New answers tagged

7

The probability that a $n\times n$ real matrix (with elements that are independent random variables with standard normal distributions) has only real eigenvalues is given by $$ 2^{-n(n-1)/4}$$ Reference: A. Edelman, The Probability that a Random Real Gaussian Matrix has $k$ Real Eigenvalues, Related Distributions, and the Circular Law. Journal of ...


0

In the linked answer, the inequality sign $\le$ in $\frac12\,I_n\le EYY^T\le I_n$ is not meant in the sense of the entrywise comparison. Rather, it is meant in this sense: for any two symmetric matrices $A$ and $B$, we write $A\le B$ if $B-A$ is positive semidefinite. In this case, we have $Y=X1_{\|X\|\le4C\sqrt{n}}$, where $EXX^T=I_n$ and $$E\langle X,\...


1

$\newcommand\al\alpha\newcommand\R{\mathbb R}$The domains of attraction to multidimensional stable distributions are characterized by Rvačeva's Theorems 4.1 (p. 194, for $\al=2$) and 4.2 (p. 196, for $\al<2$). The domains of strict attraction to multidimensional strictly stable distributions are characterized by Shimura's Theorems 3.1 (p. 354, for $\al=1$)...


2

Even $c=1-\varepsilon$ works when $n$ is large enough. This question is about bounding the diagonal bipartite Ramsey, see the recent achievement of David Conlon here http://people.maths.ox.ac.uk/~conlond/Bipartite.pdf


1

Let $A_t$ be the event that $S_1$ is a substring of $S_2$, $S_2=pS_1q$, where the length of $p$ is $t$. Then the probability of $\cup_t A_t$ can be found by inclusion-exclusion as $$\sum P(A_t)-\sum P(A_{t_1}\cap A_{t_2}) + \sum_{t_1,t_2,t_3} P(A_{t_1}\cap A_{t_2}\cap A_{t_3})-\dots$$ Terms like $P(A_{t_1}\cap A_{t_2})$ have probabilities that depend on how ...


2

One way to construct the thermodynamic limit of the states $\mu_{\Lambda,\beta,h}^\varnothing$ is to observe that, for any local function $f$ and any increasing sequence of sets $\Lambda_n\uparrow\mathbb{Z}^d$, the support of $f$ will be included inside $\Lambda_n$ for all large enough $n$. In particular, for any local function $f$, one can prove that the ...


2

Let $k_n:=\kappa_n$, $a_n:=E(X-EX)^n$, $b_n:=E|X-EX|^n$, so that $|a_n|\le b_n$. We have to show that $$|k_n|\le n^n b_n$$ for natural $n$. For $n=1,2$ this is obvious. The key is the recursion $$k_n=a_n-\sum_{m=1}^{n-1}\binom{n-1}{m-1}k_m a_{n-m}$$ at the end of this section , which implies $$|k_n|\le b_n+\sum_{m=1}^{n-1}\binom{n-1}{m-1}|k_m| b_{n-m}.$$ ...


1

It looks fine to me. If we let $I_k = [k t^\ast_M, (k+1) t^\ast_M]$ be the relevant subintervals of $[0,T]$, then the supremum of $|u_n|$ over $[0,T]$ must be almost attained along some sequence of points, and by pigeonhole infinitely many of them must be in one of the $I_k$, call it $I_{k_0}$, so that $\sup_{I_{k_0}} |u_n| = \sup_{[0,T]} |u_n|$. So if the ...


3

No, not even for sequences of countably additive measures. Take $X = \mathbb{N} = \{0,1,2,\dots\}$ with its discrete $\sigma$-algebra, and let $\mu_n$ put mass $1/n$ at the point $n$ and mass $1-1/n$ at $0$. Let $\mu$ put mass $1$ at $0$. Then it is clear that $\mu_n(A) \to \mu(A)$ for every set $A$ (consider the cases $0 \in A$ and $0 \notin A$). Set $f(n)...


