User adrien hardy - MathOverflow

A sufficient condition for a probability measure to have compact support

2013-05-05T00:35:57Z

Consider a probability measure $\mu$ on, let's say, $\mathbb R$.

Is there a necessary and sufficient condition so that $\mu$ has compact support $Supp(\mu)$ ?

I agree this question is too vague, and it may tempting to answer it by quoting the definition of compactness for the support, so let me be more precise.

Imagine you know the Fourrier transform $F_\mu$ of $\mu$ (which contains all the information concerning $\mu$), that is $$ F_\mu(t)=\int e^{itx}\mu(dx),\qquad t\in\mathbb R, $$ or its Cauchy-Stieltjes transform $C_\mu$ (similar story), i.e. $$ C_\mu(z)=\int \frac{\mu(dx)}{z-x},\qquad z\in {\mathbb C }\setminus Supp(\mu), $$ is there any (necessary and) sufficient condition on $F_\mu$ or $C_\mu$ to force $Supp(\mu)$ to be a compact set ?

Answer by Adrien Hardy for Is there a (standard) name for $\bar{A}\setminus A$?

2013-01-11T14:13:29Z

The english word for the common french expression used for this is "frontier".

Answer by Adrien Hardy for New grand projects in contemporary math

2013-01-03T23:07:12Z

Universality phenomena for determinantal point processes and relatives.

After the deep results obtained by many great researchers concerning independent random variables, lot of attention has been recently paid to a certain kind of interacting random variables, arising from several (a priori non related) fields of mathematics, which behaves in a same way as the number of such random variables goes to infinity (appearance of the Sine kernel, Tracy-Widom distribution ...) ; the so-called universality phenomenon. This class of interacting random variables is not yet identified but includes

the eigenvalues of many random matrix models
the lengths of the rows of Young diagrams distributed according to the Plancherel measure
models from statistical physics like (T)ASEP, polynuclear growth models, random tilings of geometric shapes, ...)
the zeros of the Riemann Zeta function, once assumed the RH

and many others.

For further information, see e.g. the nice (although not exhaustive) overview of Deift http://arxiv.org/abs/math-ph/0603038

Because of the diversity of the mathematics involved a huge community, including a few Fields medals, is now working on a better understanding of such a class of random variables.

Answer by Adrien Hardy for Eigenvalues of infinite matrices

2012-10-14T09:22:50Z

For me, infinite matrix means an operator on $\ell^2(\mathbb N)$ (or sometimes $\ell^2(\mathbb Z)$, but usually referred to bi-infinite matrix). Concerning the eigenvalues, you thus may just look at the general theory concerning operator on Hilbert spaces, as already pointed out in the comments above.

Answer by Adrien Hardy for Countably many random vectors and related problems.

2012-09-25T00:14:45Z

1) Yes, you can define properly the first expectation, see e.g. http://planetmath.org/encyclopedia/TotallyFiniteMeasure.html

2) Then you have with your notations $$ \mathbb E_{X_i : i\in\mathbb N}\int_{[0,1]^k}\inf_{i\in\mathbb N}\|X_i-y\|^2dy\leq \mathbb E_{X_i : i\in\mathbb N}\int_{[0,1]^k}\min_{1\leq i \leq N}\|X_i-y\|^2dy = \mathbb E_{X_1,\ldots,X_N}\int_{[0,1]^k}\min_{1\leq i \leq N}\|X_i-y\|^2dy $$ and use what you know.

Anyway, questions like this should be first asked on math stack exchange, they are not "research level questions".

What do we actually know about logarithmic energy ?

2011-07-19T09:53:49Z

In potential theory, the $\textit{logarithmic energy}$ of a Radon measure $\mu$ acting on $\mathbb{C}$ is defined by $$I(\mu)=\iint\log\frac{1}{|x-y|}\mu(dx)\mu(dy).$$ Of course it is not well defined for all measures and may takes values in $[-\infty,+\infty]$. To avoid this annoying fact, one typically restrict to measures which integrate the logarithm around infinity, that is which satisfy the condition (C) $$ \int \log(1+|x|)\mu(dx)<+\infty,$$ so that $I(\mu)>-\infty$ thanks to $|x-y|\leq (1+|x|)(1+|y|) $. A fondamental fact is that if $\mu,\nu$ both satisfy (C), have finite logarithmic energy and $\mu(\mathbb{C})=\nu(\mathbb{C})$, then $I(\mu-\nu)\geq0$ and equality holds iff $\mu=\nu$ (we extend naturally the definition of $I$ to signed measures). I understand the condition (C) to be convenient, but maybe not sharp (one can imagine $\mu$ which does not satisfies (C) but with finite logarithmic energy). This leads to my first question :

What can we says about $I(\mu-\nu)$ when (at least one of) the measures do not satisfy (C) ?

