6

$\newcommand{\N}{\mathcal N}\newcommand{\vpi}{\varphi}$For $\mu=\sum_{i=1}^n w_i\delta_{x_i}$ with $0\le w_i\le 1$ and $\sum_{i=1}^n w_i=1$, we have the following (see e.g. page 4): \begin{align*} W_p(\mu,\N)^p&=\int_0^1 du\,|F^{-1}(u)-G^{-1}(u)|^p \\ &=S(x,u):=\sum_{j=1}^n\int_{u_j}^{u_{j+1}} du\,|x_j-g(u)|^p, \end{align*} where $F$ is the ...


5

Complementing the answer by Carlo, if you take all $k$'s equal you have $$f_{\rm GUE(d)}(k,...,k)\propto \int dX e^{ik{\rm Tr}(X)}e^{-\frac{d}{2}{\rm Tr}(X^2)}.$$ Taking $x$ to be any real diagonal element from $X$, this is $$f_{\rm GUE(d)}(k,...,k)\propto \left(\int dx e^{ikx}e^{-\frac{d}{2}x^2}\right)^d.$$ I think in the end you have simply $f_{\rm GUE(d)}(...


4

The Fourier transform of the marginal distribution of a single eigenvalue in the GUE is known, $$f_{{\rm GUE}(d)}(k,0,0,\ldots,0)=e^{-\tfrac{1}{2}k^2/d}\sum_{j=0}^{d-1}(-1)^jk^{2j}\frac{(d-1)(d-2)\cdots(d-j)}{j!(j+1)!d^j},$$ see these lecture notes. A curiosity: the $d=2$ result is given in this publication in terms of the confluent hypergeometric function $...


4

The classical theorem of Darmois-Skitovich says that two linear forms (with all non-zero coefficients, and of at least two variables) of independent random variables are independent only if the random variables are normal. Combined with the comment of @Gerald Edgar, this completely settles the question. Ref. A. Kagan, Yu. Linnik, C. Rao, Characterization ...


3

If the cardinality of the set $S_f$ is $\ge1$, then the value of the integral $$ I:=\int_0^1\Big(\frac{f'(x)}{f(x)}\bigg)^2 dx$$ will be $\infty$ with nonzero probability, because with nonzero probability the polynomial $f$ will have a non-multiple root in the interval $[0,1]$. This will then also imply that the expectation of $I$ is $\infty$.


3

$\newcommand{\de}{\delta}\newcommand{\ep}{\varepsilon}$The covariance of two random variables (r.v.'s) does not change if one of them is shifted by a constant. So, without loss of generality $f(0)=0$. Let $n:=d$. To compute the asymptotics, we need to assume that \begin{equation*} f(x)=Ax+Bx^2+Cx^3+Dx^4+e_1x^5+O(x^6) \tag{1} \end{equation*} for some real ...


3

We prove the weaker bound $$ \mathbf{P} \left[ \|O \mathbb{x}\|_1 \leq \frac{cd}{\sqrt{\log d}} \right] \leq 2^{-Cd} $$ for some constants $C, c$. Define the Gaussian mean width of a compact subset $A \subset \mathbf{R}^d$ as $$ w(A) = \mathbf{E} \sup_{x \in A} \langle G,x \rangle $$ where $G$ is a standard Gaussian vector in $\mathbf{R}^d$. We use the ...


3

Here is an attempt to the problem for a worst-case $O$, with worse constants. So fix $O$, letting $o_i$ denote its $i$th row, and take $X$ random in $\{0,1\}^d$. We claim that $E |\langle o_i, X\rangle| \ge cst$. To see this, write $$\langle o_i, X\rangle = \langle o_i, \frac{{\bf 1}}{2}\rangle + \langle o_i, (X - \frac{{\bf 1}}{2})\rangle$$ and assume ...


2

I think this is straightforward by what I call the "lexicographic" method: fix a countable dense sequence $(f_n)$ in $𝐶_{0,1}$. Then as you point out, $\omega\mapsto|\mu_\omega|=\sup\int f_n\,d\mu_\omega$. Let $N(\omega)=\min\{n:\int f_n\,d\mu_\omega\ge\frac12|\mu_\omega|\}$. This is measurable because $N^{-1}\{1,\ldots,k\}=\bigcup_{j=1}^k \{\...


1

You did not let us know what $S_f$ is. Since the title of your post is "Multiple roots of Random Polynomials", it seems reasonable to assume that $S_f$ is the set of integers $n_0,\dots,n_k$ such that $0\le n_0<\dots<n_k$. One of the following cases must occur: Case 1: $n_0\ge2$. Then $0$ is a multiple root of $f$ with probability $1$. Case 2:...


1

It turns out that one can get a stronger result than demanded in the question: compute $\Delta(a,b)$ for any $a,b \in S_{d-1}$, perpendicular or not. Indeed, Claim. If $f(\rho)=a_0 + a_1 \rho + a_2 \rho^2 + a_3 \rho^3 + \mathcal O(\rho^4)$, then for every $u,v \in S_{d-1}$, and $x$ be uniform on the sphere, then we have the following approximation \begin{...


1

Adding more detail to Mikael's point, the result seems to hold on average over $O$ because of the following: Using a Chernoff bound, we can see that the probability that for any constant $\epsilon$, with probability at least $1- e^{-C d}$ the random vector $x$ has at least $\frac{(1-\epsilon)d}{2}$ 1's. Consider a fixed vector $x \in \{0,1\}^d$. For a ...


1

$\newcommand{\si}{\sigma}\newcommand{\R}{\mathbb{R}}\newcommand{\ch}{\operatorname{ch}}$Let $X$ be a $\sigma$-sub-Gaussian random variable (r.v.) for some real $\sigma>0$, that is, \begin{equation*} M_X(t):=Ee^{tX}\le e^{\sigma^2t^2/2}\quad\text{for all real $t$}. \tag{1} \end{equation*} Then the standard upper bound on $P(X\ge u)$ is $e^{-u^2/(2\si^2)...


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