New answers tagged

2

You want to maximize \begin{equation*} R(E):=\frac{\int_E\ m}{\int_E\ M} \end{equation*} over all admissible sets $E$, that is, over all Lebesgue-measurable sets $E$ such that $\int_E\ M>0$, where \begin{equation*} m(x):=\min(p(x),q(x)),\quad M(x):=\max(p(x),q(x)). \end{equation*} For \begin{equation*} r(x):=\frac{m(x)}{M(x)} \end{equation*} ...


3

This question was completely settled by J.A. Fill here: https://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.8.641


4

I expect the answer is $(2+o(1))n$. As Peter Taylor says: We have a Markov process on the set $[n]:=\{ 1,2, \ldots, n \}$ where the transition probability from $i \to j$ is the probability that a randomly selected function from $[i] \to [n]$ will have image of size $j$. Fix some $m \geq 2$ and consider running this Markov process starting at $m$. I will ...


6

We have a Markov process where the state after $i$ steps is given by the size of the codomain of $g_i$. If at time $i$ we are in a state with $j$ surviving values, we can ignore the other values and consider $f_{i+1}$ as a function from the codomain of $g_{i}$ to $[n]$. Then the probability of a transition to state $k$ is given by $\binom{n}{k} \{{j \atop k}\...


2

The inequality does not always hold. Let $D \sim$ Bernoulli$(p)$, independent of $Y$. Let $Y'$ and $Y''$ be two independent copies of $Y$ (also independent of $Y$). Let $X=D Y+(1-D) Y'$ and $Z=DY'' + (1-D)Y$. Define also $m_k=E(Y^k)$ for $k=1,2$. Then it is easy to check that Cov$(X,Y) = p V(Y)$ and Cov$(Z,Y) = (1-p) V(Y)$. Because $X$ and $Z$ have the same ...


3

Yes, the pdf of this distribution is \begin{equation} u\mapsto 2u^2 f(u)\,1(u>0) \tag{1} \end{equation} where $f$ is the standard normal pdf. Indeed, by Proposition 2, \begin{equation} G(m,b):=P(M_1>m,B_1\le b)=P(B_1>2m-b)=1-F(2m-b) \end{equation} for real $m,b$ such that $m>b_+:=\max(0,b)$, where $F$ and $f$ are the standard normal cdf and ...


3

You added a proof that a certain category satisfies the "yanking axiom", and I think I can infer that it simply says that every object in your symmetric monoidal category is dualizable. I'm not sure, though. If the "yanking axiom" means something different, then the following may be moot. Anyway, if that is what the "yanking axiom&...


3

The post asks: "Are there ever cases where formally defining physical phenomena in mathematical language is unnecessary?" The key clarification is: necessary for what? Mathematical rigor may require formal definitions -- but this may be the only goal where the formal definitions are necessary. A successful application does not require formal ...


22

The theory of classical independence and classical convolution can be generalised to noncommutative settings in several ways. The most famous one is that of free independence and free convolution (introduced by Voiculescu), but there is also boolean independence and boolean convolution (introduced by Speicher and Woroudi); monotone independence and monotone ...


0

It turns our the problem has a simple answer, once the easy case has been solved (see the OP). Indeed, we write $\overline{f} = f + \zeta_0(f)$, so that $\zeta_0(f) = 0$. Now, one has $$ T = \overline{T} + v1^\top+(v1_N^\top)^\top + \zeta_0(f)^2 1_N1_N^\top, $$ where $v = (\zeta_0(\overline{f}_i))_{i \in [N]} = (\zeta_0(f_i)-\zeta_0)_{i \in [N]}$. By the ...


3

Note that such a universal constant $c$ does not exist; by the law of large numbers ($\operatorname{Binomial}(n, p)$ being a sum of $n$ i.i.d $\operatorname{B}(p)$ r.v.), for any $p\in (0, 1)$ we have $\mathbb{E}[Y]/n\rightarrow \min\{p, 1-p\}$, so for $p$ close to $1$ constant $c$ must be taken to be at most $(1-p)/p$ to accommodate all large $n$; this ...