This question is moreover motivated by the appearance of the logarithmic energies in random matrix theory (in large deviations rate functions) and in free probability (reinterpreted up to a sign as a non-commutative entropy by Voiculescu). In this setting, $I(\mu-\nu)$ is a natural candidate for a relative free entropy, and questions of geometric nature are bothering me : Let $A_{c}$ be the set of signed measures $\mu$ with finite logarithmic energy acting on $\mathbb{C}$ with total mass $\mu(\mathbb{C})=c$ such that if $\mu$ has Jordan decomposition $\mu^+-\mu^-$, then both $\mu^+$ and $\mu^-$ satisfy (C). Then the previous fact yields that $A_0$ is a pre-Hilbert space with scalar product $$ I(\mu,\nu)=\iint\log\frac{1}{|x-y|}\mu(dx)\nu(dy).$$ Note that $A_0$ is not complete since it is clearly not closed. Moreover, it is not hard to check that $A_0$ acts by translations on $A_1$ which inherits of a structure of affine space, leading to a metric on probability measures with finite logarithmic energy satisfying (C). It is now time for my second question :

What relations may exist between this metric and metrics compatible with the weak topology ? (e.g Prohorov's ? Levy / Bounded Lipshitz ? Wasserstein's ? ...) Or total variation norm ?

Answer by Adrien Hardy for What is the name for a non-normalized distribution?

2012-09-17T09:59:28Z

For an absolutely continuous finite Borel measure $\mu(dx)=f(x)dx$ on $\mathbb R$, if $\mu(\mathbb R)\neq 1$ then $f$ is sometimes called the "intensity" of the measure.

Extracting particles from a determinantal point process

2012-02-06T16:52:21Z

Consider $N$ real random particles $x_1,\cdots, x_N$ distributed according to a density $\rho(x_1,\ldots,x_N)$ with respect to the Lebesgue measure on $\mathbb R^N$, which is assumed to be invariant under permutations : $$ \rho(x_{\sigma(1)},\ldots,x_{\sigma(N)})=\rho(x_1,\ldots,x_N),\qquad \sigma\in\mathfrak S_N. $$ We moreover assume the particles to interact as a determinantal point process with a kernel $K:\mathbb R \times \mathbb{R}\rightarrow\mathbb R$ which is, seen as a kernel operator on $L^2(dx)$, a projection operator.

This means roughly that the density distribution is given by

$$ \frac{1}{N!}\det\Big[ K(x_i,x_j) \Big] \prod_{i=1}^N dx_i. $$

Questions :

If we order the particles $x_1<\cdots< x_N$ and then extract a new system of particles $\{x_i\}_{i\in I}$, where $I\subset \{1,\ldots,N\}$, do we keep a determinantal structure (once forgetting the ordering on the $x_i$'s, $i\in I$ )?

And if Yes, what would be the new kernel ?

This question is motivated from the following particular case : Once the particles ordered, namely $x_1<\ldots< x_N$, it is well known that one has the Fredholm determinant representation : $$ \mathbb P(x_N\leq s)=\det(I-K)_{L^2(s,+\infty)}, \qquad x_1,\ldots,x_N\in\mathbb R. $$ I'm looking for a similar formula involving the operator $K$ for $ \mathbb P(x_k\leq s)$ when $1\leq k < N.$

What do we get from an euclidian affine structure ?

2011-10-22T16:57:13Z

Imagine you investigate a set of objects $\mathcal{E}$, and you just realize this that $\mathcal{E}$ possesses an affine structure with respect to some real vector space $\mathcal{V}$ having a scalar product $\langle \cdot , \cdot \rangle$ (namely an euclidian affine space). My first thought was "Nice, I can equip $\mathcal{E}$ with a structure of metric space and explore the meaning of the associated convergence". But is there moreover some generic advantage to have an euclidian structure ?

One may formulate alternatively : "What informations you get for $\mathcal{E}$ knowing that there exists a notion of orthogonality on ?"

This is a very broad question, but this is motivated by a question I posted few months ago : http://mathoverflow.net/questions/70724/what-do-we-actually-know-about-logarithmic-energy

I guess I'm looking for an answer pointing some Theorems characterizing euclidian affine spaces between metric affine spaces, but also for your first reaction, or direction for investigation, after such a discovery.

Maybe it is of interest to precise that $\mathcal{E}$ is in fact a subspace of signed measures on $\mathbb{C}$ having total mass equals to $1$, and thus is an infinite dimensional real space.

EDIT : I precise my question,

Is there some geometric theory developed for infinite dimensional real pre-Hilbert spaces (which are not complete) ?