2

A book that may contain answers to all the questions you have is 'Order Statistics' by David and Nagaraja. For you first question, the easiest way is to rely on the fact that order statistics can be written as the inverse cdf of the corresponding order statistic of a uniform sample, i.e. $$X_{(1)}, \ldots, X_{(n)} = (F^{-1}(U_{(1)}),\ldots, F^{-1}(U_{(n)})),$...


3

Looking at your "variational" $\delta$-notation, it appears that it will be enough for you to consider the Gateaux derivative, which is a weaker version of the Fréchet derivative. For $F(p):=\int_X p\ln p\,dx$, the Gateaux derivative of $F$ at a point $p$ in the direction $h\in L^1(X)$ is $f'(0)$, where $$f(t):=F(p+th).$$ For variational purposes, ...


1

The Skorohod space $D$ has $C([0,1])$ as a Banach subspace that does not have smooth bump functions by results of Bonic and Frampton from 1965. If $D$ had smooth partitions of unity, it would posses smooth bump functions, hence also $C([0,1])$ which is false. So the answer is no. See Section 14 (pp. 152−158, in particular 14.11(1)) in Kriegl and Michor's The ...


2

We have $Var\,z=m\,Var\,y_1$, $$Var\,y_1=P(x_1\ge K)(1-P(x_1\ge K))=P(x_1\ge K)P(x_1<K))=(1-F(K-1))F(K-1) =(1-I_{1-p}(n-K+1,K))I_{1-p}(n-K+1,K)$$ according to this, where $F$ is the cdf of the binomial distribution with parameters $n$ and $p$, and $I_\cdot(\cdot,\cdot)$ is the regularized incomplete beta function. Thus, $$Var\,z=m\,(1-I_{1-p}(n-K+1,K))I_{...


0

The following article is what you are looking for: Akritas, M., Arnold, S. (2000). Asymptotics for Analysis of Variance When the Number of Levels is Large, Journal of the American Statistical Association, 95:449, 212-226.


1

Simple random walk (and non-backtracking walk) on random regular graphs exhibit the cutoff phenomenon [1]. The extension to graphs with degree sequences came later; see [2] for nonbacktracking walks and [3] for simple random walk where backtrackings cause additional difficulties. In another direction, component structure at criticality was described in a ...


1

(Answering your comment) Off the top of my head, I'd look in vol.2 of Probabilités et Potentiel (Dellacherie & Meyer) or in Limit Theorems for Stochastic Processes (Jacod & Shiryaev). Another convenient resource is the blog https://almostsure.wordpress.com of Geo. Lowther. The key is that for a bounded rc martingale $M$, the predictable projection ${}...


6

Using (say) decimal notation, ASCII encoding, and a delimiter symbol such as a space or comma, as well as the law of large numbers, one can almost surely encode $N$ independent copies of $\lfloor X \rfloor$ using $O( N {\bf E} \log( 2 + |X| ) ) + o(N)$ bits. Applying the Shannon source coding theorem, we conclude that $$ {\bf H}( \lfloor X \rfloor ) \ll {\...


0

There is also the (old) book of Seneta: Non-negative Matrices and Markov Chains. First edition: 1973. See chapters 3-4.


0

If $z$ is an odd integer, we have $$R'_S(z) = \sum_{k=0}^z X_k X_{z-k}=2\sum_{k=0}^{(z-1)/2}X_k X_{z-k}=2\sum_{k=0}^{(z-1)/2}Y_k$$ where $X_k$ is Bernouilli of parameter $f(k)$ The $Y_k$'s ($k=0,\cdots, \frac{z-1}{2}$) are independent, non-identically distributed Bernouilli variables of parameter $p_k=f(k)f(z-k)$ It follows (see here and here) that $$E[R'...


6

$\newcommand{\fx}{\lfloor X\rfloor}$ $\newcommand\Z{\mathbb{Z}}$ We shall prove more than requested: that $H(\fx)<\infty$ if $E\ln(1+|X|)<\infty$. Indeed, let $$p_n:=P(\fx=n),$$ so that $$H(\fx)=-\sum_{n\in\Z}p_n\ln p_n.$$ Let $q\colon\mathbb R\to(0,\infty)$ be any function such that $$\sum_{n\in\Z}q(n)=1\tag{1}$$ and $$q(x)\le cq(\lfloor x\rfloor)\tag{...