0

This not an answer to my question (I accepted Iosif's answer) but additional information based on my most recent research, that may be helpful to the reader interested in these processes. It is too long for an update of my question. For simplicity, we use the notation $F(x)=F_0(x)$ and $F(x-k)=F_k(x)$. Theorem A Regardless of the distribution $F_k$, if $b-a$ ...


3

Since $\min(a,b)=(a+b-|a-b|)/2$, your question is really about upper bounding $E|2X-n|$, or, equivalently, $E|X-n/2|$: $$E\min(X,n-X)=n/2-2E|X-n/2|.$$ You can upper bound $E|X-n/2|$ using Jensen's inequality: $E|X-n/2|\le\sqrt{E(X-n/2)^2}$. The latter, if I'm not mistaken, evaluates to $$ n\sqrt{ p(1-p)/n+p^2-p+1/4 } =:nF(p).$$ For $n$ sufficiently large and ...


0

I claim this is not an answer but some calculus following Thomas's observation. As I cannot write it as a comment, I write it down here. The stochastic integral $$\left(M_t:=\int_0^t\frac{\sigma}{1+m(s)}dW_s\right)_{t\ge 0}$$ is a continuous martingale starting from zero. Hence, there exists some Brownian motion, denoted by $B$, s.t. $M_t=B_{\langle M\...


14

Your question is Are there ever cases where formally defining physical phenomena in mathematical language is unnecessary? It is never possible to define physical phenomena directly in mathematical language. Mathematics (and even sciences) can deal with real phenomena only through models. To deal with a model mathematically, it must be a mathematical model, ...


4

What fails for other lattices is that there seems to be no parafermionic observable with properties as nice as for hexagonal lattice; specifically, there is no analog of Lemma 1. In the definition of the parafermionic observable, one has two parameters to play with: $\sigma$ and $x$. The condition that the two configurations on the left in the picture on ...


1

Here's another, perhaps more probabilistic, approach. It's known that $\pi$ can be represented in terms of the mean occupation times as follows. Fix a state, for convenience $r$. Then, for any state $j$, $$ \pi_j=m_{rr}^{-1} E^r\left[ \sum_{n=0}^{T_r-1}1_{\{X_n=j\}}\right], $$ where $E^r$ denotes expectation for the chain started in $r$, and $T_r$ is the ...


2

This is not true. Indeed, suppose that $X_k=X_{s;k}=k+sZ_k$, where $s\downarrow0$ and the $Z_k$'s are any iid random variables (r.v.'s). To obtain a contradiction, suppose that, for the random Borel measure $\mu_s$ over $\mathbb R$ defined by $\mu_s(B):=\sum_{k\in\mathbb Z}1(X_{s;k}\in B)$, the distribution of the random variable (r.v.) $\mu_s(B)$ is Poisson ...


0

I believe I found an answer. Note that in a similar way that we constructed $P_\mu,$ we may define $P_x$ as the unique Borel probability on $M^{\mathbb N}$, such that given $A_0,\ldots,A_n \in M,$ then $$P_{x}\left(\{\omega_n\}_{n\in\mathbb N} \in M^{\mathbb N}; x_i\in A_i, \ \forall \ i\in\{0,1,\ldots,n\}\right) = \int_{A_0}\int_{A_1} \ldots \int_{A_{n-1}} ...


1

The problem is related to the following. Assume you have $K$ i.i.d. Uniform$[0,1]$ points put in the increasing order: $X_1,X_2,.., X_{K-1}$. Let $X_0=0$ and $X_K=1$. Define $D=D_K$ as the minimum distance amongst these $K+1$ points. What is the distribution of $D$? To answer this, Let $0<Y_1<Y_2<...<Y_K$ be the increasing sequence of points of a ...


