28
$begingroup$
Fix $p<q$. Then $|a|_q/|a|_p$ has the extrema that you want
$endgroup$
– Federico
Dec 4 '18 at 20:57
$begingroup$
Fix $p<q$. Then $|a|_q/|a|_p$ has the extrema that you want
$endgroup$
– Federico
Dec 4 '18 at 20:57
$begingroup$
Alternatively, you could consider $p_i=a_i^2/|a|_2^2$ and take the Shannon entropy of the $p_i$'s
$endgroup$
– Federico
Dec 4 '18 at 20:59
$begingroup$
Alternatively, you could consider $p_i=a_i^2/|a|_2^2$ and take the Shannon entropy of the $p_i$'s
$endgroup$
– Federico
Dec 4 '18 at 20:59
$begingroup$
After asking this question, I found this paper: ieeexplore.ieee.org/document/5238742
$endgroup$
– user14717
Dec 5 '18 at 0:27
$begingroup$
After asking this question, I found this paper: ieeexplore.ieee.org/document/5238742
$endgroup$
– user14717
Dec 5 '18 at 0:27
$begingroup$
@Federico: Your answer is basically summarized in the linked paper, so it's good. Do you wish to move your comments to an answer?
$endgroup$
– user14717
Dec 5 '18 at 0:44
$begingroup$
@Federico: Your answer is basically summarized in the linked paper, so it's good. Do you wish to move your comments to an answer?
$endgroup$
– user14717
Dec 5 '18 at 0:44
$begingroup$
Unfortunately I don't have access to the paper. I don't mind if you post your own answer and accept it. By the way, I'm curious to know why the Shannon entropy is an inferior choice, based on their "Robin Hood, Scaling, Rising Tide, Cloning, Bill Gates, and Babies" properties
$endgroup$
– Federico
Dec 5 '18 at 14:28
$begingroup$
Unfortunately I don't have access to the paper. I don't mind if you post your own answer and accept it. By the way, I'm curious to know why the Shannon entropy is an inferior choice, based on their "Robin Hood, Scaling, Rising Tide, Cloning, Bill Gates, and Babies" properties
$endgroup$
– Federico
Dec 5 '18 at 14:28
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
The Gini coefficient is an excellent way to measurement sparsity. Here, we use the following definition of Gini coefficient:
begin{align*}
G(v) := frac{1}{n-1}left[ frac{2}{left| vright|_{1} } sum_{i=0}^{n-1} (i+1)|v_i^{(s)}| - n - 1 right]
end{align*}
where $v_{i}^{(s)}$ is the $i$th coefficient of the vector $v$ sorted by increasing magnitude. Note that this definition differs from (say) Wikipedia's definition, which only applies it to wealth inequality, where each element is real and non-negative.
We have some nice properties of $G$:
If $v_{j} mapsto e^{itheta_j}v_{j}$, we feel intuitively that the sparsity is unchanged, and indeed $G(v)$ is invariant under these phase changes.
If $0 ne lambda in mathbb{C}$, it is easy to see that $G(lambda v) = G(v)$ (changing units doesn't change sparsity).
If $v_{j} = 0$ for all but one $j$, then $G(v) = 1$, which is the maxima of $G$. If all coefficients are of equal magnitude, then $G(v) = 0$, which is the minima of $G$.
If $Pi$ is a permutation of the elements of $v$, the $G(Pi v) = G(v)$. (Permutations don't affect sparsity.) The particular way that the Gini coefficient achieves this is via sorting, which is not ideal due to it requiring $mathcal{O}(nlog n)$ operations, but the invariance is indeed manifest.
The property of the Gini coefficient not shared by the Hoyer measure (a normalized ratio of $ell_1$ and $ell_2$ norms) is its invariance under concatenation: $G(voplus v) = G(v)$. How much weight should be given to this property is unclear, especially given the sorting requirement.
answered Dec 7 '18 at 6:05
user14717user14717
3,8281020
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The Gini coefficient is an excellent way to measurement sparsity. Here, we use the following definition of Gini coefficient:
begin{align*}
G(v) := frac{1}{n-1}left[ frac{2}{left| vright|_{1} } sum_{i=0}^{n-1} (i+1)|v_i^{(s)}| - n - 1 right]
end{align*}
where $v_{i}^{(s)}$ is the $i$th coefficient of the vector $v$ sorted by increasing magnitude. Note that this definition differs from (say) Wikipedia's definition, which only applies it to wealth inequality, where each element is real and non-negative.
