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$\newcommand{\ep}{\varepsilon} $ Let $X$ be any nonnegative random variable (r.v.) with finite mean $\mu>0$ and variance $\sigma^2<\infty$. For any real $u>0$, we have $\ln\frac xu\le\frac xu-1$ for a …

This inequality cannot be true. Let us rewrite it in the more common form $$P(R_n\ge x)\le e^{-x^2/2} \tag{1} $$ for $x\ge0$, where $R_n:=S_n/b_n$, $S_n:=\sum_1^n c_iB_i$, $b_n:=\sqrt{\sum_1^n c_i^2} …

Exponential inequalities for sums of independent random variables (r.v.'s) can be extended to martingales in a standard and completely general manner; see Theorem 8.5 or Theorem 8.1 for real-valued ma …

Let $n:=N$. Let us show that for all natural $n$ and all $p\in(0,1)$ $$P(A_n>\sqrt p)\le\frac{\sqrt p+p}{1+p},\tag{1}$$ so that $P(A_n>\sqrt p)\to0$ whenever $p\downarrow0$. Consider first the case wh …

Let $X_i=(X_{i,1},\dots,X_{i,d})$, $S:=(S_1,\dots,S_d)$, $S_j:=\sum_{i=1}^d X_{i,j}/\sqrt n$. Then, by Hoeffding's inequality, for $s\ge0$ $$P(|S_j|\ge s)\le2e^{-s^2/2},$$ whence $$E\|S\|_\infty=\int …

In general, the answer is no. Moreover, the answer is no even if \begin{equation} \phi(t)=t\ln(1+t). \tag{1} \end{equation} Indeed, suppose that $P(Z_i=0)=1-2p$ and $P(Z_i=b)=p=P(Z_i=-b)$ for all …

$\newcommand{\R}{\mathbb{R}} \renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\Ga}{\Gamma} \newcommand{\de}{\delta}$ In view of the spherical symmetry of the distribution of the $l$-dimensiona …

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} …

By the Tsirel’son--Ibragimov--Sudakov argument, reviewed on the first page in Bobkov, pushing the measure forward from the cube to the canonical Gaussian on $\mathbb R^n$ and using the Gaussian isope …

Without loss of generality, $R=1$. Let $Z_1,\ldots,Z_n$ be iid standard normal random variables (r.v.'s). Then \begin{equation} \sqrt n\, X_1\overset{\text{D}}=\frac{\sqrt n\,Z_1}{\sqrt{Z_1^2+\cdot …

Your inequality is trivial and useless as written. On its left-hand side we have a probability which is $\le1$ and goes to $0$ as $t\to\infty$, whereas on the right-hand side we have an expression whi …

The desired statement will not hold. E.g., suppose that $n\ge2$; $X_1,\dots,X_n,Y_1,\dots,Y_n$ are independent; $p=1/2$; $T=1_{X_1\ne Y_1}+n1_{X_1=Y_1}$; and $\delta=1/2$. Then $\mu:=p\,ET>n/4\to\inft …

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 …

Actually, Bernstein's inequality does not really require boundedness of the i.i.d. random summands; a finite exponential moment of the absolute value of a random summand will suffice. However, here we …

By the main result of the paper Exact Rosenthal-type bounds, we have $$E|X-np|^r\le c^r E|\Pi_\lambda-\lambda|^r $$ for real $r\in(2,\infty)\setminus(3,4)\setminus(4,5)$, where real $c>0$ and $\lambd …