# Existence of moments and slowly varying function at infinity

I have a somewhat advanced question involving the role of slowly varying functions and their relation to moments. I want to use them to derive certain results for domains of attraction.

My problem is the following.

Let $L(x) = E[Y^2; |Y|\le x]$ and assume that L(x) is slowly varying to $\infty.$ This means that for any $t > 0,$ we have

$$\lim_{x\to \infty} \frac{L(tx)}{L(x)} = 1.$$

Fix $p\in [0, 2).$ I can prove (it is some work, but not a major task) that $$\lim_{x\to\infty} \frac{x^{2-p}E[|Y|^p; |Y|>x]}{L(x)} = 0 \hskip 10pt (*)$$

I would like to prove that $E[|Y|^p] < \infty$ for $p \in [0, 2).$

I am aware that there are certain theorems (Karamata type theorems) that use arguments related to the survival function $G(x) = Pr[Y> x] = 1-F(x)$ varying slowly to infinity to derive the existence of certain moments. But, I wonder if what I know, i.e., (*), is enough.

My line of attack is the traditional one, i.e., $$E[|Y|^p] = E[|Y|^p; |Y|\le x] + E[|Y|^p; |Y|>x]$$ Now the first term on the right-hand side is clearly bounded and I was hoping to use what I learned above to prove that also the second term on the right-hand side is bounded, but I seem to run in some problems. For instance, I could use (*), and write $$E[|Y|^p; |Y|>x] = x^{2-p}E[|Y|^p; |Y|>x]/L(x) \cdot \frac{L(x)}{x^{2-p}}.$$ The first term on the right goes to 0 as $x \to \infty,$ however, I have an $\frac{\infty}{\infty}$ issue on the second term. Am I approaching the issue in the wrong way or am I missing something else?

As usual thank you to whoever could chip in some words of wisdom.

Maurice

• One other thing I thought about is that since $L(x)$ is slowly varying, it must grow much slower than any positive power of x. L(tx)/L(x) ~ x^0.. Aug 18, 2015 at 23:02

The fact that $Y$ belongs to each $\mathbb L^p$ for $p<2$ can be deduced directly. Using the definition of a slowly varying function, we have that for $n$ large enough, $L(2^{n+1})/L(2^n)\leqslant 3/2$. As a consequence, $$2^{2n}\mu\left\{2^n<|Y|\leqslant 2^{n+1}\right\} \leqslant \mathbb E\left[Y^2;2^n<|Y|\leqslant 2^{n+1} \right]\leqslant\frac{L(2^n)}2.$$ To conclude that $Y$ belongs to $\mathbb L^p$ for each $p<2$, we have to prove the convergence of the series $\sum_{n\geqslant 1}2^{(p-2)n}L(2^n)$ for each $p\lt 2$. This can be easily done using the ratio test, noticing that $L(2^{n+1})/L(2^n)\to 1$.
We can also conclude by (*) in the opening post using the fact that for each positive $\delta$, $L(x)x^{-\delta}\to 0$ as $x$ goes to $+\infty$. Indeed, for each positive $t$, $$\sup_{2^n\leqslant x\lt 2^{n+1}} L(x)/x^\delta\leqslant L(2^{n+1})2^{-n\delta}$$
and the series $\sum_nL(2^{n+1})2^{-n\delta}$ converges by the ratio test (here we used the fact that $L$ is non-decreasing, but in general, we can use the fact that $$\lim_{t\to\infty}\frac{\sup_{t\leqslant s\lt 2t }L(s)}{\inf_{t\leqslant s\lt 2t }L(s)}=1).$$