We have some nice properties of $G$:
If $v_{j} mapsto e^{itheta_j}v_{j}$, we feel intuitively that the sparsity is unchanged, and indeed $G(v)$ is invariant under these phase changes.
If $0 ne lambda in mathbb{C}$, it is easy to see that $G(lambda v) = G(v)$ (changing units doesn't change sparsity).
If $v_{j} = 0$ for all but one $j$, then $G(v) = 1$, which is the maxima of $G$. If all coefficients are of equal magnitude, then $G(v) = 0$, which is the minima of $G$.
If $Pi$ is a permutation of the elements of $v$, the $G(Pi v) = G(v)$. (Permutations don't affect sparsity.) The particular way that the Gini coefficient achieves this is via sorting, which is not ideal due to it requiring $mathcal{O}(nlog n)$ operations, but the invariance is indeed manifest.
The property of the Gini coefficient not shared by the Hoyer measure (a normalized ratio of $ell_1$ and $ell_2$ norms) is its invariance under concatenation: $G(voplus v) = G(v)$. How much weight should be given to this property is unclear, especially given the sorting requirement.
answered Dec 7 '18 at 6:05
user14717user14717
3,8281020
$endgroup$
add a comment |
$begingroup$
The Gini coefficient is an excellent way to measurement sparsity. Here, we use the following definition of Gini coefficient:
begin{align*}
G(v) := frac{1}{n-1}left[ frac{2}{left| vright|_{1} } sum_{i=0}^{n-1} (i+1)|v_i^{(s)}| - n - 1 right]
end{align*}
where $v_{i}^{(s)}$ is the $i$th coefficient of the vector $v$ sorted by increasing magnitude. Note that this definition differs from (say) Wikipedia's definition, which only applies it to wealth inequality, where each element is real and non-negative.
We have some nice properties of $G$:
If $v_{j} mapsto e^{itheta_j}v_{j}$, we feel intuitively that the sparsity is unchanged, and indeed $G(v)$ is invariant under these phase changes.
If $0 ne lambda in mathbb{C}$, it is easy to see that $G(lambda v) = G(v)$ (changing units doesn't change sparsity).
If $v_{j} = 0$ for all but one $j$, then $G(v) = 1$, which is the maxima of $G$. If all coefficients are of equal magnitude, then $G(v) = 0$, which is the minima of $G$.
If $Pi$ is a permutation of the elements of $v$, the $G(Pi v) = G(v)$. (Permutations don't affect sparsity.) The particular way that the Gini coefficient achieves this is via sorting, which is not ideal due to it requiring $mathcal{O}(nlog n)$ operations, but the invariance is indeed manifest.
The property of the Gini coefficient not shared by the Hoyer measure (a normalized ratio of $ell_1$ and $ell_2$ norms) is its invariance under concatenation: $G(voplus v) = G(v)$. How much weight should be given to this property is unclear, especially given the sorting requirement.
answered Dec 7 '18 at 6:05
user14717user14717
3,8281020
$endgroup$
add a comment |
$begingroup$
The Gini coefficient is an excellent way to measurement sparsity. Here, we use the following definition of Gini coefficient:
begin{align*}
G(v) := frac{1}{n-1}left[ frac{2}{left| vright|_{1} } sum_{i=0}^{n-1} (i+1)|v_i^{(s)}| - n - 1 right]
end{align*}
where $v_{i}^{(s)}$ is the $i$th coefficient of the vector $v$ sorted by increasing magnitude. Note that this definition differs from (say) Wikipedia's definition, which only applies it to wealth inequality, where each element is real and non-negative.
We have some nice properties of $G$:
If $v_{j} mapsto e^{itheta_j}v_{j}$, we feel intuitively that the sparsity is unchanged, and indeed $G(v)$ is invariant under these phase changes.
