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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 …

$\newcommand{\ep}{\varepsilon}\newcommand{\vpi}{\varphi}\newcommand{\de}{\delta}$Yes, this is true: By the reflection principle (see e.g. Proposition 2, for $M:=\max_{0\le t\le1}W_t$, \begin{equation} …

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 …

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 …

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 …

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., …

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 $Let $u:=\eta>0$, so that the probability in question is $P(\ln X>E\ln X+u)$. Note that this probability will not change if we replace there $X$ by $tX$ for any real $t>0$. So, …

$\newcommand\Ga\Gamma\newcommand\Z{\mathbb Z}\DeclareMathOperator\Pois{Pois}$Let $t\mathrel{:=}\lambda$ and $k\mathrel{:=}x\in\Z\cap[0,t)$. Then for $X_t\sim \Pois(t)$ we have $$P(X_t\ge k)=1-\frac{\G …

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\ …

Indeed, if the probability mass function of an integer-valued random variable is log concave as a function on $\mathbb Z$, then the corresponding cdf is also log concave as a function on $\mathbb Z$. …

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

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.

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 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 …