TITLE: What does $\sim$ in $X\sim \mathcal{N}(\mu,\sigma^{2})$ really mean?
QUESTION [1 upvotes]: This is a bit of a silly question, but I can't seem to find the answer anywhere.
I feel like $X\sim \mathcal{N}(\mu,\sigma^{2})$ means that $\sim$ is a relation, but if it is a relation, what precisely is this relation?
If this is a relation, could I instead write $X\in \mathcal{N}(\mu,\sigma^{2})/ \sim$?
One definition I think would be reasonable is to say that $X \sim Y$ if $X$ and $Y$ have the same characteristic or moment generating function, but it seems a little heavy handed to be the "right" definition.
Another definition I would guess is to put a topology on something which characterizes the random variables and say $X\sim Y$ if they satisfy a homeomorphism?
From the homeomorphism viewpoint, I would want to conclude that if $X$ is a random variable whose image is always non-negative, then $X^{2}$ could be normally distributed since you can take the positive square root and exhibit a bijection between $X$ and $X^{2}$ without going into $\mathbb{C}$, but I know this to not be true since it is Chi-Squared, so this can't be a reasonable definition.

REPLY [7 votes]: I agree with Emily that most people just understand this notation to be shorthand for the corresponding sentence in words.
But if you want to think of it as having formal meaning, I would say that $X \sim \nu$ means:

$X$ is a random variable on some probability space $(\Omega, \mathcal{F}, \mathbb{P})$, taking its values in some measurable space $(E, \mathcal{B})$;
$\nu$ is a probability measure on $(E, \mathcal{B})$;
For every $A \in \mathcal{B}$, we have $\mathbb{P}(X \in A) = \nu(A)$.  That is, $\nu$ is the distribution (or law) of $X$.

So under that interpretation, you should interpret $\mathcal{N}(\mu, \sigma^2)$ as meaning "the probability measure $\nu$ on $(\mathbb{R}, \mathcal{B}_\mathbb{R})$ defined by $\nu(A) = \int_A \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-(x-\mu)^2/2\sigma^2}\,dx$", where $\mathcal{B}_\mathbb{R}$ is the Borel $\sigma$-algebra on $\mathbb{R}$.


TITLE: Gaussian curvature radius
QUESTION [3 upvotes]: In the paper Surface sampling and the intrinsic Voronoi diagram (2008), Ramsay Dyer defines the Gaussian curvature radius at a point $x$ of a surface $S$ to be $\rho_K(x) = 1/\sqrt{K(x)}$ where $K(x)=\kappa_1(x) \kappa_2(x)$ is the Gaussian curvature at $x$.
Trying to track back the notion in Berger's A panoramic view of Riemannian geometry, and in Lee's Riemannian manifolds and in Chavel's Riemannian Geometry yielded nothing.
My question is two-folded:

Where can I find more information about this notion?
Is there a reason not to define it as $\rho_K(x) = 1/|K(x)|$? Otherwise, this definition is only valid for non-negatively curved surfaces.

EDIT As pointed out by Deane Yang, there is no sense in the definition I suggested. Nevertheless, if one wants to relate the Gaussian curvature to a radius (for either negatively or positively curved surfaces) how about this alternative: $\rho_{K}(x)=1/\sqrt{|K(x)|}$?

REPLY [0 votes]: Deane's answer is similar to what I would have tried to say if I'd got here on time. I don't recall seeing the "Gaussian curvature radius" defined before, so I can't point you to other references. The definition is natural. On the one hand the bound on the distance to a conjugate point (Morse-Schönberg lemma) is given in terms of a bound on the Gaussian curvature radius, and on the other hand the Gaussian curvature radius provides an upper bound to the "maximal curvature radius"(reciprocal of the maximum of the absolute values of the principal curvatures). As Deane pointed out, these two curvature radii coincide on the sphere.
Since we are only using it as an upper bound, we just define it to be infinite if the curvature is non-positive. In flat or negatively curved spaces, conjugate points are not an issue; geodesics diverge.
As to your alternative, $\rho_K(x) = 1/\sqrt{|K(x)|}$, I guess it depends on what you want to do. You are making a smaller sizing function, but why? The spirit of the Morse-Schönberg lemma is better captured without the absolute value signs. If the infinite values disturb you, you have not avoided them when the Gaussian curvature vanishes.


TITLE: multiplication with limits
QUESTION [1 upvotes]: I am wondering if the following can always be said, or if there are any restrictions:
$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=a,\,|a|<\infty$$
then its true that:
$$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}a.g(x)$$
My thoughts are that this can be said, but is not relevant if:
$$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}g(x)\to0,-\infty,\infty$$
Thanks in advance.