If $0 ne lambda in mathbb{C}$, it is easy to see that $G(lambda v) = G(v)$ (changing units doesn't change sparsity).
If $v_{j} = 0$ for all but one $j$, then $G(v) = 1$, which is the maxima of $G$. If all coefficients are of equal magnitude, then $G(v) = 0$, which is the minima of $G$.
If $Pi$ is a permutation of the elements of $v$, the $G(Pi v) = G(v)$. (Permutations don't affect sparsity.) The particular way that the Gini coefficient achieves this is via sorting, which is not ideal due to it requiring $mathcal{O}(nlog n)$ operations, but the invariance is indeed manifest.
The property of the Gini coefficient not shared by the Hoyer measure (a normalized ratio of $ell_1$ and $ell_2$ norms) is its invariance under concatenation: $G(voplus v) = G(v)$. How much weight should be given to this property is unclear, especially given the sorting requirement.
answered Dec 7 '18 at 6:05
user14717user14717
3,8281020
$endgroup$
The Gini coefficient is an excellent way to measurement sparsity. Here, we use the following definition of Gini coefficient:
begin{align*}
G(v) := frac{1}{n-1}left[ frac{2}{left| vright|_{1} } sum_{i=0}^{n-1} (i+1)|v_i^{(s)}| - n - 1 right]
end{align*}
where $v_{i}^{(s)}$ is the $i$th coefficient of the vector $v$ sorted by increasing magnitude. Note that this definition differs from (say) Wikipedia's definition, which only applies it to wealth inequality, where each element is real and non-negative.
We have some nice properties of $G$:
If $v_{j} mapsto e^{itheta_j}v_{j}$, we feel intuitively that the sparsity is unchanged, and indeed $G(v)$ is invariant under these phase changes.
If $0 ne lambda in mathbb{C}$, it is easy to see that $G(lambda v) = G(v)$ (changing units doesn't change sparsity).
If $v_{j} = 0$ for all but one $j$, then $G(v) = 1$, which is the maxima of $G$. If all coefficients are of equal magnitude, then $G(v) = 0$, which is the minima of $G$.
If $Pi$ is a permutation of the elements of $v$, the $G(Pi v) = G(v)$. (Permutations don't affect sparsity.) The particular way that the Gini coefficient achieves this is via sorting, which is not ideal due to it requiring $mathcal{O}(nlog n)$ operations, but the invariance is indeed manifest.
The property of the Gini coefficient not shared by the Hoyer measure (a normalized ratio of $ell_1$ and $ell_2$ norms) is its invariance under concatenation: $G(voplus v) = G(v)$. How much weight should be given to this property is unclear, especially given the sorting requirement.
answered Dec 7 '18 at 6:05
user14717user14717
3,8281020
answered Dec 7 '18 at 6:05
user14717user14717
3,8281020
answered Dec 7 '18 at 6:05
user14717user14717
3,8281020
answered Dec 7 '18 at 6:05
user14717user14717
3,8281020
3,8281020
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$begingroup$
Fix $p<q$. Then $|a|_q/|a|_p$ has the extrema that you want
$endgroup$
– Federico
Dec 4 '18 at 20:57
$begingroup$
Alternatively, you could consider $p_i=a_i^2/|a|_2^2$ and take the Shannon entropy of the $p_i$'s
$endgroup$
– Federico
Dec 4 '18 at 20:59
$begingroup$
After asking this question, I found this paper: ieeexplore.ieee.org/document/5238742
$endgroup$
– user14717
Dec 5 '18 at 0:27
$begingroup$
@Federico: Your answer is basically summarized in the linked paper, so it's good. Do you wish to move your comments to an answer?
$endgroup$
– user14717
Dec 5 '18 at 0:44
$begingroup$
Unfortunately I don't have access to the paper. I don't mind if you post your own answer and accept it. By the way, I'm curious to know why the Shannon entropy is an inferior choice, based on their "Robin Hood, Scaling, Rising Tide, Cloning, Bill Gates, and Babies" properties
$endgroup$
– Federico
Dec 5 '18 at 14:28