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Since $G$ is bounded and $\sigma W\ge0$, for all large enough real $M$ we have $A=\emptyset$, $\int_A\cdots=0$, and $\log\int_A\cdots=-\infty$, which yields what you wanted.

The conditions $x>EX/\sqrt{EX^2}$ and $EX\ge0$ imply $x>0$. So, $t[bX_i -x(X_i^2+b^2)/2]$ is a quadratic polynomial in $X_i$ whose leading coefficient $-tx/2$ is less than $0$ (except for the trivial …

$\newcommand{\De}{\Delta} \newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \renewcommand{\th}{\theta} \newcommand{\R}{\ma …

By the de Moivre–Laplace theorem, \begin{equation} P(S_n=k)=P(B_n=(n+k)/2)\sim\frac1{\sqrt{\pi n/2}}\,\exp\{-k^2/(2n)\}, \end{equation} where $B_n$ is a random variable with the binomial distributio …

The exact asymptotics of $f(n)$ for large $n$ follows by Theorem 2.1, more specifically formula (2.4). However, to use that formula (2.4), you will have to compute lots of asymptotics regarding the …

The asymptotics of the ratio $$r_n(x,y):=\frac1{f_n(\sqrt nx)}f_{n-1}\left(\frac{nx + y}{\sqrt{n-1}}\right) $$ will very much depend on the tail asymptotics of the density (say $f$) of $X$. E.g., …

$\newcommand{\La}{\Lambda}$ The Legendre transform $\La^*_X$ of the log-moment generating function of a random variable $X$ is given by the formula $$\La^*_X(x):=\inf_{t\ge0}(-tx+\La_X(t)), $$ where …

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 …

If (say) $s:=\sqrt{EX_1^2}<\infty$ then, by the central limit theorem, $Z_n:=S_n/n^{1/2}\to sZ\sim N(0,s^2)$ (as $n\to\infty$ in distribution). Also, $EZ_n^2=s^2<\infty$ and hence the sequence $(|Z_n| …

$\newcommand{\ep}{\epsilon}$Somehow, I have only now recalled about Latala's inequalities for moments of the sums of positive independent random variables (r.v.'s), which, in particular, allow one to …

Suppose e.g. that $X=X_1+\dots+X_n$, where the $X_i$' are independent zero-mean random vectors with $\|X_i\|\le1$ for all $i$. Then the Hoeffding--Azuma inequality (see e.g. Wikipedia) yields $$P(\|X\ …

Here it is more convenient to consider the order of magnitude of $S_n:=\sum_1^n X_i$, rather than that of $S_n/n$. Take any real $c>0$. Let $x:=cn^{1/p}$, $Y_i:=X_i\,1(X_i<x)$, $T_n:=\sum_1^n Y_i$, $M …

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 …

The answer to your question is contained in the following local limit theorem for large deviations, due to V. Petrov, Theorem 6: Suppose that $X,X_1,X_2,\dots$ are iid random variables such that …

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 …