Meaning of Delta Notation
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I was reading a paper (https://arxiv.org/pdf/1609.03891.pdf), and on the first page the author's write the following line:
$$
mu^sigma = frac{1}{n} sum_{i = 1}^n delta_{left(frac{2i}{n} - 1, frac{2sigma(i)}{n} -1right)}.
$$
Here, $sigma$ is in the symmetric group on $n$ letters. The authors write that $mu^sigma$ is supposed to be a probability measure on $[-1, 1]^2$. I'm having trouble parsing the expression. Is the delta supposed to be the Kroenecker delta? I don't think so, because the subscript ${left(frac{2i}{n} - 1, frac{2sigma(i) }{n} -1right)}$ doesn't seem to make sense (it would be equivalent to saying $i = sigma(i)?$). Also, why is the domain $[-1, 1]^2$?
Could someone help me understand this expression?
probability probability-theory notation
edited Dec 26 '18 at 23:55
MJD
47.3k29213396
asked Dec 26 '18 at 23:47
ProbablyProbably
111
$endgroup$
add a comment |
$begingroup$
I was reading a paper (https://arxiv.org/pdf/1609.03891.pdf), and on the first page the author's write the following line:
$$
mu^sigma = frac{1}{n} sum_{i = 1}^n delta_{left(frac{2i}{n} - 1, frac{2sigma(i)}{n} -1right)}.
$$
Here, $sigma$ is in the symmetric group on $n$ letters. The authors write that $mu^sigma$ is supposed to be a probability measure on $[-1, 1]^2$. I'm having trouble parsing the expression. Is the delta supposed to be the Kroenecker delta? I don't think so, because the subscript ${left(frac{2i}{n} - 1, frac{2sigma(i) }{n} -1right)}$ doesn't seem to make sense (it would be equivalent to saying $i = sigma(i)?$). Also, why is the domain $[-1, 1]^2$?
Could someone help me understand this expression?
probability probability-theory notation
edited Dec 26 '18 at 23:55
MJD
47.3k29213396
asked Dec 26 '18 at 23:47
ProbablyProbably
111
$endgroup$
add a comment |
$begingroup$
I was reading a paper (https://arxiv.org/pdf/1609.03891.pdf), and on the first page the author's write the following line:
$$
mu^sigma = frac{1}{n} sum_{i = 1}^n delta_{left(frac{2i}{n} - 1, frac{2sigma(i)}{n} -1right)}.
$$
Here, $sigma$ is in the symmetric group on $n$ letters. The authors write that $mu^sigma$ is supposed to be a probability measure on $[-1, 1]^2$. I'm having trouble parsing the expression. Is the delta supposed to be the Kroenecker delta? I don't think so, because the subscript ${left(frac{2i}{n} - 1, frac{2sigma(i) }{n} -1right)}$ doesn't seem to make sense (it would be equivalent to saying $i = sigma(i)?$). Also, why is the domain $[-1, 1]^2$?
Could someone help me understand this expression?
probability probability-theory notation
edited Dec 26 '18 at 23:55
MJD
47.3k29213396
asked Dec 26 '18 at 23:47
ProbablyProbably
111
$endgroup$
I was reading a paper (https://arxiv.org/pdf/1609.03891.pdf), and on the first page the author's write the following line:
$$
mu^sigma = frac{1}{n} sum_{i = 1}^n delta_{left(frac{2i}{n} - 1, frac{2sigma(i)}{n} -1right)}.
$$
Here, $sigma$ is in the symmetric group on $n$ letters. The authors write that $mu^sigma$ is supposed to be a probability measure on $[-1, 1]^2$. I'm having trouble parsing the expression. Is the delta supposed to be the Kroenecker delta? I don't think so, because the subscript ${left(frac{2i}{n} - 1, frac{2sigma(i) }{n} -1right)}$ doesn't seem to make sense (it would be equivalent to saying $i = sigma(i)?$). Also, why is the domain $[-1, 1]^2$?
Could someone help me understand this expression?
probability probability-theory notation
probability probability-theory notation
edited Dec 26 '18 at 23:55
MJD
47.3k29213396
asked Dec 26 '18 at 23:47
ProbablyProbably
111
edited Dec 26 '18 at 23:55
MJD
47.3k29213396
asked Dec 26 '18 at 23:47
ProbablyProbably
111
edited Dec 26 '18 at 23:55
MJD
47.3k29213396
edited Dec 26 '18 at 23:55
MJD
47.3k29213396
edited Dec 26 '18 at 23:55
MJD
47.3k29213396
47.3k29213396
asked Dec 26 '18 at 23:47
ProbablyProbably
111
asked Dec 26 '18 at 23:47
ProbablyProbably
111
asked Dec 26 '18 at 23:47
ProbablyProbably
111
111
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that
$$
{left(frac{2i}{n} - 1, frac{2sigma(i) }{n} -1right)}
$$
is a point in $[-1,1]^2$.
So $delta$ with that subscript is a unit point mass at that point.
Add $n$ point masses and divide by $n$, we get a probability measure.
edited Dec 26 '18 at 23:59
MJD
47.3k29213396
answered Dec 26 '18 at 23:55
GEdgarGEdgar
62.5k267171
$endgroup$
$begingroup$
That seems to be what the diagrams in the middle of page 4 are about; they are maps of where the point masses are.
$endgroup$
– MJD
Dec 26 '18 at 23:58
$begingroup$
@MJD So $mu^sigma$ places $n$ masses of $1/n$ in the square $[-1, 1]^2$ and assigns measure $0$ to every other point. Is that the right way to interpret the expression?
$endgroup$
– Probably
Dec 27 '18 at 0:01
$begingroup$
Yes. A measure is a function on sets. The pdf is a distribution. That's the Dirac delta in $mathbb{R}^2$ : $delta(x,y) = delta(x)delta(y)$ the latter being the usual Dirac delta in $mathbb{R}$ @Probably
$endgroup$
– reuns
Dec 27 '18 at 1:00
add a comment |
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1 Answer
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oldest
votes
1 Answer
1
active
oldest
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votes
$begingroup$
Note that
$$
{left(frac{2i}{n} - 1, frac{2sigma(i) }{n} -1right)}
$$
is a point in $[-1,1]^2$.
So $delta$ with that subscript is a unit point mass at that point.
Add $n$ point masses and divide by $n$, we get a probability measure.
edited Dec 26 '18 at 23:59
MJD
47.3k29213396
answered Dec 26 '18 at 23:55
GEdgarGEdgar
62.5k267171
$endgroup$
$begingroup$
That seems to be what the diagrams in the middle of page 4 are about; they are maps of where the point masses are.
$endgroup$
– MJD
Dec 26 '18 at 23:58
$begingroup$
@MJD So $mu^sigma$ places $n$ masses of $1/n$ in the square $[-1, 1]^2$ and assigns measure $0$ to every other point. Is that the right way to interpret the expression?
$endgroup$
– Probably
Dec 27 '18 at 0:01
$begingroup$
Yes. A measure is a function on sets. The pdf is a distribution. That's the Dirac delta in $mathbb{R}^2$ : $delta(x,y) = delta(x)delta(y)$ the latter being the usual Dirac delta in $mathbb{R}$ @Probably
$endgroup$
– reuns
Dec 27 '18 at 1:00
add a comment |
$begingroup$
Note that
$$
{left(frac{2i}{n} - 1, frac{2sigma(i) }{n} -1right)}
$$
is a point in $[-1,1]^2$.
So $delta$ with that subscript is a unit point mass at that point.
Add $n$ point masses and divide by $n$, we get a probability measure.
edited Dec 26 '18 at 23:59
MJD
47.3k29213396
answered Dec 26 '18 at 23:55
GEdgarGEdgar
62.5k267171
$endgroup$
$begingroup$
That seems to be what the diagrams in the middle of page 4 are about; they are maps of where the point masses are.
$endgroup$
– MJD
Dec 26 '18 at 23:58
$begingroup$
@MJD So $mu^sigma$ places $n$ masses of $1/n$ in the square $[-1, 1]^2$ and assigns measure $0$ to every other point. Is that the right way to interpret the expression?
$endgroup$
– Probably
Dec 27 '18 at 0:01
$begingroup$
Yes. A measure is a function on sets. The pdf is a distribution. That's the Dirac delta in $mathbb{R}^2$ : $delta(x,y) = delta(x)delta(y)$ the latter being the usual Dirac delta in $mathbb{R}$ @Probably
$endgroup$
– reuns
Dec 27 '18 at 1:00
add a comment |
$begingroup$
Note that
$$
{left(frac{2i}{n} - 1, frac{2sigma(i) }{n} -1right)}
$$
is a point in $[-1,1]^2$.
