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Without loss of generality, the three points on the unit circle are $1,e^{is},e^{it}$, where $s$ and $t$ are uniformly and independently distributed on the interval $[0,2\pi]$, so that $a=|e^{is}-1|$, …

We have $f=e^g$, $g$ is concave, $\int f=1$, $\int x f(x)\,dx=0$, and $\int x^2 f(x)\,dx=1$. As you noted, then $f(0)\ge 1/8$ and hence $$g(0)\ge-a,\tag{0}\label{0}$$ where $a:=\ln8$. We have to show …

The answer is no. E.g., if $p_1=1/2$, $p_2=1/2$, $q_1=1/100$, $q_2=99/100$, and $\beta=1/10$, then the ratio of the left-hand side of the conjectured inequality to its right-hand is $0.00877\ldots<1$. …

$\newcommand{\R}{\mathbb R}\newcommand{\tr}{\operatorname{tr}}$The answer is yes. Indeed, it is easy to see (cf. e.g. Proposition 7 or the beginning of its proof) that the Wasserstein distance between …

Take any $x\in X$ with $m:=\mu(\{x\})\in(0,\infty)$ (since $X$ is finite and $\mu$ is a probability measure, such a point $x$ exists). Let $R:=\min\{d(y,x)\colon y\in X\setminus\{x\}\}$. Then $R\in(0, …

$\newcommand\la\lambda\newcommand\ep\varepsilon\newcommand{\de}{\delta}\newcommand\R{\mathbb R}\newcommand{\Si}{\Sigma}$The answer is negative, even for $k=3$. A counterexample is as follows: $$\lambd …

The probability that, of the $m$ random points on the $n$-sphere (that is, on the unit sphere in $\mathbb R^n$), $m/2$ points lie in the all-positive orthant and the other $m/2$ lie in the all-negativ …

$\newcommand\ep\epsilon\newcommand{\de}{\delta}\newcommand{\R}{\mathbb R}$We have \begin{equation*} P((1-\ep)\|x\|\le\|Ax\|\le(1+\ep)\|x\|)\ge\de \tag{1}\label{1} \end{equation*} for some $\ep,\de …

$\newcommand\R{\mathbb R}$Let $f$ be a radial pdf on $\R^n$, so that $$f(x)=g(|x|)\tag{1}\label{1}$$ for some function $g\colon[0,\infty)\to\R$ and all $x\in\R^n$, where $|x|:=|x|_2$. Then your desire …

No. E.g., suppose that $n=1$, $m=2$, and the pdf $f$ of each of the $m=2$ iid sample points $X_i:=\mathbf c^{(i)}$ ($i=1,\dots,m$) is given by the formula $f(x)=e^{-x-1}1(x>-1)$ for real $x$. Then, by …

The answer is no. E.g., let $n=3$ and $z_j=e^{i(j-1)2\pi/3}$ for $j=1,2,3$. Then $f(z)>|z|+15/100>|z|$ if $|z|\le1$. This counterexample generalizes to any $n\ge3$. Indeed, take any $n\ge3$ and let $ …

$\newcommand{\De}{\Delta}\newcommand{\R}{\mathbb R}\newcommand{\ga}{\gamma}\newcommand{\Ga}{\Gamma}$This is to provide a detalization on Will Sawin's comment. Specifically, let us show that the best u …

$\newcommand\R{\mathbb R}\newcommand\la{\lambda}$A measure $\mu$ is Ahlfors regular (according to your definition) iff it has a density (with respect to the Lebesgue measure $\la$) bounded away from $ …

$\newcommand\R{\mathbb R}\newcommand{\Si}{\Sigma}\newcommand{\si}{\sigma}$The answer is no: in general (and usually) there are no positive absolute constants $a$ and $b$ such that for some unit vector …

$\newcommand{\ep}{\varepsilon}\newcommand\R{\mathbb R}\newcommand{\de}{\delta}\newcommand{\vpi}{\varphi}\newcommand{\Ga}{\Gamma}$In Question 2 you wanted a good lower bound on $E_V P(X\in A_V^\ep)$, w …