I emphasis that the elements of the space I have in mind are measures, and the scalar product has the form $$ \langle \mu, \nu \rangle = \iint K(x,y)d\mu(x)d\mu(y) $$ with a kernel $K : \mathbb{C}^2\rightarrow \mathbb{R}$ given by $$ K(x,y)=\log|x-y|. $$ Maybe there also exists some references for general kernels which are less singular ?

A generalization of the Sanov Theorem

2012-03-02T12:51:25Z

Let $(X_n)_{n\in\mathbb{N}}$ be a sequence of i.i.d random variables with law $\mu$. The Sanov Theorem then states that the empirical measures $$ \mu^N =\frac{1}{N} \sum _{n=1}^N\delta _{X_n} $$ satisfy a large deviation principle at speed $N$ with good rate function $H(.\mid\mu)$, $H$ being the relative entropy.

I was wondering, what is known if we consider a sequence of independent random variables $(X_n)_{n\in\mathbb N}$ but with different laws $(\mu_n)_{n\in\mathbb N}$ ? For example, what if you take $X_n=Y_n+a_n$, where $(Y_n)_{n\in\mathbb n}$ is a sequence of i.i.d random variables with law $\mu$, and $(a_n)_{n\in\mathbb N}$ is a sequence of real numbers such that $$ \frac{1}{N} \sum _{n=1}^N\delta _{a_n}\rightarrow \nu \qquad \mbox{(weakly)} $$ for some probability measure $\nu$ as $N\rightarrow \infty$ ?

EDIT : (after the comment of Anthony Quas) Let's say we may assume the that the convergence rate for the $a_n$'s is as you want, for example at the rate $\exp(-N.)$. My interest is more about what would be the rate function.

Classical convolution VS Free Convolution

2012-04-14T13:40:56Z

We denote $\varphi:\mathbb R^2\rightarrow\mathbb R$ the addition of real numbers, and $\varphi_*:M_1(\mathbb R^2)\rightarrow M_1(\mathbb R)$ the induced push-forward map (where $M_1(\Delta)$ stands for the set of probability measures on $\Delta$).

Note that given $\mu$, $\nu\in M_1(\mathbb R)$, we have by definition of the (classical) convolution of measures $\varphi_*(\mu\otimes\nu)=\mu *\nu$.

Now, consider two random variables $X$ and $Y$ taking values in $\mathbb R$, and denote $\mu_X$, $\mu_Y\in M_1(\mathbb R)$ their laws. If $X$ and $Y$ independent, then the random variables $(X,Y)$ taking values in $\mathbb R^2$ has for law $\mu_{(X,Y)}=\mu_X\otimes\mu_Y$ and $\varphi_*(\mu_{(X,Y)})=\mu_X*\mu_Y$.

I was wondering if it is possible to describe the free (additive) convolution in the same setting :

Consider two auto-adjoint non-commutative random variables $a$ and $b$ which are free, and denote $\mu_a$, $\mu_b\in M_1(\mathbb R)$ their laws. Is there a (universal) bilinear map $\star:M_1(\mathbb R)\times M_1(\mathbb R)\rightarrow M_1(\mathbb R^2)$ such that $\mu_{(a,b)}=\mu_a\star\mu_b$, and moreover $\varphi_*(\mu_{a}\star\mu_b)=\mu_a\boxplus\mu_b$ ?

Somehow the vague question is "what is the analogue of the free product when one describe the elements of an operator algebra through their spectral measures ?"

And what about the free multiplicative convolution ? and the rectangular one ?

EDIT (after the comments). As Mikael de la Salle explained, there is no hope to obtain such operation $\star:M_1(\mathbb R)\times M_1(\mathbb R)\rightarrow M_1(\mathbb R^2)$ because of lack of bilinearity of $\boxplus$. Terry Tao also emphasis that $M_1(\mathbb R^2)$ is certainly not the good space to consider (there is no "space below" once we deal with non-commutativity !).

This motivates the following question :

Is there exists some "space" (let us stay vague) $\mathcal E$ which may represent the joint laws of two non-commutative random variables, equipped with $\star : \mathcal E\rightarrow M_1(\mathbb R)$, and a map $\varphi_*:\mathcal E\rightarrow M_1(\mathbb R)$ which "looks like the $\varphi_*$", such that we have the spliting $$ \varphi_{*} \circ \star = \boxplus \qquad ? $$

Growth of the recurrence coefficients of orthogonal polynomials

2012-03-18T18:40:11Z

Consider the sequence of measures $$d\mu_N(x)=e^{-NV(x)}dx$$ on the real axis, where $V$ is continuous and satisfies the growth assumption $$\lim_{|x|\rightarrow\infty}(V(x)-2\log|x|)=+\infty.$$

