# Inequality with slowly varying functions

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.

• 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.
• From the paper $\ell$ is slowly varying means that $\lim_{t\to \infty}\frac{\ell(tx)}{\ell(t)}=1$ for all $x>0$. Apr 27, 2022 at 9:07
• Yes, perhaps I should specify: $\ell$ is slowly varying in the Karamata sense. (See: en.wikipedia.org/wiki/Slowly_varying_function) Apr 27, 2022 at 9:17
• It might be useful to try to express the last ratio you mention as $\ell\left( n \left( n^{1-\alpha} f(n)\ell(n) \right)^{1/\alpha} \right)/\ell(n)$, in particular $\left( n^{1-\alpha} f(n)\ell(n) \right)^{1/\alpha}$ might not be too weird of a function. Apr 27, 2022 at 9:51