1

Set $J$ the number of groups, and $N_j$ the number of observations for the group $j$, $1\le j\le N$. The test statistic is the ratio $$ T_{n,J}:=\frac{A}{B}, $$ with $$ A:=\frac{\sum_{j=1}^J N_j (\bar X- \bar{X}_{\cdot j})^2}{J-1} $$ and $$ B:=\frac{\sum_{j=1}^J \sum_{i=1}^{N_j}(X_{ij}- \bar{X}_{\cdot j})}{N-J}, $$ where $\bar{X}_{\cdot j}$ is the mean of ...


6

Since $\lfloor X\rfloor$ has finite entropy if and only if $|\lfloor X\rfloor|$ has finite entropy, it suffices to consider random variables taking values in the natural numbers. Write $p_n$ for $\mathbb P(X=n)$ (so that $\sum_n p_n=1$). We have $X\in L^q$ if and only if $\sum p_n n^q<\infty$. Suppose $X\in L^q$ so that $\sum p_n n^q<\infty$. Then let $...


3

Smooth random functions, random ODEs, and Gaussian processes (2018) describes an approach that takes a finite Fourier series on the interval $(0,1)$ with randomly chosen coefficients. The integral of this function approaches Brownian motion in the limit that the number $M$ of Fourier coefficients tends to infinity. The plot shows three such functions, for $M=...


1

I wonder if the probability is dependent only on $r_i$, or also dependent on the placement of $R$ within $S$? In these two examples,        it takes an average of $2.70$ steps to reach slicing $R$ on the left, but $3.16$ steps on the right. I realize I'm ignoring your condition that $s_i \gg r_i$. Added 4Aug2020. I include ...


0

A sufficient (and necessary) condition is that each bounded right-continuous martingale is continuous. This is true, for example, if the filtration is that of a Brownian motion. The condition you suggest is too weak. Example: Let $U$ be uniformly distributed on $(0,1)$ and let $\xi$ be an independent random variable taking the two values $1$ and $-1$ with ...


5

Example 2 of arXiv:0704.2826 considers the analogous problem for the continuous-time random walk, in the more general case that the curve has the form $g(t)=a+b\sqrt{T-t}$ with $a+b\sqrt T\geq 0$. The random walk starting at the origin stays below that curve for all $t<T$ with probability $$P(T,a,b)=1-\frac{\int_{-\infty}^{-a/\sqrt T} e^{-y^2/2}dy}{\int_{-...


2

This is also too long for a comment but it shows where the real problem lies. With the new formulation, the complementary expectation of the sum of two cut sides is just of the form $$ \frac {\sum_i a_i^2(|\cos\theta_i|+|\sin\theta_i|)}{\max_i(a_i|\cos\theta_i|)+\max_i(a_i|\sin\theta_i|)} $$ where $a_i$ are the sides and the angles $\theta_i$ that the sides $...


2

Let the vertices of the triangle be $A$, $B$, and $C$, which we also use for the angle measures, opposite the sides of lengths $a$, $b$ and $c$ respectively. Suppose we know that in the ideal configuration, a horizontal line cuts the triangle at $A$, and a vertical line cuts the triangle at $B$. Let $\theta$ be the angle between the horizontal line and side $...


4

Coincidentally, a day after you asked this question, Alex Simpson gave a nice talk (video, slides) where he gave a synthetic formulation of probability theory. In this formulation, random variables are a primitive notion, not maps from a sample space to a measurable space. Hence there's no need to keep track (or even mention) sample spaces at all. That's ...


1

A bit too long for a comment, here's one suggestion: take $a\lt b\lt c$ and use $(0,0)$ and $(c,0)$ as two points of the triangle. The third point $(x,y)$ (taking $y\gt 0$ WLOG) can be found in whatever usual way you prefer. Now, instead of rotating the triangle, rotate the lines: we can parametrize the pencils of lines as being in the directions $(\cos\...