Then, denote $P_{k,N}$ the $k$-th orthonormal polynomial associated to $\mu_N$, which is known to satisfy a three-terms recurrence relation $$xP_{k,N}=a_{k+1,N}P_{k+1,N}+b_{k,N}P_{k,N}+a_{k,N}P_{k-1,N}.$$

If we take for example the ($N$-independent) Gaussian case, $\mu_N=e^{-x^2/2}$, where is known that $$a_{k,N}=\sqrt{k},\qquad b_{k,N}=0,$$

it then follows by change of variable that for its renormalized version, $\mu_N=e^{-Nx^2/2}$, we have $$a_{k,N}=\sqrt{k/N}, \qquad b_{k,N}=0.$$ Thus, there exists $\epsilon>0$ such that as $N\rightarrow\infty$

$$ \max_{k\geq 0 \; :\;\left|\frac{k}{N}-1\right|\leq \epsilon} |a_{k,N}|=O(1),$$ and of course a similar statement holds for $b_{k,N}$.

The same observation can be done when considering renormalized Laguerre weights, where this time the $b_{k,N}$'s are not identically zero.

I'm looking for a (relatively simple) proof of such statement for general $\mu_N$ as introduced above.

In fact, it can be proved as a consequence of a Riemann-Hilbert asymptotic analysis that there exists functions $a$ and $b$ such that $$ \lim_{k/N\rightarrow s}a_{k,N}=a(s), \qquad \lim_{k/N\rightarrow s}b_{k,N}=b(s),$$ but it would be nice to be able to establish the more modest statement of the "boudedness" with a simpler proof.

The log kernel and Bochner Theorem

2011-11-10T07:15:47Z

I was wondering if it possible to find a measure $\eta$ on $\mathbb{R}$ such that

$$ L(x):=\log\frac{1}{|x|}=\int e^{itx}\;d\eta(t) $$ for every $x\in [0,1/2]$.

On a structural ground, this question looks to be related to the Bochner Theorem which states that a continuous function $Q:\mathbb{R}\rightarrow\mathbb{C}$ is the Fourier transform of some $finite$ measure iff $Q$ is positive definite (that is for all $n\geq 0$ and all $x_1,\ldots,x_n\in\mathbb{R}$ we have $\det \big[L(x_i-x_j)\big]_{i,j=1}^n\geq 0$).

Note that here $L(0)=+\infty$, and thus doesn't fit with the setting of Bochner Theorem. Nevertheless, one could allow $\eta$ to be an infinite Radon measure, so that $$ \int e^{it 0}d\eta(t)=\eta(\mathbb{R})=+\infty. $$

An other question is : Is $L$ positive definite on [0,1/2] in some sense ? Note that by restricting $L$ to $[0,1/2]$ we have $L(x-y)\geq 0$ for any $x,y\in[0,1/2]$.

Do you know any generalizations of the Bochner theorem which deal with functions which may take the value $+\infty$ ?

Thanks for your help in advance !

Answer by Adrien Hardy for Using slides in math classroom

2011-11-04T23:53:03Z

It also depends on how do you think it is the best for your students to learn : By listening (hopefully carefully) to the course, and then reading notes you'll provide them, OR by letting them write themselves the content.

I don't like to much the first option, certainly because I've not been used too, and I believe it is a huge advantage to write yourself everything at the moment, because of obvious memorization advantages (it was important for me to have my own notations, a kind of taming procedure) and, once you read your notes again, you usually remember where was the parts the teacher got enthusiastic.

Considering then the second option, it is for me an evidence that blackboard win :

you give the time to the students to write since you do it yourself
the statements stay longer (at least if you have enough blackboards, or just keep the main Theorem on !)
there is more interactions content-author-students
your eyes are not constantly dried by this terrible white light
it allows improvisation
it is more classy (personal point of view, I agree)

Against :

it is suicidal (that is terribly soporific for the students) to NOT prepare a lot your presentation, at least as long as you should spend time one slides
it requires a good handwriting from the teacher
its not convenient for drawing complex pictures

My conclusion is then the same than André Henriques !

Answer by Adrien Hardy for analogue of GUE and Ginibre in higher dimensions

2011-10-23T11:26:05Z

By $\|z_1\|$, I understand you mean the maximal modulus of the $z_i$'s.

If you are interested in the process of the $\|z_i\|$'s, you have no chance for a determinantal structure since you may have two different $z_i$'s with same modulus with non zero probability.

