-1
$\begingroup$

I need help to understand the following :

For any non-negative random variable $\zeta$: $\sum_{k=1}^{\infty}\mathbb{P}(\zeta\geq k)\leq\mathbb{E}(\zeta)\leq 1+\sum_{k=1}^{\infty}\mathbb{P}(\zeta\geq k)$.

Therefore, if $x_0,\dotsc,x_n$ are iid random variables and $\mathbb{E}(\ln{(1+|x_0|))}<\infty$, it follows that : $\sum_{k=1}^{\infty}\mathbb{P}(|x_k|\geq e^{\gamma k})<\infty$ for any positive constant $\gamma$.

I don't understand where the $\gamma>0$ is coming from? Isn't it for $\gamma\geq 1$?

Improve this question
$\endgroup$
1
  • 3
    $\begingroup$ Looks more appropriate for Mathematics Stack Exchange math.stackexchange.com than MathOverflow as this is not research level. $\endgroup$
    – Alexander Pruss
    Commented Aug 20, 2022 at 17:36

1 Answer 1

Reset to default
4
$\begingroup$

Every $\gamma>0$ seems ok. We have $$ \mathbb{P}(|x_k|\geqslant e^{\gamma k})\leqslant \mathbb{P}(1+|x_k|\geqslant e^{\gamma k})= \mathbb{P}(\gamma^{-1}\log(1+|x_k|)\geqslant k), $$ and the sum of these probabilities is finite.

Improve this answer
$\endgroup$
2
  • $\begingroup$ $\mathbb{P}(\gamma^{-1}\ln{(1+|x|)}\geq k)\leq \mathbb{E}(\gamma^{-1}\ln{(1+|x|)})=\gamma^{-1}\mathbb{E}(\ln{(1+|x|)})$ but $\frac{1}{\gamma}$ tends to infinity when gamma tends to 0 so that product isn't < infinity ? $\endgroup$
    – Maxou
    Commented Aug 20, 2022 at 17:31
  • 2
    $\begingroup$ $\gamma$ is constant, so the sum is finite for every $\gamma>0$. It is not said that the sum is uniformly bounded for $\gamma>0$. $\endgroup$
    – Fedor Petrov
    Commented Aug 20, 2022 at 18:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .