Meaning of Delta Notation












2














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?










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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?










    share|cite|improve this question









    New contributor




    Probably is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      2












      2








      2







      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?










      share|cite|improve this question









      New contributor




      Probably is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      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






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      share|cite|improve this question









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      share|cite|improve this question








      edited Dec 26 '18 at 23:55









      MJD

      46.9k28208392




      46.9k28208392






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      asked Dec 26 '18 at 23:47









      Probably

      111




      111




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          1 Answer
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          3














          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.






          share|cite|improve this answer























          • That seems to be what the diagrams in the middle of page 4 are about; they are maps of where the point masses are.
            – MJD
            Dec 26 '18 at 23:58










          • @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?
            – Probably
            Dec 27 '18 at 0:01












          • 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
            – reuns
            Dec 27 '18 at 1:00













          Your Answer





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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          3














          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.






          share|cite|improve this answer























          • That seems to be what the diagrams in the middle of page 4 are about; they are maps of where the point masses are.
            – MJD
            Dec 26 '18 at 23:58










          • @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?
            – Probably
            Dec 27 '18 at 0:01












          • 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
            – reuns
            Dec 27 '18 at 1:00


















          3














          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.






          share|cite|improve this answer























          • That seems to be what the diagrams in the middle of page 4 are about; they are maps of where the point masses are.
            – MJD
            Dec 26 '18 at 23:58










          • @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?
            – Probably
            Dec 27 '18 at 0:01












          • 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
            – reuns
            Dec 27 '18 at 1:00
















          3












          3








          3






          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.






          share|cite|improve this answer














          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 26 '18 at 23:59









          MJD

          46.9k28208392




          46.9k28208392










          answered Dec 26 '18 at 23:55









          GEdgar

          61.7k267168




          61.7k267168












          • That seems to be what the diagrams in the middle of page 4 are about; they are maps of where the point masses are.
            – MJD
            Dec 26 '18 at 23:58










          • @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?
            – Probably
            Dec 27 '18 at 0:01












          • 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
            – reuns
            Dec 27 '18 at 1:00




















          • That seems to be what the diagrams in the middle of page 4 are about; they are maps of where the point masses are.
            – MJD
            Dec 26 '18 at 23:58










          • @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?
            – Probably
            Dec 27 '18 at 0:01












          • 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
            – reuns
            Dec 27 '18 at 1:00


















          That seems to be what the diagrams in the middle of page 4 are about; they are maps of where the point masses are.
          – MJD
          Dec 26 '18 at 23:58




          That seems to be what the diagrams in the middle of page 4 are about; they are maps of where the point masses are.
          – MJD
          Dec 26 '18 at 23:58












          @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?
          – Probably
          Dec 27 '18 at 0:01






          @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?
          – Probably
          Dec 27 '18 at 0:01














          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
          – reuns
          Dec 27 '18 at 1:00






          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
          – reuns
          Dec 27 '18 at 1:00












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