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$\newcommand\si\sigma$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 …

$\newcommand\si\sigma$After the clarification by the OP, my previous answer should be modified as follows. Suppose that for some real $C$, some real $h>0$, and all $t\in[-h,h]$ we have $$M(t):=Ee^{tX} …

$\newcommand\si\sigma$Your condition on the moment generating function of $X$ implies that for such $|t|<1/b$ and $m=1,2,\dots$, $$\frac{t^{2m}}{(2m)!}\,E X^{2m}\le E\cosh tX\le\exp\frac{t^2 \si^2}{2( …

$\newcommand\ol\overline$In accordance with the comment by the OP, consider \begin{equation*} p_{N,n}:=P\Big(\max_{k\in\ol{N,n}}\frac{|S_k|}{\sqrt{k/2}}\le1\Big), \end{equation*} where $\ol{A,B}:= …

$\newcommand\ep\epsilon$Let $X_i:=c^{i-1}(x_i-Ex_i)$, so that the $X_i$'s are independent zero-mean random variables. The condition $c=1+1/m$ for $m\ge n$ implies that $$c=1+b/n,$$ where $0<b=O(1)$. L …

Of course not. E.g., let $T$ be the set of all natural numbers, let $U$ be any random variable (r.v.), let the $Y(t)$'s be iid standard normal, and let $X(t):=\min(U,Y(t))$ for all $t$. Then the condi …

$\newcommand\th\theta\newcommand\R{\Bbb R}$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 …

$\newcommand\de\delta$If $X$ is the sum of $n$ random variables each of them with values in $[0,1]$ and $\mu=EX$, then clearly $\mu\le n$. So, if $\de\in[0,1]$ and $k\le\lfloor\de^2\mu e^{-1/3}\rfloor …

$\newcommand\E{\operatorname{E}}\newcommand\var{\operatorname{Var}}\newcommand\si{\sigma}$This will not work. E.g., if $N=10$, $\{c_1,\dots,c_{10}\}=\{-1, -1, -1, -1, -1, 1, 1, 1, 1, 1\}$, $n=5$, and …

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 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, $ …

Simple and explicit upper and lower bounds on the binomial tail that are asymptotically sharp (as $n\to\infty$) and improve the Chernoff upper bound by a factor $\asymp1/\sqrt n$ were rather recently …

Your upper bound on $\mathbf{E} \left[ \exp\left(\theta R\left(X+X'\right)\right)\right]$ is $$f(a,b;t):=(\cosh at+bt^2)^2+\sinh^2 at,$$ where $$a:=EX\ge0,\quad b:=Ee^X-e^{EX}\ge0,\quad t:=\theta.$$ Y …

Let $X:=X_n$. For your probability $$p_{a,t}:=P(|\sqrt X-\sqrt a|>t)$$ to make sense, we need to assume that $X\ge0$ and $a\ge0$. Also, if $t<0$, then $p_{a,t}=1$. So, without loss of generality $t\ge …

$\newcommand{\la}{\lambda}$Up to the equality in distribution, \begin{equation*} Y=\sqrt{\sum_1^k(Z_i+L_k)^2}, \end{equation*} where $Z,Z_1,\dots,Z_k$ are independent standard normal random variab …