Note: I am reposting this question from Math Stack Exchange, which failed to receive an answer after several weeks and a bounty. Also, I believe it fits the requirements for this website, as it relates to a research paper.


Let $X$ be a random variable with distribution function $F$ on a probability space $(\Omega, \mathcal F, P)$.

Suppose that there exist $\alpha \in (0,2)$ and a slowly varying function $\ell(\cdot)$ such that $$ \bar F(x) := 1 - F(x) = \frac{C_1(x)}{x^\alpha} \ell(x) \quad \text{and} \quad F(-x) = \frac{C_2(x)}{x^\alpha} \ell(x) \quad \text{for $x > 0$,} $$ where $C_1(\cdot), C_2(\cdot)$ are non-negative functions with $C_i := \lim_{x \to \infty} C_i(x)$, and $C_1 + C_2 > 0$.

Why then does the following hold?

There exist $C, \tilde C > 0$ such that $$ \sum_{n=1}^\infty P\big( |X| > a_n \big) \leq \sum_{n=1}^\infty \frac{C}{nf(n)} \leq \tilde C \int_1^\infty \frac{dt}{t f(t)}, $$ where we define $a_n := [n f(n) \ell(n) ]^{1/\alpha}$ for an arbitrary positive function $f$ with the properties $$ \limsup_{t \to \infty} \sup_{0 \leq t \leq x} \frac{f(t)}{f(x)} < \infty \quad \text{and} \quad \int_1^\infty \frac{dt}{tf(t)}< \infty. $$

If not, what if we also required that $f$ be slowly varying, too?

Background and Thoughts

This comes from Cai's 2006 paper "Chover-Type Laws of the Iterated Logarithm for Weighted sums of $\rho^*$-Mixing Sequences". Cai writes on page 5 that this is "easily seen" based on the representation of $F$ above. I don't see why. What follows below is my attempt so far.

$$ \begin{aligned} \sum_{n=1}^\infty P\big( |X| > a_n \big) &= \sum_{n=1}^\infty \Big[ P\big( X > a_n \big) + P\big( X < - a_n \big) \Big] \\ &\leq \sum_{n=1}^\infty \Big[ \bar F(a_n) + F(-a_n) \Big] \\ &= \sum_{n=1}^\infty \frac{C_1(a_n) + C_2(a_n)}{a_n^\alpha} \ell(a_n) \\ &\leq \sum_{n=1}^\infty \frac{C}{a_n^\alpha} \ell(a_n) = C \sum_{n=1}^\infty \frac{1}{nf(n)} \cdot \frac{\ell(a_n)}{\ell(n)}, \end{aligned} $$ where the inequality on the last line holds for some $C>0$, since $C_1(\cdot)$ and $C_2(\cdot)$ are convergent.

If $\frac{\ell(a_n)}{\ell(n)} = \ell\Big( \big[ n f(n) \ell(n) \big]^{1/\alpha} \Big) \Big/ \ell(n)$ were bounded, then the first desired inequality would follow. However, it's not clear why this would have to hold.


  • Indeed, I think the author implicitly uses the fact (?) that $\left\{ \frac{\ell(a_n)}{\ell(n)} \right\}$ is a bounded sequence several times in the paper. But I still don't know why that's the case.

  • Although not stated in the paper, maybe we need some more restrictions on $f$, such that it is also a slowly varying function. If $f$ were slowly varying, then $u(x) := x^{1/\alpha}\cdot[f(x) \ell(x)]^{1/\alpha}$ would be regularly varying with coefficient $1/\alpha$. And $\frac{\ell(a_n)}{\ell(n)} = \frac{\ell\big(u(n)\big)}{\ell(n)}$. Since $u(x) \to \infty$ and is regularly varying, $\ell \circ u$ is slowly varying. Hence, $\frac{\ell(a_n)}{\ell(n)} = \frac{ \ell \big(u(n) \big)}{\ell(n)}$ is slowly varying. But, of course, slowly varying functions aren't necessarily bounded.

  • $\begingroup$ It all depends on the exact definition of "slowly varying". So, what is it in the paper? $\endgroup$
    – fedja
    May 29, 2022 at 16:14
  • $\begingroup$ @fedja : The definition of a slowly varying function $\ell$ in the paper is the standard one: $\ell(tx)\sim\ell(t)$ as $t\to\infty$ for each $x>0$. This is formula (1.6) in the paper. $\endgroup$ May 29, 2022 at 16:57
  • $\begingroup$ @IosifPinelis Then you fried the author :-) $\endgroup$
    – fedja
    May 29, 2022 at 17:30
  • 1
    $\begingroup$ @zxmkn : All right. Please let me know if some of the steps need further details. $\endgroup$ May 30, 2022 at 14:09
  • 1
    $\begingroup$ @zxmkn : This follows because $f$ and $\ell$ are slowly varying functions, and for any slowly varying function $L$ we have $L(n)=n^{o(1)}$, by the Karamata representation theorem (en.wikipedia.org/wiki/… ). You can also directly show that $\ln(f(n)\ell(n))=o(\ln n)$, which is equivalent to what you are asking here about. $\endgroup$ May 30, 2022 at 16:55

1 Answer 1


$\newcommand{\al}{\alpha}$This inequality is false, for any $\al\in(0,2)$.

Indeed, consider first the case $\al\ne1$. Suppose that $C_1(x)+C_2(x)=1$ for all $x>1$, \begin{equation} \ell(x)=e^{b\sqrt{\ln x}} \tag{1}\label{1} \end{equation} for some real $b$ and all $x>1$, and $f(t)=\ln^2 t$ for all $t>2$.

Then, reasoning as in your post, for all large enough $n$ we have \begin{equation} P(|X|>a_n)\asymp\frac1{nf(n)}\frac{\ell(a_n)}{\ell(n)} =\frac1{n\ln^2 n}\frac{\ell(a_n)}{\ell(n)}, \end{equation} $a_n=n^{1/\al+o(1)}$, \begin{equation} \frac{\ell(a_n)}{\ell(n)}=\exp\{b(\sqrt{1/\al}-1+o(1))\sqrt{\ln n}\}. \end{equation} Letting now $b=1$ if $\al\in(0,1)$ and $b=-1$ if $\al\in(1,2)$, we see that for all large enough $n$ \begin{equation} \frac{\ell(a_n)}{\ell(n)}>\ln n \end{equation} and hence \begin{equation} \sum_n P(|X|>a_n)=\infty, \end{equation} whereas $\int_1^\infty \frac{dt}{t f(t)}<\infty$, so that the inequality \begin{equation} \sum_{n=1}^\infty P\big( |X| > a_n \big) \le \tilde C \int_1^\infty \frac{dt}{t f(t)} \end{equation} cannot hold for any real $\tilde C$.

In the case $\al=1$, instead of \eqref{1} similarly consider \begin{equation} \ell(x)=\exp\frac{\ln x}{\ln\ln x} \end{equation} for $x>e$.


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