Relationship between two random variables? What is the relationship between a random variable obeying the subexponential distribution defined here and a random variable $X$
satisfying $P\left(\left|X\right|>t\right)\le\alpha e^{-\beta t}$
for all $t>0$ and some $\alpha,\beta>0$?
Thanks a lot for any helpful answers.
 A: As explained in the comments, if the distribution of a nonnegative random variable $Y$ is subexponential, then $\mathrm P(Y\gt x)\gg\mathrm e^{-cx}$ when $x\to+\infty$, for every positive $c$. Hence, if $\mathrm P(X\gt x)\leqslant \alpha\mathrm e^{-\beta x}$ when $x\to+\infty$ for some positive $\beta$, then $\mathrm P(X\gt x)\ll\mathrm P(Y\gt x)$ when $x\to+\infty$. And?
Edit The fact that every nonnegative random variable $X$ with subexponential distribution is such that $G(x)\gg\mathrm e^{-cx}$ when $x\to+\infty$, for every positive $c$, where $G(x)=\mathrm P(X\gt x)$, is a basic result of the field. Let us give a hands-on proof of a very restricted version of this fact, namely, that $G(x)\sim a\mathrm e^{-bx}$ is impossible.
By hypothesis, $\mathrm P(X+X'\gt x)\sim2G(x)$ when $x\to+\infty$, where $X'$ is independent of $X$ with the same distribution. Note that 
$$
[X+X'\gt x]=[X\gt x\vee X'\gt x]\cup A_x,\quad A_x=[X+X'\gt x\geqslant X,x\geqslant X'],
$$ 
and that the two events on the RHS are disjoint. In particular,
$$
\mathrm P(X+X'\gt x)=\mathrm P(A_x)+1-\mathrm P(X\leqslant x,X'\leqslant x).
$$
By independence, $\mathrm P(X\leqslant x,X'\leqslant x)=\mathrm P(X\leqslant x)^2=(1-G(x))^2$, hence
$$
\mathrm P(X+X'\gt x)=\mathrm P(A_x)+2G(x)-G(x)^2.
$$
Since $G(x)^2\ll G(x)$, the property of being subexponential is equivalent to 
$$
G(x)\gg\mathrm P(A_x).
$$
Now, $A_{2x}\supset[x\lt X\leqslant 2x,x\lt X'\leqslant 2x]$ hence, if the distribution is subexponential, then
$$
G(2x)\gg\mathrm P(x\lt X\leqslant 2x)^2=G(x)^2-2G(x)G(2x)+G(2x)^2.
$$
Since $G(x)G(2x)\ll G(2x)$ and $G(2x)^2\ll G(2x)$, one gets $G(2x)\gg G(x)^2$.
But if $G(x)\sim a\mathrm e^{-bx}$, then $G(2x)\sim a^2\mathrm e^{-2bx}\sim a G(x)^2$, hence $G(2x)\gg G(x)^2$ is impossible.