Concerning the process of the $z_i$'s, note that even if the Ginibre Ensemble (that is the case $\mathbb{R}^2$) is indeed determinantal, its kernel is related to the polynomials orthogonal with $\exp(-\|x\|^2/2)$, that is the $(z^k)_{k\geq 0}$ ... which have trivial zeros ! My point is that except on $\mathbb{R}$ you won't get so much information concerning $\|z_1\|$ from a determinantal structure.

I don't know how prove the convergence of $\|z_1\|$, but note that from your density expression, once renormalized $z_i\rightarrow z_i/\sqrt{N}$, you still can use the Coulomb-gaz approach to characterize the global distribution of the $z_i$'s (for example by proving a large deviation principle for the empirical measure) : It is given by the unique minimizer $\mu^*$ of the functional $$ \iint\log\frac{1}{\| x-y\|}d\mu(x)d\mu(y) +\frac{1}{2}\int \|x\|^2d\mu(x) $$ over probability measures $\mu$ on $\mathbb{R}^3$ (or higher). I guess that $\|z_1\|$ should converge towards $\max \big(Supp(\mu^*)\cap \mathbb{R}\big)$...

Answer by Adrien Hardy for Relationship between free probability and deterministic graphs?

2011-10-23T11:08:03Z

I believe the relation between deterministic graphs and free probability you mentioned is not something generic. In fact, the main property of your matrix $M$ which makes connection with free probability (at the best of my knowledge) is not to be the adjacency matrix of some graph, but a Jacobi matrix related to some orthogonal polynomials, which themselves come from random matrix models.

Let me try to develop : We first need some computations.

We have to assume $N$ even to make things properly. The Chebychev (monic) polynomials $(T_k)_{k\geq0}$ of the first kind satisfy $T_k(x)=x^k+\ldots$ and are orthogonal for the weight $$ w(x)=\frac{1}{\sqrt{1-x^2}}, $$ defined on $[-1,1]$, namey for any $k\neq l$ $$ \int_{-1}^1T_k(x)T_{l}(x)w(x)=0. $$ Their Jacobi matrix (associated with its recurrent coefficients) is actually $M/2$. For our purpose its enough to know that the zeros of $T_N$ are actually the eigenvalues of $M/2$. Moreover, a formula due to Heine yields $$ T_N(x)=\int_{-1}^1\ldots\int_{-1}^1\prod_{i=1}^N(x-x_i)\prod_{1\leq i < j \leq N}|x_i-x_j|^2\prod_{i=1}^Nw(x_i)dx_i. $$ The change of variables $x_i=\cos\theta_i$ gives $$ T_N(x)=\int_{0}^{2\pi}\ldots \int_0^{2\pi}\prod_{i=1}^N(x-\cos\theta_i)\prod_{1\leq i < j \leq N}|\cos\theta_i - \cos\theta_j|^2\prod_{i=1}^Nd\theta_i $$ and by Weyl formula $$ T_N(x)=\int_{\mathcal{U}_N}\det(xI_N-\frac{U+U^*}{2}) dU = \mathbb{E}_{Haar}\Big(\det(xI_N-\frac{U+U^*}{2})\Big) $$ where $dU$ stands for the Haar measure of the unitary group $\mathcal{U}_N$.

Conclusion : The random matrix $U+U^*$, with $U$ distributed according to Haar, has for mean eigenvalues the zeros of $T_N(x/2)$, and equivalently the eigenvalues of $M$. Thus they should have the same limiting distribution as $N\rightarrow\infty$ as soon as that the limiting distribution of $U+U^*$ is deterministic.

One one hand, the limiting distribution of $M$ is indeed known to be the arcsine distribution (note it is also the limiting distribution of the zeros of $T_N$ as $N\rightarrow\infty$, which is known to minimize the logarithmic energy $$ \iint \log\frac{1}{|x-y|}d\mu(x)d\mu(y) $$ over all probability measure $\mu$ on $[-1,1]$, a classical statement in potential theory).

On an other hand, by the invariance property of the Haar measure, the distribution of $U+U^* $ is the same than $A+VAV^*$, with $V$ also distributed according to Haar, which is known to converge by Voiculescu Theorem towards $\mu_A\boxplus\mu_A$, where $\mu_A$ is the limiting distribution of your matrix $A$, namely $\mu_A=\frac{1}{2}(\delta_1+\delta_{-1})$.