0

Here is an example, too long for a comment. Consider the torus, $[-\pi,\pi]^2$, with coordinates $t$ and $u$ and metric \begin{pmatrix} 1 & 0\\ 0 & (9+\sin t)^2 \end{pmatrix} Then the scalar curvature is $$\frac{2\sin t}{9+\sin t}$$ and the pdf of that scalar curvature is $$\frac{18}{\pi(2-w)^2\sqrt{1+4w}\sqrt{1-5w}}\ \text{ on }\ \frac{-1}{4}\le w\...


2

Let $f$ be a pdf on $[0,\infty)$. Let $\hat f$ be the Laplace transform of $f$, so that \begin{equation} \hat f(s)=\int_0^\infty f(x)e^{-sx}\,dx \end{equation} for real $s\ge0$. Suppose that \begin{equation} |\hat f(s)-\hat f(0)|\le Cs^a \end{equation} for some real $a,C>0$ and all real $s\ge0$. Then for all real $s>0$ \begin{equation} Cs^a\...


2

No, this cannot be done in general. Indeed, let $A:=\mathcal A$. You want to express $$P((X,Y)\in A)=Ef(X,Y)$$ as $$\sum_{k=1}^K c_k F(a_k,b_k)=Eg(X,Y),$$ where $$f(x,y):=1(x\le2,x-y\le3)$$ and $$g(x,y):=\sum_{k=1}^K c_k 1(x\le a_k,y\le b_k).$$ However, for any choice of the numbers $a_k,b_k,c_k$, there will be some $(x_*,y_*)\in\mathbb R^2$ such that $f(x_*...


2

In general, the answer is no. Moreover, the answer is no even if \begin{equation} \phi(t)=t\ln(1+t). \tag{1} \end{equation} Indeed, suppose that $P(Z_i=0)=1-2p$ and $P(Z_i=b)=p=P(Z_i=-b)$ for all $i$, where \begin{equation*} p:=\frac1{2\phi(b)}, \end{equation*} $\phi$ is as given by (1), and $b$ is a large enough positive real number so that $p\in(0,...


1

no, the second term is basically $B_t/t$ which (proabably) does go to 0, but the first is governed by the law of the iterated logarithm and does not. In fact, you know that limsup of that term is 1.


2

Assume that $\det\Sigma\ne0$. Then the random matrix $X$ is of rank $m$ almost surely (a.s.). So, a.s. the Moore--Penrose inverse of $X$ is $X^+=X^\top(XX^\top)^{-1}$ and hence $$X^+X=X^\top(XX^\top)^{-1}X.$$ It appears that $EX^+X=EX^\top(XX^\top)^{-1}X$ cannot be expressed in closed form, even in the fully specified case when $m=2$, $n=3$, and $\Sigma=\...


2

This is an open question. The difficulty is that Monge's cost function is very degenerate, so doesn't satisfy the assumptions of the standard regularity theory of optimal transport. The first difficulty is that the solutions to this problem need not be unique. However, the transport will occur along disjoint line segments (called transport rays) and in order ...


0

This is just a comment on a related issue. You don't have stochastic comparison of processes at a given time but you DO have stochastic comparison of hitting times. To see this, you can couple the two processes $j=1,2$ AFTER having performed the (respective) time change transformations into standard BM. As a consequence, the two processes will have the SAME ...


1

Update: The bounds below are not sharp after all, and I now think that the only sharp bounds can be combinatorial. For example, in the notation below, let $p_{ac}=p_{ad}=p_{bc}=p_{bd}=1/4$, and: let $X_a$ have probability $1/4$ for each of $\pm1, \pm2$ let $X_b$ have probability $1/4$ for each of $\pm3, \pm4$ let $X_c$ have probability $1/4$ for each of $\...


0

There exists some recent work dealing with it, see Lemma 2.1 in https://arxiv.org/abs/1705.01299 for the quadratic case and theorem 2.6 in https://arxiv.org/abs/2102.06379 for the general one.


4

For real $a$ and $x\in(0,1)$, let \begin{equation*} p_a(x):=e^{ax+h(x)}/C(a), \end{equation*} where $h(x):=-x\ln x-(1-x)\ln(1-x)$ and $C(a):=\int_0^1 e^{ax+h(x)}\,dx$, so that $p_a$ is a pdf on $(0,1)$ and \begin{equation*} V(p)=\int_0^1(p(x)\ln p(x)-h(x)p(x))\,dx \tag{1} \end{equation*} and \begin{equation*} \int_0^1 xp_a(x)\,dx=L'(a), \quad L(...