2

From the context, it appears that $b\in(0,\infty)$ and $c\in[0,\infty)$ (and, likely, $c\le b$). Anyway, let $c_1:=\min(b,c)$. Note that With probability $1$, either $Y=0$ or $c_1<|Y|\le b$; The conditional distribution of $X$ given $Y=0$ is the uniform distribution on the interval $[-c_1,c_1]$ and hence $d(X|Y=0)=1$. The conditional distribution of $X$ ...


7

This is not a true "no pen or paper" solution requested by fedja, but at least it avoids integrals. :-) Let $X$ and $Y$ be independent random vectors on the unit sphere. Write $E = (X - Y) / |X - Y|$ for the unit vector parallel to the chord $XY$, and $Z = \tfrac{1}{2} (X + Y)$. Claim: Conditionally on $E = e$, the projection $Z = \tfrac{1}{2}(X + ...


3

If $\Sigma=I_p$, then the distribution of $\sum_{j=1}^p\xi_j^2$ for $(\xi_1,\cdots,\xi_p)^\top\sim N(\vec{\mu},\Sigma)$ is the non-central chi-square distribution with $p$ degrees of freedom and non-centrality parameter $\vec{\mu}^\top\vec{\mu}$. If $\Sigma\ne I_p$, then the distribution of $\sum_{j=1}^p\xi_j^2$ for $(\xi_1,\cdots,\xi_p)^\top\sim N(\vec{\mu},...


1

You just need to use a higher precision, using exact numbers when possible, as shown in the Mathematica work below (with $k:=\kappa$ and $t:=\theta$). (However, this question is indeed better suited for Mathematica SE.)


2

The accuracy issue with the evaluation of the hypergeometric function can be avoided for integer $\kappa$, since then the full expression reduces to an error function (see my answer to your previous question). I tried this out for $\kappa=5$. Then $$\mathbb{E}\left[ a \mathcal{Q} \left( \sqrt{b } \gamma \right) \right]=\frac{a}{48 b^4 {\theta}^8} \left[\sqrt{...


1

What happens if $n = k = 1$ and $X = Y$? In this case $\rho = 1$. Let $M_X = 0$ if $X < 0$ and 1 otherwise. Let $M_Y = 0$ if $Y < 0$ and 1 otherwise. Then $I(M_X;M_Y) = H(M_X) - H(M_X|M_Y) = 1$. This seems to contradict your wanted inequality.


1

Mathematica 12.0 does the job by Integrate[Exp[\[Lambda]*t^4],{t, \[Alpha], 0}, Assumptions->\[Alpha]<0 && \[Lambda] >= 1]* Integrate[Exp[-\[Lambda]*t^4],{t,-Infinity,\[Alpha]},Assumptions->\[Alpha]<0&&\[Lambda]>=1] $$-\frac{(-1)^{3/4} \alpha E_{\frac{3}{4}}\left(\alpha ^4 \lambda \right) \left(\Gamma \left(\frac{1}{4},-\...


4

Let $a:=\alpha$ and $x:=\lambda$. Using now the substitution $t=x^{-1/4}s$ and introducing $b:=x^{1/4}a$, we see that the problem is to show that $\sup_{b\le0,x\ge1}x^{-1/2}I(b)J(b)=\sup_{b\le0}I(b)J(b)<\infty$, where $$I(b):=\int_b^0 e^{s^4}\,ds=\int_0^{-b}e^{s^4}\,ds,\quad J(b):=\int_{-\infty}^b e^{-s^4}\,ds$$ for $b\le0$. By l'Hospital's rule, $$I(b)\...


2

The answer is yes: https://fabricebaudoin.wordpress.com/2012/10/02/lecture-25-girsanov-theore/ Here is a good link. I think it is a better link than the currently most upvoted answer.


1

Assume the random walk starts at the origin. Let $T_0=0$ and for $i\geq 1$ let $T_i$ be the time of the $i$th visit of the walk to the origin after time $0$. Let $X_i=T_i-T_{i-1}$. Then the $X_i$ are i.i.d. (you may like to think of this in terms of the strong Markov property, for example), and they take even positive integer values. Let $Y_i=X_i/2$, so that ...


