# Inequality with slowly varying functions

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.

Question

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.

Updates

• 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.

• It all depends on the exact definition of "slowly varying". So, what is it in the paper? May 29, 2022 at 16:14
• @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. May 29, 2022 at 16:57
• @IosifPinelis Then you fried the author :-) May 29, 2022 at 17:30
• @zxmkn : All right. Please let me know if some of the steps need further details. May 30, 2022 at 14:09
• @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. May 30, 2022 at 16:55

## 1 Answer

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

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