23

The entry from May 24, 1796 is worked out in a more general form on February 16, 1797 [reproduced below from this scan]
$$1-a+a^3-a^6+a^{10}+\cdots=\frac{1}{\displaystyle 1+\frac{\strut a}{\displaystyle 1+\frac{\strut a^2-a}{\displaystyle 1+\frac{\strut a^3}{\displaystyle 1+\frac{\strut a^4-a^2}{\displaystyle 1+\frac{\strut a^5}{1+\cdots}}}}}}$$
so the ...

answered Jun 26 '19 at 20:19

Carlo Beenakker

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12

After you diagonalize the covariance matrix, you have $||X||^2=\sum_{k=1}^n X_i^2$, right? And $X_1^2,\ldots,X_n^2$ are independent $\chi(1)$-distributed r.v. (up to a scaling with $\frac{1}{\sqrt{\lambda_i}})$, i.e., they are gamma-distributed, but with different scale parameter. So you can apply the results of this paper:
P. G. Moschopoulos (1985) The ...

12

I don't know the answers to your questions, i.e., I don't know whether there has been any work already done on this problem, nor do I know whether there is any kind of 'heat flow' argument for constructing solutions.
However, I suspect that the latter, if possible, is not going to be completely straightforward. After all, in the example I mention in my ...

answered Oct 4 '12 at 12:29

Robert Bryant

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12

Yes, the conjectured lower bound is true and can be proved using fairly simple, if somewhat tedious, analysis of derivatives.
First define
$$
b := f - N^2 = x(xN + n) - (xN + n)^2 + N(1-N)\>.
$$
The plan is to show that $b$ is a decreasing function bounded below by zero.
Let $u := x N + n$, so that $b = (x-u)u + (1-N)N = (x-u)u + (1-u')u'$. Since $u(-x) ...

11

A recursion formula for the moments of the Gaussian orthogonal ensemble, M. Ledoux (2009).
The desired recursion formula for the moment $b_p^N\equiv E\,[\,{\rm tr}\,(S_N^{2p})]$ is
I notice a difference in normalization, you'll want to divide $b_p^N$ by $2^p$.

answered Feb 6 '18 at 17:08

Carlo Beenakker

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8

Well, there are some easy points on $C_R$. The point $((1+R)/2, (1-R)/(2i))$, for example.
More generally, the equation $x^2+y^2=R$ is equivalent to $(x+iy)(x-iy) = R$. So finding a rational point on $C_R$ is exactly equivalent to finding $u$ and $v$ in $\mathbb{Q}(i)$ with $uv=R$. You can do that by choosing $u$ at random, and computing $v=R/u$, $x=(u+v)/...

answered Sep 11 '12 at 17:59

David E Speyer

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8

Interesting! Put another way, the fractional part of the standard Gaussian variable closely approximates the standard uniform variable.
You can also closely approximate standard Gaussian from standard uniform: Add 12 independent uniform variables before subtracting 6. This approximation is also shockingly good; see John D. Cook's comment here. (12 isn't "...

8

There always exists at least one solution: If $x >\mu_i$ for all $i$, then each of the terms in the sum is positive and if $x < \mu_i$ for all $i$, then each of the terms in the sum is negative. Thus, the function on the left hand side has at least one zero and all the zeroes lie strictly between the lowest of the $\mu_i$ and the highest.
Finding ...

answered Dec 14 '18 at 10:31

Robert Bryant

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7

In the notation of the question,
$$
\int f(x,y)\nu(y)dy=W_\sigma\ast \nu(x)
$$
where $W_\sigma(x)=e^{-|x|^2/2\sigma^2}$ and $\ast$ denotes convolution. Thus, if
$$
\int f(x,y)\nu(y)dy=\lambda\nu(x)
$$
then by taking Fourier transforms:
$$
\lambda \hat{\nu}= \widehat{\lambda \nu}
=\widehat{W_\sigma\ast \nu}
=\widehat{W_\sigma}\widehat{\nu}
=c_1W_{c_2}\...