3

Arbitrary metric spaces, no. A classic text discussing such "derivation" is: Hayes, C. A.; Pauc, C. Y., Derivation and martingales, Ergebnisse der Mathematik und ihrer Grenzgebiete. 49. Berlin-Heidelberg-New York: Springer-Verlag. VII, 203 p. (1970). ZBL0192.40604. I think you get trouble even in $\mathbb R^2$ if you define your metric so that the $...


13

Let \begin{equation*} N:=\inf\{n\ge2\colon X_{n-1}>X_n\}, \end{equation*} where $X_1,X_2,\dots$ are independent random variables uniformly distributed on $[0,1]$. We want to find \begin{equation*} EX_N=\sum_{n=2}^\infty EX_n\,1(X_1\le\cdots\le X_{n-1}>X_n). \tag{1} \end{equation*} We have \begin{equation*} \begin{aligned} &EX_n\,1(X_1\...


2

As in the comment by leo monsaingeon, let $$p_*(x):=e^{h(x)}/c,$$ where $h(x):=-x\ln x-(1-x)\ln(1-x)$ and $c:=\int_0^1 e^{h(x)}\,dx$, so that $p$ is a pdf on $(0,1)$ with mean $1/2$, and $$V(p)=\int_0^1(p(x)\ln p(x)-h(x)p(x))\,dx.$$ For any pdf $q$ on $(0,1)$ with $V(q)<\infty$, the directional derivative of $V$ at $p_*$ in the direction of $q-p_*$ is $$\...


1

$\newcommand{\ep}{\varepsilon}\newcommand\R{\mathbb R}$Yes, the minimizer $\mu$ is absolutely continuous (w.r. to the Lebesgue measure $|\cdot|$). Indeed, you showed that \begin{equation*} F(\rho_n)\to m:=\inf_\rho F(\rho) \tag{-1} \end{equation*} and \begin{equation*} \mu_{\rho_n}\to\mu \tag{-0.5} \end{equation*} weakly for some sequence $(\rho_n)$ ...


0

Even if we accept your proposal to model a player with numbers $W$ and $D$, it seems unlikely to me that we can get very far without further assumptions. Focus on drawn positions for a moment. Say that a position is a "tightrope" if only a tiny fraction of the legal moves save the draw, and say that a position is an "easy street" if ...


2

Indeed, the condition $\det M\ne0$ can be expressed as a certain non-independence condition, as follows. For $i$ and $j$ in $[J]:=\{1,\dots,J\}$, let \begin{equation*} p_{i|j}:=P(X=i|Y=j), \end{equation*} so that $M=[p_{i|j}]_{i,j\in[J]}$. Suppose that $\det M=0$. Then for some real $c_1,\dots,c_J$ not all of which are $0$ and for all $i\in[J]$ we have \...


1

As several others have pointed out, $\pi$ obviously isn't actually random, since its digits are fixed in place forever. But are the digits "as good as random"? The short answer is that, as far as anyone can tell, the answer seems to be empirically yes (See Marsaglia's On the Randomness of Pi and Other Decimal Expansions). That is, there are no ...


1

Core concepts: i) Co-tangent bundle of a given manifold, ii) canonical one form $\theta$ on any co-tangent bundle, associated symplectic form $d \theta$, associated canonical Liouville (phase-space) volume form (=measure) $d \theta \wedge \ldots \wedge d \theta$. The latter equals $dq dp$ in any coordinate system, iii) conservation of this symplectic form ...


1

$\newcommand\R{\mathbb R}$You want to find $\ln M_W(t)$, where $M_W$ is the moment generating function (mgf) of $W$, given by \begin{equation*} M_W(t)=E e^{tW}=\sum_{n=0}^\infty\frac{t^n EW^n}{n!} \end{equation*} for real $t$. By obvious rescaling, without loss of generality $\sigma=1$. For natural $n$, \begin{equation*} W^n=|v|^{2(n-1)} vv^\top, \...