Answer by Adrien Hardy for Eigenvalue densities of sample covariance matrices when the population covariance matrix is a perturbed identity matrix

2011-10-22T19:24:58Z

You can rewrite $$ S=\frac{1}{M}C^{1/2}XX^T(C^{1/2})^T $$ where $X_{.j}\sim \mathcal{N}(\boldsymbol 0, \boldsymbol I)$. It is then classical that the limiting eigenvalue distribution of $S$ (in the almost sure weak convergence) is given by $MP_\gamma\boxtimes\nu$, the free multiplicative convolution of the Marchenko-Pastur distribution $MP_\gamma$ of paramter $\gamma$ with the limiting eigenvalue distribution $\nu$ of $C$. Thus, you indeed obtain $MP_\gamma$ if and only if $\nu=\delta_1$, namely if and only if $$ \frac{1}{N}\sum_{i=1}^N\delta_{v_i} \longrightarrow \delta_0\qquad \mbox{weakly as $N\rightarrow\infty$ }. $$ For a reference, you can look at the book of Anderson, Guionnet and Zeitouni, "Introduction to random matrices" (Chapter 5), but also any introduction to free probability I guess.

As you can see, free probability provides a powerful language to describe such perturbed random matrix models!

NB : As an example, a spiked model is the situation where only a fixed finite number of $v_i$'s are non zero, which fit with the above characterization.

Convergence of Fredholm determinants

2011-08-12T07:24:52Z

Let $(X_N)_N$ be a sequence of trace class operators acting on, say, $L^2(\mathbb{R})$. What are the minimal assumptions in order to have the convergence of their Fredholm determinant $$ \lim_N\det(I+X_N) ? $$ I know $X\mapsto \det(I+X_N)$ is continuous for the trace class norm topology (once restricted to trace class), ok. Let's say that $X_N$ converges weakly to $X$, it is enough to have the convergence of Fredholm determinants ? What conditions should we add ? I guess I just need a good reference on the topic. Any ideas ? Thanks in advance.

When should we expect Tracy-Widom ?

2011-07-26T12:19:11Z

The Tracy-Widom law describes, among other things, the fluctuations of maximal eigenvalues of many random large matrix models. Because of its universal character, it obtained his position on the podium of very famous laws in probability theory. I'd like to discuss what are the ingredients to be present in order expect his apparition.

More precisely, the Tracy-Widom law has for cumulative distribution the Fredholm determinant $$ F(s)=\det(I-A_s) $$ where the operator $A_s$ acts on $L^2(s,+\infty)$ by $$ A_sf(x)=\int A(x,y)dy,\qquad A(x,y)=\frac{Ai(x)Ai'(y)-Ai(y)Ai'(x)}{x-y}, $$ $Ai$ being the Airy function. It is moreover possible to rewrite $F$ in a more explicit (?) form, involving a solution of the Painlevé II equation. It is known that this distribution describes the fluctuations of the maximal value of the GUE, and actually of a large class of Wigner Matrices. It curiously also appears in many interacting particle processes, such as ASEP, TASEP, longest increasing subsequence of uniformly random permutations, polynuclear growth models ... (For an introduction, see http://arxiv.org/abs/math-ph/0603038 and references inside. You may jump at (30) if you are in a hurry, and read more about particles models in Section 3). A natural (but ambitious) question is

You have $N$ interacting random points $(x_1,\ldots,x_N)$ on $\mathbb{R}$, when can you predict that $x_{\max}^{(N)}=\max_{i=1}^N x_i$ will fluctuate (up to a rescaling) according to Tracy-Widom law around its large $N$ limiting value ?

Assume that the limiting distribution of the $x_i$'s $$ \mu(dx)=\lim_{N\rightarrow\infty}\frac{1}{N}\sum_{i=1}^N\delta_{x_i}\qquad \mbox{(in the weak topology)} $$ admits a density $f$ on a compact support $S(\mu)$, and note $x_\max=\max S(\mu)$ (which can be assumed to be positive by translation). I have the impression that a necessary condition for the appearance of Tracy-Widom is to satisfy the three following points :

1) (strong repulsion) There exists a strong repulsion between the $x_i$'s (typically, the joint density of the $x_i$'s has a term like $\prod_{i\neq j}|x_i-x_j|$, or at least the $x_i$'s form a determinantal point process).

2) (no jump for $x_\max^{(N)} $) $x_\max^{(N)}\rightarrow x_\max$ a.s. when $N\rightarrow\infty$.

3) (soft edge) The density of $\mu$ vanishes like a square root around $x_\max$, i.e. $f(x)\sim (x_\max-x)^{1/2}$ when $x\rightarrow x_\max$.

For TASEP and longest increasing subsequence models, one can see that 1), 2) and 3) hold [since these models are somehow discretizations of random matrix models where everything is explicit (Wishart and GUE respectively)]. For the Wigner matrices, 2) and 3) clearly hold [Wigner's semicircular law], and I guess 1) is ok [because of the local semicircular law]. For ASEP, 1) clearly holds [because of the E of ASEP], 2) and 3) are not so clear to me, but sound reasonable.

Do you know any interacting particle model where Tracy-Widom holds but where one of the previous points is cruelly violated ?