7

Both the Gaussian maximum entropy distribution and the Gaussian solution of the diffusion equation (heat equation) follow from the central limit theorem, that the limiting distribution of the sum of i.i.d. random variables with given average and variance is a Gaussian. The connection between the central limit theorem and the diffusion equation, which ...

answered Sep 2 '17 at 12:26

Carlo Beenakker

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7

Here is a another approach.
For convenience I write $n$ instead of $N$, and $A_n$ for $A$.
By definition
$$\det(A_n) = \sum_{\pi\in S_n} \operatorname{sign}(\pi) \prod_{i=1}^n a_{i,\pi(i)}$$
By assumption $A_n$ is symmetric and the $a_{i,j}$ are (for $j\geq i$) mutually independent random variables $X_{i,j}$
with $\mathbb{E}(X_{i,i})=p=:a$, $\mathbb{E}...

6

This problem reduces quickly to Holder continuity of the operator square root. That is, there exists a $C > 0$ such that
$$
\begin{align}
\lVert\sqrt{A}-\sqrt{B}\rVert\le C\lVert A-B\rVert^{1/2}&&{\rm(1)}
\end{align}
$$
for any positive semidefinite operators $A,B$.[1]
Assuming (1), the proof of continuity with Holder exponent $1/2$ as mentioned ...

6

A well-known probabilist once told me that the 3 main classes of stochastic processes are Gaussian, Markov, and martingales.
Martingales are definitely useful in finance and also with respect to other betting games, but it seems to me that they are essential mostly because we have such powerful tools to study them. These tools have often proved useful in ...

6

They are not equivalent.
For an explicit counterexample, let $\{e_1, e_2, \dots\}$ be an orthonormal basis for $H$, and let $C$ be the diagonal operator $C e_n = \frac{1}{n^2} e_n$. Let $D =2C$. Then $R_v = 1/2$ for every $v$ so (2) is satisfied.
Define random variables $X_n$ on $H$ by $X_n(v) = {n} \langle v, e_n \rangle$. Then under $N(0,C)$, the $X_n$...

answered Jan 20 '15 at 17:30

Nate Eldredge

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6

It is true, and it follows from the following fact, sometimes referred to as Sudakov-Fernique inequality, sometimes as Slepian-Fernique lemma:
Assume that $(X_t)$ and $(Y_t)$ are two centered Gaussian processes. If $\|Y_s - Y_t\|_2 \leqslant \|X_s - X_t\|_2$ then $\mathbb{E} \sup Y_t \leqslant \mathbb{E} \sup X_t$.
In this inequality the index set $T$ is ...

6

EDITED: As pointed out by Anthony and John, my 2am solution was anything but. In summary, the conjecture is TRUE for $C$ smaller than approximately 0.6880137 and false for larger $C$.
The exact value is
$$ S(C,n) = \sum_{k=0}^n 2^{-n}\binom{n}{k} \exp\bigl(Cn^{-3}(n-2k)^4\bigr)$$
since the product of the first two terms is the probability that $\sum_i X_i=n-...

6

For all the similarities, there's a significant difference between both parts of the analogy.
In the IUT side of the analogy, we are trying to compare the "mutually alien copies" at each end of the $\Theta$-link
$$\{\underline{\underline{q^{j^2}}}\} \longleftrightarrow \underline{\underline{q}}\tag{1}$$
But in the case of the Gaussian integral, we are ...

6

Here is why solutions do exist. $\newcommand{\bR}{\mathbb{R}}$ Consider the function $f:\bR\to\bR$
$$
f(x)=\sum_{i=1}^n e^{-(x-\mu_i)^2}.
$$
Note that
$$
0\leq f(x)\leq n, \;\;\forall x\in\bR
$$
and
$$
\lim_{\vert x\vert\to\infty} f(x)= 0.
\tag{$1$}
$$
Set
$$
M:=\sup_{x\in \bR} f(x).
$$
If $(x_\nu)$ is a sequence of real numbers such that
$$
\lim_{\nu\to\...

answered Dec 14 '18 at 10:49

Liviu Nicolaescu

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6

It is simple to reduce to the case where $\sum_i\mu_i=0$. Then the simplest nontrivial case is $(x-\mu)e^{-(x-\mu)^2}+(x+\mu)e^{-(x+\mu)^2}=0$, which is equivalent to $(x+\mu)/(x-\mu)=e^{4x\mu}$. Here there is a solution of the form
$$ x = \frac{1}{\sqrt{2}} \left(1 + \frac{1}{3}\mu^2 + \frac{11}{90}\mu^4 + \frac{17}{630}\mu^6 - \frac{281}{37800}\mu^8 - \...

answered Dec 14 '18 at 12:22

Neil Strickland

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6

First, we need to fix the notation a bit. Let $X_1,X_2,\dots$ be iid zero-mean unit-variance random variables (r.v.'s). For each natural $n$, let the $n$-tuple $(J_1,\dots,J_n):=(J_{n,1},\dots,J_{n,n})$ of r.v.'s be independent of the $X_k$'s and have the multinomial distribution with parameters $n,1/n,\dots,1/n$.
For each $k\in[n]:=\{1,\dots,n\}$, the ...

5

Denote the Gaussian random vectors by $X(n)$.
Clearly, $\mathrm{tr} \, S(n) = \mathsf{E} \, \Vert X(n) \Vert^2$, so $\mathrm{tr} \, S(n) \to 0$ is certainly sufficient for weak convergence to $0$. And in fact it's also necessary. Indeed, $\Vert X(n) \Vert^2$ is a quadratic functional of a Gaussian, and for those convergence in probability is known to be ...

5

Here is a complete solution. The idea is to kill the entries of $N$ in two steps, by applying two appropriately constructed first-order differential operators, which will result in a simple elementary expression:
Let $b:=f-N^2$. As noted by cardinal, $b$ is an even function. So, it is enough to show that $b>0$ on $[0,\infty)$. Let
$$
b_0(x):=\frac{b(x)}{...

5

Too long for a comment...
I'm not sure why this is so surprising. If one sums translates by $\varepsilon \mathbb{Z}$ of the standard Gaussian then for $\varepsilon$ sufficiently small the result ought to be roughly constant. More generally, one can take any sufficiently nice function in lieu of the Gaussian, albeit with the awareness that this will affect $\...

5

$\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}
\newcommand{\ep}{\epsilon}
\newcommand{\ga}{\gamma}
\newcommand{\Ga}{\Gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\newcommand{\thh}{\theta}
\newcommand{\R}{\mathbb{R}}
\newcommand{\E}{\operatorname{\mathsf E}}
\newcommand{\PP}{\operatorname{\mathsf P}}$
Let $M:=\|X\|_{\infty}$. Then for $x>...

5

There is no closed form expression, for approximations see:
Lower and upper bounds for approximation of the Kullback-Leibler divergence between Gaussian mixture models (2012)
A lower and an upper bound for the Kullback-Leibler divergence between
two Gaussian mixtures are proposed. The mean of these bounds provides an
approximation to the KL ...

answered Aug 10 '18 at 16:43

Carlo Beenakker

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5

OK, here are my 2 cents. I'll try to outline the logic and then make a conclusion. We'll start with the Gaussian case.
Suppose that we have a Gaussian peak $f(t)=e^{-t^2/2}$. We want to de-convolve it to the Dirac delta-measure. Since the Fourier transform $\widehat f(s)=\int f(t)e^{-ist}\,dx$ satisfies $\widehat{(f*g)}=\widehat f \widehat {g{,}}$ and the ...

4

To some extent this depends on the model of computation. Some remarks:
The problem must always involve an accuracy parameter $\epsilon$, since in general the answer will not be rational
Even in the case of a standard 1-dimensional normal, the question is not so easy. I believe this can be done to accuracy $\epsilon$ in $\widetilde{O}(\log^2 (1/\epsilon))$ ...

4

The magic words are "Kac-Rice formula". This is exactly the question which comes up in analyzing zeros of random polynomials.

answered Jul 24 '13 at 21:30

Igor Rivin

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4

Suppose your process is continuous. (By the Kolmogorov-Centsov continuity theorem, a sufficient condition for this is that $c(s,s) + c(t,t) - 2 c(s,t) \le C |s-t|^\gamma$ for some $C, \gamma > 0$.) Then your process induces a Gaussian measure on $C([0,T])$ and a theorem due to Fernique gives an asymptotic result:
Theorem. There exists $C$ and $\...

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