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Iosif Pinelis
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The exact lower bound on $k$ is $0$.

Indeed, let $a:=\alpha$. By Jensen's inequality, $\ln Ee^{ax}\ge0$ for all real $a$. So, $k\ge0$ when $k$ is defined.

On the other hand, let $y$ be any random variable (r.v.) such that $Ee^{ay}<\infty=Ee^y$ for $a\in(0,1)$. One can easily construct a zero-mean r.v. $x$ taking only finitely many values such that $Ee^{ax}$ be however close to $Ee^{ay}<\infty$ and $Ee^x$ be however close to $Ee^y=\infty$, and then the $k$ for $x$ wil be however close to $0$. $\quad\Box$

Iosif Pinelis
  • 127.7k
  • 8
  • 107
  • 229