Assume $X_1, X_2$ are iid. distributed in a way so that they have a survival function of the form $P(X_i>x)=L(x)*x^{-\alpha}$. Where $L(x)$ slowly varying function (defined by: $\lim_{x\to\infty} \frac{L(kx)}{L(x)}=1$ for all $k > 0$).
I want to investigate if the product $X_1X_2$ follows a power-law distribution (so has the same structure in the survival function).
I have seen a few people note that this seems to be true, but I can't find a proof and am currently unable to work it out myself. I have started thinking about some ways to prove it, but I always run into trouble.
For example: Is the derivative of $L$ again a slowly varying function? (I believe it is, but I can not prove it). (This would give a way to describe the density of the product).
Edit: For further clarification. I am looking for the proof of Lemma 4.1 (1) in https://web.math.ku.dk/~mikosch/Preprint/Anders/n094p171.pdf (I do not understand the proof cited)
A proof is given by ON CLOSURE AND FACTORIZATION PROPERTIES OF SUBEXPONENTIAL AND RELATED DISTRIBUTIONS.
I will first provide a more expansive version of the proof of the corollary, but won't touch on the proof of Theorem 3, which can be found in the document linked.
In the following let $RV_\alpha$ denote the set of regular varying functions and $\mathcal{L}_\alpha$ the set of functions so that:
$$ F \in \mathcal{L}_\alpha \Leftrightarrow \forall y \in \mathbb{R}: \lim_{x \to \infty} \frac{1-F(x+y)}{1-F(x)} = e^{-\alpha y}$$
For the survival function of a distribution function $F$ write: $\overline{F} = 1 - F$.
Let $X^*$ and $Y^*$ be two independent non-zero random variables with distribution $F^*$ and $G^*$ respectively. Write $H^*$ for the distribution function of the product of $X^*Y^*$
I want to proof: $\overline{F^*}, \overline{G^*} \in RV_\alpha \Rightarrow \overline{H^*} \in RV_\alpha$
Define $X := ln(X^*), Y := ln(Y^*)$
For the distribution functions of $X$ and $Y$ write $F$ and $G$. $F(x) = F^*(e^x)$ and $G(x) = G^*(e^x)$
$$\lim_{x \to \infty} \frac{\overline{F}(x+y)}{\overline{F}(x)} = \lim_{x \to \infty} \frac{\overline{F^*}(e^xe^y)}{\overline{F^*}(e^x)} = \lim_{x \to \infty} \frac{e^{-x\alpha}e^{-y\alpha} L(e^xe^y)}{e^{-x\alpha}L(e^x)} \quad \text{ for some slowly varying function } L$$
$$\lim_{x \to \infty} \frac{e^{-x\alpha}e^{-y\alpha} L(e^xe^y)}{e^{-x\alpha}L(e^x)} = e^{-y\alpha} \lim_{x \to \infty} \frac{L(e^xe^y)}{L(e^x)} = e^{-y\alpha} \quad \Rightarrow \quad F \in \mathcal{L}_\alpha$$
And using the same argument $G \in \mathcal{L}_\alpha$.
With theorem 3: $(F * G) \in \mathcal{L}_\alpha$
$$(F * G)(k) = P(X+Y \leq k) = P(ln(X^*)+ln(Y^*) \leq k) = P(ln(X^*Y^*) \leq k) = P(X^*Y^* \leq e^k) = H^*(e^k) \\ \Rightarrow H^*(x) = (F * G)(ln(x)) \quad \forall x > 0$$
$$\Rightarrow \lim_{t \to \infty} \frac{\overline{H^*}(tx)}{\overline{H^*}(t)} = e^{-\alpha ln(x)} = x^{-\alpha} \Rightarrow \overline{H^*} \in RV_\alpha$$