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For each real $k>0$, \begin{equation}E\psi_\infty(|X|/k)=\infty\,P(|X|>k)+P(|X|=k) \\ =\left\{\begin{aligned}\infty\text{ if } P(|X|>k)>0,\\ P(|X|=k)\le1\text{ if } P(|X|>k)=0. \end{aligned}\right. \ …

Here is a much simpler proof, actually of the more general fact that $$f(x):=x^p(1+\ln^+ x)^s$$ is convex in $x\ge0$ for any real $p\ge1$ and $s\ge0$, where $\ln^+ x:=\ln\max(1,x)$. For the left and r …

Here is how to remove the assumption that $p-2+\delta\ge0$. Let \begin{equation*} s:=p-1+\delta. \end{equation*} The conditions $p>1$ and $\delta>0$ imply $s\ge0$. No other conditions on $p$ and …

$\newcommand{\ep}{\varepsilon}\newcommand{\Si}{\Sigma}\newcommand{\R}{\mathbb R}$A natural generalization of the fact that the $p$-norm converges to the $\infty$-norm as $p\to\infty$ is as follows. Le …

Yes, we can say so. Indeed, let us show that the conditions $f\in L^r$ and $g\in L^s\ln L$ imply $fg\in L\ln^t L$ for $t:=1/s$. Moreover, we shall show that the value $t=1/s$ here is optimal, as it ca …

For the equality $\|S_n\|_{L^2}=\|X_i\|_{L^2}$ you need the zero-mean condition -- that $EX_i=0$. Let $X,X_1,\dots,X_n$ be any random variables with the same norm: $\|X\|=\|X_1\|=\cdots=\|X_n\|$. Th …

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 …