Is $X+Y$ regularly varying at zero if $X$ and $Y$ are and are independent?

Question

Imagine it holds that

$$\lim\limits_{t \to 0}F(tx)/F(t)=x^\alpha,$$

where $F$ is the cdf of $X$ respectively $Y$ and $\alpha>0$. Does it then also hold that

$$\lim\limits_{t \to 0}F_{X+Y}(tx)/F_{X+Y}(t)=x^\beta,$$

for some $\beta>0$?

It is a well known fact that this is true if $X$ and $Y$ are regularly varying at infinity(Then X+Y is regularly varying at infinity as well with the same $\alpha$), however it is not possible to easily adapt this result to this problem

Answer 1

Yes, there is a result at zero, however, it is very different from the one at infinity you mentioned. Namely, the indices add up:

If $X_i$, $i=1,2$, are independent with cdfs $F_{X_i}$, which are regulatly varying at zero with indices $\alpha_i$, $i=1,2$, then $F_{X_1 +X_2}$ is regularly varying at zero of index $\alpha_1+\alpha_2$.

Proof follows from the corresponding Tauberian theorem (see e.g. Bingham, Goldie, Teugels, Theorem 1.7.1'):

A non-decreasing function $U\colon (0,\infty)\to (0,\infty)$, whose Laplace-Stieltjes transform $\hat U(s) = \int_0^\infty e^{-st} dU(s)$ is finite for large $s$, is regularly varying at zero of index $\rho\ge 0$ iff $\hat U(s)$ is regularly varying on infinity of index $-\rho$.

That said, $\hat F_{X_i}$ are regularly varying at infinity with indices $-\alpha_i$, $i=1,2$, so $\hat F_{X_1+X_2} = \hat F_{X_1}\hat F_{X_2}$ is regularly varying at infinity of index $-\alpha_1-\alpha_2$. Hence, $F_{X_1+X_2}$ is regularly varying at zero of index $\alpha_1+\alpha_2$.