Of course the condition 1) is pretty vague, and would deserve to be defined precisely. It is a part of the question !

NB : I have a pretty weak physical background, so if by any chance a physicist was lost on MO, I'd love to hear his/her criteria for Tracy-Widom...

Answer by Adrien Hardy for Why only three classical matrix ensembles in RMT? (Newbie question)

2011-07-26T00:31:15Z

These three ensembles are hermitian matrices over a (finite dimensional real) field of numbers, and it is known that the only finite dimensional real fields are the real numbers, the complex numbers ($2$-dimensional) and the quaternionic numbers ($4$-dimensional). Octonions are not a field of number since you do not have associativity. The motivation in physics comes from the fact that an hermitian matrix represents a finite dimentional Hamiltonian (an Hermitian operator) in quantum mechanics (then you add randomness, in order to take in account the lack of information about your system, and you let the size of the matrix, that is the dimension of your state space where your Hamiltionian is acting on, going to infinity). In this setting, $N\times N$ quaternionic matrices have to be seen as subclasses of complex hermitian matrices (but of size $2N\times 2N$) and both real symmetric and quaternionic hermitian matrices are a subclass of complex hermitian matrices, with extra symmetries. Anyway, you may imagine many different matrix models relevant for studying (look for Wigner matrices, the answer of Beenakker about other symmetries in physics, the generalized $\beta$-ensemble of Edelman, etc ...)

Answer by Adrien Hardy for Product rules are local and covariance identities are global

2011-06-16T22:01:51Z

I understand that you consider a "parametrized" identity on a space of differentiable functions $$ \mathcal{L}_{a,b}(f,g)=[f-a][g-b]+a[g-b]+b[f-a]-fg+ab=0 $$ and two different linear forms acting on that space, namely $\phi : f\mapsto f'(x_0)$ and $\psi : f\mapsto\mathbb{E}(f)$, which induce the identities $\phi [\mathcal{L}_{x_0,\;x_0}(f,g)]=0$ (the product rule) and $\psi[\mathcal{L}_{\mathbb{E}(f),\;\mathbb{E}(g)}(f,g)]=0$ (the definition of the covariance).

Of course you can evaluate this identity along all the (linear) maps you want playing with the parameters $a$ and $b$, but maybe your question was about a potential more subtle relation between "derivative" and "variance" of functions ? In this case, I have the impression that these two identities are different in nature, since the parameters $a$ and $b$ have to be chosen in a very different way for each form.

Answer by Adrien Hardy for Estimating the probability that one Poisson RV is larger than another

2011-06-01T18:38:24Z

By simple computations : The definition of the modified Bessel function of the first kind yields $$ I_k(\lambda)=\sum_{n\geq 0}\frac{1}{n!(n+k)!}\left(\frac{\lambda}{2}\right)^{2n+k} $$ so that we get (the sums transpositions are clearly allowed) $$F(\lambda)=e^{-\lambda-1}\sum_{k\geq 1}\sum_{n\geq 0}\frac{\lambda^{k+n}}{n!(k+n)!}=e^{-\lambda-1}\sum_{n\geq 1}a_n\lambda^n \qquad \mbox{where}\qquad a_n=\frac{1}{n!}\sum_{k=0}^{n-1}\frac{1}{k!}.$$ Thus, deriving under the sign sum

$$ F'(\lambda) = e^{-\lambda-1}\Big(1+\sum_{n\geq 1 }[(n+1)a_{n+1}-a_n)]\lambda^n\Big) = e^{-\lambda-1}\sum_{n\geq 0}\frac{1}{(n!)^2}\lambda^n $$ we obtain the closed form $$ F'(\lambda)=e^{-\lambda-1}I_0(2\sqrt{\lambda}). $$ One finally get $$F'(0)=e^{-1}, \quad F'(1)=e^{-2}I_0(2)=e^{-2}\sum_{n\geq0}\frac{1}{(n!)^2}$$ and, using the asymptotic formula when $\lambda\rightarrow+\infty$ for all $k$ $$ I_k(\lambda)=\frac{e^{\lambda}}{\sqrt{2\pi\lambda}}\Big(1+O(\lambda^{-1})\Big), $$ that $$ F'(\lambda)=\frac{e^{2\sqrt{\lambda}-\lambda-1}}{2\sqrt{\pi\sqrt{\lambda}}}\Big(1+O(\lambda^{-1/2})\Big) $$ when $\lambda\rightarrow+\infty$.

Compact sets of the complex plane having the K-property ?

2011-06-24T15:22:16Z

I would like to have a better understanding of a notion I've met in the beautiful book of Nikishin and Sorokin "Rational approximation and Orthogonality", since they do not provide examples.