REPLY [2 votes]: No, This is trying to will limits into existance.
$\lim 1 = 1$ but $\lim \frac {f(x)}{f(x)} = 1$.  Just let $f(x)$ be a function without a defined limit.
However if you are given that $\lim f(x)$ and $\lim g(x)$ exist then
if $\lim f(x) = m; \lim g(x) = k\ne 0$ then $\lim \frac{f(x)}{g(x)}=\frac mk = a$ so $\lim f(x) = m = ak = \lim a*g(x)$.
If $\lim f(x) = m\ne 0$ or $\lim f(x) =\pm \infty$; while $\lim g(x) = 0$ then $\lim \frac {f(x)}{g(x)}\ne a$.
If $\lim f(x) =0$ and $\lim g(x) = 0$ then $\lim f(x) = \lim a*g(x)$ for all $a$.
If $\lim f(x) = m; \lim g(x) =\pm \infty$ then $\lim\frac {f(x)}{g(x)} = 0 = a$ and $0*g(x) = 0$ so $\lim ag(x)= 0$ which is not necessarily equal to $m$.
So say  $f(x) =1$ and $g(x) = x$ would be an exception.  $\lim f(x) = 1\ne 0 =\lim 0*g(x)$.
If $\lim f(x) = \infty$ and $\lim g(x)\ne \infty$ then $\lim\frac{f(x)}{g(x)}\ne a$.
$\lim f(x) = \pm \infty$ and $\lim g(x) = \pm \infty$ but $\lim \frac {f(x)}{g(x)} = a\ne 0$ then $\lim f(x) = \pm \infty$ and $\lim a*g=\pm \infty$.  (I'll leave the signs to be self-evident)
Final case to explore is $\lim f(x) = \pm \infty$ and $\lim g(x) = \pm \infty$ but $\lim \frac {f(x)}{g(x)} = 0$ then $\lim f(x) = \pm \infty$ but $\lim 0*g(x) = 0$.
So there are 3 exceptions:  $\lim f; \lim g$ don't exist; $\lim f\ne 0$ but $\lim g(x)=\pm \infty$; or $\lim f(x)=\pm \infty; \lim g(x) = \pm \infty$ but $\lim\frac {f(x)}{g(x)} = 0$.


TITLE: Convert into affine form
QUESTION [0 upvotes]: I've $x+y=y+z=\frac{x+z}{2}$ and want that on affine form or in $(x,y,z) - t (a,b,c)$. I know it should be really simple but I don't find anywhere how to do it...

REPLY [0 votes]: The equation $x+y=y+z$ gives $x=z$ which you may as well call another name $u$. Then the equation $y+z=(x+z)/2$ becomes $y+u=(u+u)/2=u$. This then gives $y=0$, so what is known about $x,y,z$ is only that $x=z$ and $y=0$. If you change to $t$ for the common value of $x,z$ then $(x,y,z)=(t,0,t)=t(1,0,1)$ and maybe what you're looking for is the form $(x,y,z)-t(1,0,1)$ which will be zero on (and only on) the solution set of your equations.


TITLE: Obstructions for embedding a graph on a surface of genus g
QUESTION [21 upvotes]: Kuratowski's theorem tells us the complete graph $K_5$ and the bipartite graph $K_{3,3}$ are the only obstructions to a graph being planar, ie embeddable in the plane with no edge-crossings.

Is the list of obstructions to being able to embed a graph with no edge-crossings on the surface of genus $g$ known to be finite for all $g$?

REPLY [5 votes]: For the projective plane, i.e. the nonorientable surface of genus 1, this is known. Look at "A Kuratowski Theorem for the Projective Plane" in the homepage of Dan Archdeacon here:
http://www.emba.uvm.edu/~archdeac/
This was his PhD thesis. In particular, he found that there are exactly 103 graphs such that if any graph $G$ contains one of these graphs as a subgraph, then $G$ cannot be not embedded into projective plane. You can refer to his original thesis for the list of the 103 graphs, or you can refer to the Appendix A of the book "Graphs on Surfaces" by Bojan Mohar and Carsten Thomassen.


TITLE: Prove irreducibility of $X^4+15X^3+7\in\mathbb Z[X]$
QUESTION [3 upvotes]: What technique can I apply to prove that $X^4+15X^3+7\in\mathbb Z[X]$ is irreducible?

I can't apply Eisenstein because 7 and 15 have no common prime multiple but 1; I tried to apply Eisenstein with $X+1$ as well but no success.
Can I consider the polynomial in $\mathbb F_p[X]$ for some $p$ and conclude that it is irredicible in $\mathbb Z[X]$ as well?
Or is there another trick to this?

REPLY [0 votes]: Z is a UFD together with this polynomial is a premitive polynomial. Consider a prime p=2 and reduce this polynomial in Z2[x] with both degree of polynomials are  same with irreducible in Z2[x]. Then it is irreducible in Z[x]