1

$\newcommand\Om\Omega$ $\newcommand\F{\mathcal F}$ $\newcommand\M{\mathcal M}$ With the help from Google Translate: Throughout the work $m$ is a fixed integer $\ge2$ and $j$ runs through the integers from $1$ to $m$. For each $j$, $(\Om_j,\F_j,P_j)$ is a probability space. Set $\Om=\prod\limits_j\Om_j$, $\F=\bigotimes\limits_j\F_j$, $P=\bigotimes\limits_j ...


3

The axioms don't tell you what theory you constructed. For that you need to go beyond the construction of correlation functions of the elementary field $\phi$ (the basic chapter on renormalization in QFT textbooks) and produce, e.g., by a point-splitting procedure, correlations with insertion of composite fields like $\phi^3$. You should then identify your ...


4

Your probability measure is a product measure, so by $$\text{Var}_y(\langle z,X\rangle) = \sum_{i=1}^nz_i^2\text{Var}_{y_i}(X_i)$$ everything reduces to the 1d case. Let $q_y(dx)=e^{xy-\varphi(x)-C(y)}dx$ be one of the marginals, where $y\in\mathbb{R}$ and $C(y)$ is chosen such that $q_y$ is normalized, and denote by $U_y$ the 1d r.v. with distributon $q_y$. ...


1

$\newcommand\Ga\Gamma$ Without loss of generality $a=1$. Let then $Q:=\mathcal Q$, $k:=\kappa>0$, and $t:=\theta\sqrt b>0$, so that $\sqrt b\,\gamma$ has the gamma distribution with parameters $k,t$. Let also $c:=\Ga(k)t^k$. Then, letting $f$ denote the standard normal pdf, we have $$c\,EaQ(\sqrt b\,\gamma)=\int_0^\infty dy\,y^{k-1} e^{-y/t}Q(y) =\...


1

Mathematica gives an answer in terms of a hypergeometric function: $$\frac{a 2^{-\frac{\kappa}{2}-\frac{3}{2}} b^{-\frac{\kappa}{2}} {\theta}^{-k} \, _2F_2\left(\frac{\kappa}{2}+\frac{1}{2},\frac{\kappa}{2};\frac{1}{2},\frac{\kappa}{2}+1;\frac{1}{2 b {\theta}^2}\right)}{\sqrt{\pi } \Gamma \left(\frac{\kappa}{2}+1\right)}-\frac{a 2^{-\frac{\kappa}{2}-2} \...


2

How about the following reference T. Gneiting, Criteria of Po´lya type for radial positive definite functions, Proc. Am. Math. Soc. 129 (2001), 2309–2318. I have plotted the results for 6th derivative all numeric but it seems working,


4

This is equation (4.12) in the cited paper (also at arXiv:1004.4389), which follows from theorem (4.1). The $\xi_i$'s may be independent equally probable random signs, or they may be independent standard normal variables.


2

I think Timothy Chow's comment is right that there is no result about planar graphs with your lemma as an explicit corollary. I believe the following 2007 research paper by Guido Helden might be of use to you: http://publications.rwth-aachen.de/record/62349/ It is about hamiltonicity of maximal planar graphs and planar triangulations, and starts with a very ...


11

$\newcommand\Ga\Gamma \newcommand{\R}{\mathbb{R}} \newcommand{\de}{\delta} \newcommand{\ga}{\gamma} \newcommand{\Si}{\Sigma}$ We have to show that for $a:=-\ln(2^{1/b-1/2}/v)\in\R$ and $b:=\beta\in(1,2)$, \begin{equation*} I(a):=\int_{-\infty}^{\infty} e^{-iat}R(t)\,dt>0, \tag{1} \end{equation*} where \begin{equation*} R(t):=\frac{\Ga\big((1+it)/b\...


2

Such a sequence $a_n$ does not exist even for a well studied example like returns to the origin of simple random walk in one dimension. If $X_i$ denotes the number of steps from the $i-1$ time the walk returned to the origin to the $i$'th time, then $X_i$ are i.i.d. and their sum $S_n$ is the number of steps until the $n$'th return time to the origin. In [1] ...


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