As it is classical to do in potential theory, denote for $\mu$ in $M_1(K)$, the set of probability measures on a compact set $K\subset \mathbb{C}$, its logarithmic energy by $$ I(\mu)=\iint \log\frac{1}{|x-y|}d\mu(x)d\mu(y)$$ and define the capacity of a compact set $K\subset\mathbb{C}$ as $$ Cap(K)=\exp\Big(-\inf_{\mu\in M_1(K)} I(\mu)\Big).$$ $K$ is said to satisfy the K-property at $z\in K$ if there exists $\rho_z > 0$ and $k_z> 0$ such that $$ Cap(K\cap D(z,\rho))\geq \rho^{k_z}$$ for any $0< \rho < \rho_z$, where $D(z,\rho)$ stands for the disc centered at $z$ with radius $\rho$. We say that $K$ satisfies the K-property if it satisfies the K-property at every $z\in K$.

One can show that segments, or circles, satisfy this K-property.

Questions :

Example of compact sets with positive capacity which do not have the K-property ?
More generally, do you have references about K-property for compact sets ?

Comment by Adrien Hardy

2013-05-05T09:22:39Z

Yes, it does ! By any chance do you know if there is a similar characterization involving $C_\mu$ ?

Comment by Adrien Hardy

2013-04-27T09:18:54Z

I'm not sure the $q$-Catalan numbers "count things", since they are not integers

Comment by Adrien Hardy

2013-04-19T22:52:10Z

Well, for finite dimensional spaces, like $2\times 2$ matrices, all norms are equivalent. Thus, chose your favorite norm on $2\times 2$ matrices (I'd chose the $\sup$ over the coefficients), and then take the $L^2$ norm (with respect to $z$) of it. That's what I'll use for $\|v\|_{L^2(\Sigma)}$.

Comment by Adrien Hardy

2013-03-03T23:10:54Z

Hi Tom. At least for a real Banach space $X$, one may define a Gaussian measure $\gamma$ on $X$ by duality, that is a measure such that for any $f\in X^*$, $f_*\gamma$ is a (real) Gaussian measure. Maybe it does not help to much, but my point is that, for me, this is more about duality than projections. (see e.g. en.wikipedia.org/wiki/Abstract_Wiener_space)

Comment by Adrien Hardy

2013-02-26T15:07:53Z

well, no not the dimension, that's obvious after your comment :)

Comment by Adrien Hardy

2013-02-26T14:48:33Z

I agree. but the question is to describe these isometries. For e.g. do they all have the form $\wedge^n\Phi$ for some linear map $\Phi:\mathbb R^d\rightarrow\mathbb R^d$ ? What can we say about $\Phi$ then ? What the dimension of that space Etc ...

Comment by Adrien Hardy

2013-02-26T10:10:36Z

"the genus" of a planar curve ?

Comment by Adrien Hardy

2012-12-17T21:19:34Z

First, this may be not true if your sequence of RVs are not independent. If they are independent, then your statement is indeed the CLT. If this is not clear, I advice you to ask your question on math.stackexchange, where people would certainly more interested to develop than on mathoverflow (which concerns research level questions).

Comment by Adrien Hardy

2012-10-15T16:38:40Z

Not the mathematicians which work with determinantal point processes. The distinction between "density" and "intensity" is often made when one speaks about the Radon-Nikodym derivative of a non-probability measure.

Comment by Adrien Hardy

2012-10-02T09:50:15Z

Oups, that was silly ... The question is now edited, sorry about that.

Comment by Adrien Hardy

2012-09-25T11:58:57Z

I like it too !

Comment by Adrien Hardy

2012-09-25T10:17:45Z

@roork: I apology if you felt there is some contempt my answer, that is not my point. The thing is the answer follows from the definition of this infinite product probability space, and in particular its expectation, and looking for a commonly used definition is more appropriate for MSE. Usually people on MO downvote and directly close such questions, I just wanted to warn you.

Comment by Adrien Hardy

2012-06-04T10:42:01Z

@Deane Yang : I agree, that's why I speak about "Euclidean affine structure" (with a typo, I just realize, sorry for the gallicism...). Maybe I should call it "Euclidean structure", but I preferred to emphases that we do not have a norm, but just a metric and the notion of angles.

Comment by Adrien Hardy

2012-04-26T19:06:15Z

Hi Steven ! Hope everything is fine in Sweden.

Comment by Adrien Hardy

2012-04-24T20:34:22Z

I kind of ask a similar question to a researcher doing maths for biology once, like "did someone use symmetry theories with a significant impact in biology ?" (I had in mind the Noether theorem and relatives). His answer : "it seems it is not enough to apply known theorems of math-phy to math-bio, it really needs new technology and objects".