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34

First of all this has nothing to do with the inflection point of $e^{-\alpha x^2}$. According to Poisson summation formula (see Whittaker, Watson, Modern analysis, chapter 21.51) $$\sum_{n=-\infty}^\infty e^{-\alpha (x-n)^2}=2{\sqrt{\frac{\pi}{\alpha}}}\left(1+2\sum_{n=1}^\infty e^{-\frac{\pi^2}{\alpha}n^2}\cos2\pi n x\right),\quad \text{Re}~\alpha>0.\... 23 The entry from May 24, 1796 is worked out in a more general form on February 16, 1797 [reproduced below from this scan]$$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 ... 21 The comments list many reasons why the Gaussian distribution is special, but is it "the most fundamental" among all distributions, as suggested in the OP? I would like to argue that (1) conservation laws are among the most fundamental laws of Nature, and (2) a quantity that obeys a conservation law will naturally follow an exponential -- rather ... 16 There is a whole book which addresses exactly this type of questions: suppose that a distribution has such and such properties, then it must be Gaussian (or sometimes Poisson). MR0346969 Kagan, A. M.; Linnik, Yu. V.; Rao, C. Radhakrishna Characterization problems in mathematical statistics, John Wiley & Sons, New York-London-Sydney, 1973. 14 If the random vector (X,Y) in the plane has independent coordinates and a rotation-invariant distribution, then it is Gaussian. 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 ... 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. 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 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 ... 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 ... 7 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... 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 ...

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

6

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 $\mathbb{E}(X_{i,i})=p=:a$, $\mathbb{E}... 6 Here is why solutions do exist.$\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 $$\lim_{\nu\to\... 6 It is simple to reduce to the case where \sum_i\mu_i=0. Then the simplest nontrivial case is (x-\mu)e^{-(x-\mu)^2}+(x+\mu)e^{-(x+\mu)^2}=0, which is equivalent to (x+\mu)/(x-\mu)=e^{4x\mu}. Here there is a solution of the form$$ 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 - \... 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 ... 6 There is a lot of confusion around the concept of "Wick" product. Much of it is due to the following. As you mention, there is a general formula for the Wick product of a collection of random variables. Given a collection$\{X_i\}_{i \in I}$and an$I$-valued multiindex$\alpha$, it says that$X^{\diamond \alpha}$is the unique polynomial of degree ... 6 The Normal Distribution is the limit, in the sense of distributions, of the scaled sum of$n$IID variables. This is the Central Limit Theorem. I posted an outline of a proof of the Central Limit Theorem here: In what follows, the fourier transform used is $$\widehat{f}(\xi)=\int_{-\infty}^\infty f(x)\,e^{-i2\pi x\xi}\;\mathrm{d}x$$ Suppose we have a ... 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 $$b_0(x):=\frac{b(x)}{... 5 Denote the Gaussian random vectors by X(n). Clearly, \mathrm{tr} \, S(n) = \mathsf{E} \, \Vert X(n) \Vert^2, so \mathrm{tr} \, S(n) \to 0 is certainly sufficient for weak convergence to 0. And in fact it's also necessary. Indeed, \Vert X(n) \Vert^2 is a quadratic functional of a Gaussian, and for those convergence in probability is known to be ... 5 Too long for a comment... I'm not sure why this is so surprising. If one sums translates by \varepsilon \mathbb{Z} of the standard Gaussian then for \varepsilon sufficiently small the result ought to be roughly constant. More generally, one can take any sufficiently nice function in lieu of the Gaussian, albeit with the awareness that this will affect \... 5 OK, here are my 2 cents. I'll try to outline the logic and then make a conclusion. We'll start with the Gaussian case. Suppose that we have a Gaussian peak f(t)=e^{-t^2/2}. We want to de-convolve it to the Dirac delta-measure. Since the Fourier transform \widehat f(s)=\int f(t)e^{-ist}\,dx satisfies \widehat{(f*g)}=\widehat f \widehat {g{,}} and the ... 5 Convolution with a Gaussian kernel of an n-point function has n^2 complexity, while Fourier transformation (FFT), multiplication, and inverse Fourier transformation is only of complexity n\log n. Here is a Python code for the two-dimensional case. 4 indeed, integration by parts does the trick:$$\int d\mathbf{x}\; \mathbf{x} \,f(\mathbf{x})\exp\left(-\tfrac{1}{2}\mathbf{x}\cdot \mathbf{C}^{-1}\cdot\mathbf{x}\right)=-\int d\mathbf{x}\; \,f(\mathbf{x})\,\mathbf{C}\cdot\frac{\partial}{\partial \mathbf{x}}\exp\left(-\tfrac{1}{2}\mathbf{x}\cdot \mathbf{C}^{-1}\cdot\mathbf{x}\right)=\int d\mathbf{x}\; ... 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))\$ ...

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