So $delta$ with that subscript is a unit point mass at that point.
Add $n$ point masses and divide by $n$, we get a probability measure.
edited Dec 26 '18 at 23:59
MJD
47.3k29213396
answered Dec 26 '18 at 23:55
GEdgarGEdgar
62.5k267171
$endgroup$
Note that
$$
{left(frac{2i}{n} - 1, frac{2sigma(i) }{n} -1right)}
$$
is a point in $[-1,1]^2$.
So $delta$ with that subscript is a unit point mass at that point.
Add $n$ point masses and divide by $n$, we get a probability measure.
edited Dec 26 '18 at 23:59
MJD
47.3k29213396
answered Dec 26 '18 at 23:55
GEdgarGEdgar
62.5k267171
edited Dec 26 '18 at 23:59
MJD
47.3k29213396
edited Dec 26 '18 at 23:59
MJD
47.3k29213396
edited Dec 26 '18 at 23:59
MJD
47.3k29213396
47.3k29213396
answered Dec 26 '18 at 23:55
GEdgarGEdgar
62.5k267171
answered Dec 26 '18 at 23:55
GEdgarGEdgar
62.5k267171
answered Dec 26 '18 at 23:55
GEdgarGEdgar
62.5k267171
62.5k267171
$begingroup$
That seems to be what the diagrams in the middle of page 4 are about; they are maps of where the point masses are.
$endgroup$
– MJD
Dec 26 '18 at 23:58
$begingroup$
@MJD So $mu^sigma$ places $n$ masses of $1/n$ in the square $[-1, 1]^2$ and assigns measure $0$ to every other point. Is that the right way to interpret the expression?
$endgroup$
– Probably
Dec 27 '18 at 0:01
$begingroup$
Yes. A measure is a function on sets. The pdf is a distribution. That's the Dirac delta in $mathbb{R}^2$ : $delta(x,y) = delta(x)delta(y)$ the latter being the usual Dirac delta in $mathbb{R}$ @Probably
$endgroup$
– reuns
Dec 27 '18 at 1:00
add a comment |
$begingroup$
That seems to be what the diagrams in the middle of page 4 are about; they are maps of where the point masses are.
$endgroup$
– MJD
Dec 26 '18 at 23:58
$begingroup$
@MJD So $mu^sigma$ places $n$ masses of $1/n$ in the square $[-1, 1]^2$ and assigns measure $0$ to every other point. Is that the right way to interpret the expression?
$endgroup$
– Probably
Dec 27 '18 at 0:01
$begingroup$
Yes. A measure is a function on sets. The pdf is a distribution. That's the Dirac delta in $mathbb{R}^2$ : $delta(x,y) = delta(x)delta(y)$ the latter being the usual Dirac delta in $mathbb{R}$ @Probably
$endgroup$
– reuns
Dec 27 '18 at 1:00
$begingroup$
That seems to be what the diagrams in the middle of page 4 are about; they are maps of where the point masses are.
$endgroup$
– MJD
Dec 26 '18 at 23:58
$begingroup$
That seems to be what the diagrams in the middle of page 4 are about; they are maps of where the point masses are.
$endgroup$
– MJD
Dec 26 '18 at 23:58
$begingroup$
@MJD So $mu^sigma$ places $n$ masses of $1/n$ in the square $[-1, 1]^2$ and assigns measure $0$ to every other point. Is that the right way to interpret the expression?
$endgroup$
– Probably
Dec 27 '18 at 0:01
$begingroup$
@MJD So $mu^sigma$ places $n$ masses of $1/n$ in the square $[-1, 1]^2$ and assigns measure $0$ to every other point. Is that the right way to interpret the expression?
$endgroup$
– Probably
Dec 27 '18 at 0:01
$begingroup$
Yes. A measure is a function on sets. The pdf is a distribution. That's the Dirac delta in $mathbb{R}^2$ : $delta(x,y) = delta(x)delta(y)$ the latter being the usual Dirac delta in $mathbb{R}$ @Probably
$endgroup$
– reuns
Dec 27 '18 at 1:00
$begingroup$
Yes. A measure is a function on sets. The pdf is a distribution. That's the Dirac delta in $mathbb{R}^2$ : $delta(x,y) = delta(x)delta(y)$ the latter being the usual Dirac delta in $mathbb{R}$ @Probably
$endgroup$
– reuns
Dec 27 '18 at 1:00
add a comment |
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