2

You may want to use the following two facts: Minus the logarithm of an uniform distribution on $[0,1]$ is an exponential distribution (easy). An exponential distribution is infinitely divisible as a member of the family of Gamma distributions. Simulation of Gamma distributions is extensively discussed in the literature. In particular, if $Y_i,Y_j$ are two ...


2

By the Berry--Esseen inequality, the limit distribution of $Z_n^*$ is the standard normal distribution if $a_k=1/k^{1/2}$. If $a_k=1/2^k$, then the limit distribution of $Z_n^*$ is uniform on the interval $[-\sqrt3,\sqrt3]$. This follows (i) because $(X_k+1)/2$ may be viewed as the $k$th binary digit of a random number uniformly distributed on the interval $...


1

This largest singular value is the norm of the matrix. You can use a net argument to show that there is a $C$ so that $$\mathbb{P}(\| A \|_{op} \geq C\sqrt{N}(\sqrt{\alpha} + \sqrt{1 - \alpha} + t)) \leq 2 e^{-Nt^2}$$ for all $\alpha, t$. For a reference, this appears as Theorem 4.4.5 in Vershynin's High Dimensional Probability book.


3

Yes, there are various results available in more general settings. The typical route would be to combine an upper bound on the expected distance between the law and the empirical measure (like Theorem 1.1 in Boissard and Le Gouic) with a concentration estimate around this expectation (like Proposition 20 in Weed and Bach). In particular, both of these ...


3

Let $\{H_n\}_{n \ge 0}$ be a sequence of Hilbert spaces, with $H_{n+1} \subset H_n$ (Clarification: we assume that $H_{n+1}$ is a subspace of $H_n$) for all $n\ge 0$, and denote $H_\infty=\cap_{n=1}^\infty H_n$. Claim: If $P_n$ is the orthogonal projection from $H_0$ to $H_n$, and $P_\infty$ is the orthogonal projection from $H_0$ to $H_\infty$, then for all ...


3

I will begin with a reformulation of your question which makes it not only more symmetric, by also (at least for me) more natural and interesting. I will pass from your variables $(W,H,Q)$ to new variables $(X,Y,Z)$ (which are not your original $X,Y,Z$) by putting $X=W, Y=-Q, Z=-H$. Then your condition (3) becomes just $X+Y+Z=0$. Therefore you are asking ...


2

First, a quick note on terminology: the standard term for a "reverse order pair" is an inversion. Knowing this makes it easier to search for the answer: The generating function for the number of permutations with $k$ inversions is the $q$-factorial $[n]_q!$ as shown e.g. here.


0

No, we cannot. Formally, $\varphi$ is the eigenfunction of the unbounded operator $L_s$ on $L^2(\Omega)$, defined initially by $$ L_s u(x) = (-\Delta)^s u(x) \qquad \text{for } x \in \Omega , $$ where $u \in C_c^\infty(\Omega)$ (and it is understood that $u(x) = 0$ for $x \notin \Omega$), and then extended to an appropriate domain (e.g. by means of ...


10

This isn't true. Let $V$ be the standard two dimensional representation of $U(2)$ and let $W = (\det V)^{-1} \otimes \mathrm{Sym}^2(V)$. If the eigenvalues of $g$ acting on $V$ are $z_1$ and $z_2$, then the eigenvalues of $g$ acting on $W$ are $z_1 z_2^{-1}$, $1$ and $z_1^{-1} z_2$. So $\det (\mathrm{Id} - \pi_W(g))=0$ for all $g$. Okay, general nonsense ...


0

I know that the limiting eigenvalue distribution satisfies a variational principle (see e.g. the joint work with A. Kuijlaars Large Deviations for a Non-Centered Wishart Matrix) from which you may derive a degree three algebraic equation for the Sieltjes transform, and eventually recover an analytic formula for the limiting density using the Cauchy–Plemelj ...


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