Search Results

The Wikipedia article that you are looking at is, not about the Komlós–Major–Tusnády (KMT) approximation for sums of iid r.v.'s, but about the KMT approximation for the empirical process. However, the …

Yes, there is a way. Let $$X:=\hat s_n-s,\quad h:=s_n-s,\quad a(x):=a(n,x),\quad b(x):=\limsup_{y\uparrow x}b(n,y).$$ We have $$P(|X|>x)\le a(x)\quad\text{and}\quad P(|X-h|\ge x)\le b(x)$$ for all $ …

You cannot improve the bound $L^2$ on $Var\,f(X)$ unless an additional condition on $f$ is assumed. Indeed, let $f(x)\equiv Lx_{11}$, where $x_{11}$ is the first diagonal entry of a matrix $x\in R^{n\ …

As was noted in the comments by Yuval and Kevin, even if $X_1$ is bounded, the best upper bound on the probability in question is a negative power of $\ln n$. To get such a bound (and even an asymptot …

The answer to Question 1 is yes: Shift-rescale the $w_i$'s by considering $v_i:=(1+w_i)/2$ with values almost surely in $[0,1]$, so that $w_i=2v_i-1$, for all $i$. By the Cauchy--Schwarz inequality, $ …

$\newcommand\th\theta\newcommand\R{\Bbb R}$Assume that $d\ge2$. Without loss of generality $v=(1,0,\dots,0)$. Identify then $\eta$ with $X=(X_1,X_2)$, where $X_1$ and $X_2$ are independent random elem …

This inequality follows from Theorem 2 in Hoeffding's 1963 paper, and in fact Hoeffding's result yields a better bound. Indeed, Hoeffding's inequality can be written as \begin{equation} P(\sum Z_i<t …

$\newcommand\si\sigma$The answer is no. E.g., suppose that $P(X_i=1)=2/e=1-(X_i=0)$ for $i=0,1$. Then $X_0$ and $X_1$ are sub-Gaussian with parameter $\si=1/\sqrt2$, so that we can take $\si_0=\si_1=1 …

The answer is negative. Indeed, for simplicity, let $a=-1$ and $b=1$. In the case when $\delta=1/10$, $n=1$, and $X_1$ is uniformly distributed on $[-1,1]$ (so that $\mu=0$), let $f(\mathbf X,\delta …

$\newcommand{\de}{\delta}$Your inequality (2) does hold. Actually, a better and more general bound holds. First here, let us standardize and simplify notations. Let us use $X_i$ instead of $x_i$, $x$ …

$\newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\R}{\mathbb{R}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operato …

$\newcommand{\al}{\alpha} \newcommand{\be}{\beta} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\la …

Unconditional? Certainly not. E.g., suppose that $\mu$ is the standard Gaussian measure on $\mathbb R^d$ and $B$ is the ball of radius $r>0$ centered at $0$. Then for any $\delta\in(0,1)$, by the law …

$\newcommand{\de}{\delta}$ The "dependent" version of the multiplicative Chernoff bound can be proved quite similarly to the "independent" case. Indeed, let $E_{i-1}$ denote the conditional expectatio …

Without further restrictions on $w,x$, you cannot beat the Markov bound by much for $\alpha$ close to $1$ (as in your post). Indeed, let $a:=\alpha$. Let $x_i=a$ for all $i=1,\dots,n$, and let $w_ …