2
$\begingroup$

I was reading a paper (https://arxiv.org/pdf/1609.03891.pdf), and on the first page the author's write the following line: $$ \mu^\sigma = \frac{1}{n} \sum_{i = 1}^n \delta_{\left(\frac{2i}{n} - 1, \frac{2\sigma(i)}{n} -1\right)}. $$ 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 ${\left(\frac{2i}{n} - 1, \frac{2\sigma(i) }{n} -1\right)}$ 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?

$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

Note that $$ {\left(\frac{2i}{n} - 1, \frac{2\sigma(i) }{n} -1\right)} $$ 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.

$\endgroup$
3
  • $\begingroup$ 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
    Commented Dec 26, 2018 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
    Commented Dec 27, 2018 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
    Commented Dec 27, 2018 at 1:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .