pr.probability – Exit probability on a finite interval

I have a question about the estimate of the exit probability on a finite interval. Given a $q$ function bounded and continuous, given the following SDE
begin{cases}
dX_s=(beta-q(s))X_sds+frac{1}{2}beta^2(X_s)^2dW_s \
X_t=y
end{cases}

I should estimate the probability $mathbb{P}{exists s in (t,T) : (s,X_s) in A }$, where $A$ has this form ${(t,x) in (0,T) times (0,+infty) : 0 leq y leq L}$.
Can someone help me with this estimate?

lo.logic – Are there structures in a finite signature that are recursively categorically axiomatizable in SOL but not finitely categorically axiomatizable?

Recall that a structure $mathcal{M} = langle M, I^sigma_M rangle$ in a signature $sigma$ is categorically axiomatized by a second-order theory $T$ when, for any $sigma$-structure $mathcal{N} = langle N, I^sigma_N rangle$, $langle N, mathcal{P}(N), I^sigma_N rangle vDash T$ just in case $mathcal{N}$ is isomorphic to $mathcal{M}$.

It is fairly easy to find a structure in a finite signature that is categorically second-order axiomatizable but not finitely categorically second-order axiomatizable. Add a single function symbol $f$ to the language of second-order arithmetic, and choose a non-second-order-definable $zeta: mathbb{N} rightarrow mathbb{N}$. Then consider the theory $T$ that adds to the axioms of second-order arithmetic ($mathsf{Z}^2$) the sentence $f(bar{n}) = overline{zeta(n)}$ for each natural number $n$, where $bar{m}$ is the canonical numeral for $m$. (I owe the idea for this example to Andrew Bacon.)

This theory $T$, however, is not recursively axiomatizable. Is there a structure in a finite signature that has a recursive categorical second-order axiomatization but no finite categorical second-order axiomatization?

I believe that it is possible to find a recursively axiomatizable second-order theory $T$ whose spectrum (i.e., the set ${kappa in mathsf{Card}: exists mathcal{M} (mathcal{M} vDash T$ and $vert mathscr{M} vert = kappa)}$) is shared by no finitely axiomatizable second-order theory, using partial truth predicates. (Consider the theory with $mathsf{Z}^2$ relativized to some predicate $N$ and ${$“The cardinality of the non-$N$s is not $Sigma^1_n$-characterizable”$: n in omega}$.) But I cannot see how to turn this into a categorical theory.

gt.geometric topology – Thurston measure under finite covers

Let $S= S_{g,n}$ be a finite type orientable surface of genus $g$ and $n$ punctures and let $mathcal{ML}(S)$ denote the corresponding space of measured laminations. The Thurston measure, $mu^{Th},$ is a mapping class group invariant and locally finite Borel measure on $mathcal{ML}(S)$ which is obtained as a weak-$star$ limit of (appropriately weighted and rescaled) sums of Dirac measures supported on the set of integral multi-curves.

The Thurston measure arises in Mirzakhani’s curve counting framework. Concretely, given a hyperbolic metric $rho$ on $S_{g,n}$, let $B_{rho} subset mathcal{ML}(S)$ denote the set of measured laminations with $rho$-length at most $1$. Then $mu^{Th}(B_{rho})$ controls the top coefficient of the polynomial that counts multi-curves up to a certain $rho$-length and living in a given mapping class group orbit.

Question: Fix a hyperbolic metric $rho$ on $S$ and a finite (not necessarily regular) cover $p: Y rightarrow S$. Let $rho_{p}$ denote the hyperbolic metric on $Y$ obtained by pulling $rho$ back to $Y$ via $p$. Is there a straightforward relationship between $mu^{Th}(B_{rho_{p}})$ and $mu^{Th}(B_{rho})$? For example, is the ratio
$$ frac{mu^{Th}(B_{rho})}{mu^{Th}(B_{rho_{p}})} $$
uniformly bounded away from $0$ and $infty$? Does it equal a fixed value, independent of $rho$? If so, can it be easily related to the degree of the cover $Y rightarrow S$?

It seems hard to approach the above by thinking about counting curves on $Y$ versus $S$, because “most” simple closed curves on $Y$ project to non-simple curves on $S$. But, maybe the generalizations of curve counting for non-simple curves due to Mirzakhani (https://arxiv.org/pdf/1601.03342.pdf) or Erlandsson-Souto (https://arxiv.org/pdf/1904.05091.pdf) could be useful. Of course, both apply to counting curves in a fixed mapping class group orbit, so it’s not clear (to me) how to apply these results either since multi-curves on $Y$ can project to curves on $S$ with arbitrarily large self intersection.

Thanks for reading! I appreciate any ideas or reading suggestions.

ag.algebraic geometry – Properties of pointless projective curves over finite fields?

Probably not research level, feel free to downvote.

We got construction of bounded degree projective curves
with no points over finite fields. This construction generalizes to higher dimension.

One of the curves over $GF(5)$ triggered internal Magma error, suggesting reporting the bug.

Working over $GF(p)$, let $f(x),g(y)$ be polynomials with no roots.
Let $n>1$ be integer and $s$ element of $GF(p)$ which is not
$n$-th power. Let $F$ be the homogenization of $f(x)^n-s g(y)^n=0$.
Then $F$ has no projective points. To go to higher dimension
take $(h(z)f(x))^n-s g(y)^n=0$

Q1 What properties these pointless curves have, e.g.
do they have some extremal invariants too?

Q2 Are there theoretical reasons to explain why Magma
fails to compute the Zeta function of pointless curve
in this example online

p:=5;
K<x,y>:=AffineSpace(GF(p),2);
C:=Curve(K,x^8 + 2*y^8 - x^7 + x^6 - y^6 + 2*x^5 + 2*y^5 - 2*x^4 + 2*y^3 + 2*x^2 + y^2 - x - y + 2);
pc:=ProjectiveClosure(C);
Genus(pc);
Points(pc);
IrreducibleComponents(pc);
ZetaFunction(pc);//report a bug

finite automata – Thinking about how an automaton can be created that accepts a language that switches the first and second alphabet

I have been thinking about this problem for quite a while, but to no avail.

Say I have a deterministic automaton M (you can assume it to be the 5-tuple (Q, sigma, q0, delta, F), which is given to accept a language L.

How do we create an automaton that would accept the language that has the first and second letter of the language L switched (given that the first and second letter in fact exist in a string accepted by M).

Is there an easy way to do this?

finite automata – Build an FA that accepts only the words baa, ab, and abb and no other strings longer or shorter

I have been trying solve this problem for a while now for a university assignment. I’m required to build a DFA and an NFA for the above question. So far I have been able to solve the DFA but can not find a solution for a proper NFA after multiple tries.

The solution of my DFA for the language provided above:

My attempts for the NFA are down below. I apologize for my messy handwriting but these were just rough works that I was drawing out on the go.

My first attempt at solving the NFA

My second attempt at solving the NFA

My third attempt at solving the NFA

finite automata – Prove that language is irregular

For $i ge 0$ define $w_i = b^i aa$.
For any $i,j ge 0$ with $i neq j$ you have what $b^i$ is a distinguishing extension for $w_i$ and $w_j$. Indeed $w_i b^i notin L_2$ but $w_ib^j in L_2$.

Then the number of equivalence classes of the set ${ w_i mid i ge 0}$ with respect to the equivalence relation “having a distinguishing extension” is not finite and, by the Myhill-Nerode theorem, $L_2$ is not regular.


Here is an alternative solution that uses closure properties and the pumping lemma following the hint of Hendrik Jan.

Suppose that $L_2$ is regular. Then so is $L’ = overline{L}_2 cap {b^i aa b^j} = {b^i aa b^i}$, which is easily shown to be non-regular using the pumping lemma.

Indeed, suppose that $L’$ was regular, and let $p$ be its pumping length. For some $k in {1, dots, p}$, the word $b^p aa b^p in L’$ can be written as $b^k b^{p-k} aa b^p$ such that $b^{ki} b^{p-k} aa b^p in L’$ for every $i ge 0$. Nevertheless, for $i=0$ we have $b^{p-k} aa b^p notin L’$, yielding a contradiction.

automata – What’s the difference between a Generalized Nondeterministic Finite Automaton (GNFA) and a Generalized Transition Graph (GTG)?

I’ve recently come across a few articles that talk about a “Generalized Transition Graph” (GTG), but I’ve never heard of such a thing before. This other answer to a similar question leads me to believe that GNFAs are simply GTGs with a different name where GNFA is the more standard nomenclature/name.

There is a Wikipedia article on GNFAs that describes them as:

…a variation of a nondeterministic finite automaton (NFA) where each transition is labeled with any regular expression.

which from what I understand is described by others as a GTG. Unfortunately, there’s no article on Generalized Transition Graphs on Wikipedia. The definitions for GTGs that I have come across are:

So again, are GNFAs the same as GTGs?

elementary set theory – The set of all finite subsets of $mathbb{Z}_+$ is countable. Is my proof correct?

I assume the following results:

  • A countable union of countable sets is countable.
  • A finite cartesian product of countable sets is countable.
  • If for any set $A$ there is a bijective correspondence with some countable set $B$, then $A$ is countable.

This is my proof of the statement in the title:

Proof. Let $J$ denote the set of all finite subsets of $mathbb{Z}_+$. Then $J = bigcup_{n in mathbb{Z}_+} J_n$, where $J_n$ is the set of all subsets of $mathbb{Z}_+$ with $n$ elements. We prove that $J_n$ is countable by showing that for any $n in mathbb{Z}_+$ there exists a bijective function $f : mathbb{Z}_+^n rightarrow J_n$ such that for any tuple $(k_1,k_2,…,k_n) in mathbb{Z}_+^n$, the following conditions hold:

  1. The set $f(k_1, k_2,…,k_n)$ contains $n$ elements
  2. The largest element in $f(k_1,k_2,…,k_n)$ is $sum_{i=1}^nk_i$.

We use induction. Base case $n = 1$ is trivial, since we can define a function $f(n) = {n}$, which is clearly bijective and satisfies the above conditions. Thus, assume that the result holds for some $k in mathbb{Z}_+$. Hence, there exists $f_k: mathbb{Z}^k_+ rightarrow J_k$ which is a bijection satisfying the above properties. To show that the result holds for $k+1$, define $f : mathbb{Z}_+^{k+1} rightarrow J_{k+1}$ as follows:
begin{align*}
f(n_1,n_2,…,n_{k+1}) = f_k(n_1,…n_k) cup {n_1+n_2+n_3+…+n_{k+1}}
end{align*}

By the inductive hypothesis, the set $f_k(n_1,…,n_k)$ contains $k$ distinct elements and the largest element is $sum_{i=1}^{k}n_i$. Thus, since:
begin{align*}
sum_{i=1}^{k}n_i < sum_{i=1}^{k+1}n_i
end{align*}

it follows that $f$ satisfies conditions (1) and (2). Then since $f_k$ is bijective and the last element in $f(n_1,…,n_{k+1})$ is not in $f_k(n_1,…,n_k)$, the bijectivity of $f$ follows. Hence, the result holds for $k+1$. Thus, since $mathbb{Z}_+^n$ is countable and there exists a bijection $f : mathbb{Z}_+^n rightarrow J_n$, it follows that $J_n$ is countable for all $n in mathbb{Z}_+$.

Therefore, since $J_n$ is countable, we conclude that $J = bigcup_{n in mathbb{Z}_+} J_n $ is countable.

Is this proof correct? If so, how can I improve it? Thank you!