When does the ratio X/Y of two random variables have a finite moment-generating function? Let $X$ and $Y$ be two positive random variables with $Y < X$; these may be highly correlated.  I would like a reasonable condition on $X$ and $Y$ so that the ratio $X/Y$ has a finite moment-generating function.  By this I mean that $\mathbb E e^{r X/Y} < \infty$ for all $r \in \mathbb R$.
Here's one answer.  Suppose that X and 1/Y both have super-Gaussian tail decay, that is,
$P(X > u) \le C e^{-cu^{2+\delta}},$
for positive constants $c$, $C$ and $\delta$, and similarly for $1/Y$.  Then $X/Y$ has super-exponential tail decay:
$P(X/Y > u) = P(X > Yu$ and $Y \le 1/\sqrt{u}) + P(X > Yu$ and $Y > 1/\sqrt{u})$
$\le P(1/Y > \sqrt{u}) + P(X > \sqrt{u})$
$\le 2Ce^{-cu^{(2+\delta)/2}} = 2Ce^{-cu^{1+\delta/2}}.$
This gives a finite moment-generating function by the following simple argument, which uses the fundamental theorem of calculus and Tonelli's theorem.  Let $Z = X/Y$.
$\mathbb Ee^{rZ} = \mathbb E \left[ 1 + \int_0^Z re^{ru} ~du \right] = 1 + \int_0^\infty re^{ru} \mathbb P(Z > u) ~du \le 1 + 2Cr \int_0^\infty e^{ru} e^{-cu^{1+\delta/2}} ~du$,
which is finite for all $r \in \mathbb R$.  Thus, super-Gaussian tail decay suffices, but this is a very strong condition which I'd like to weaken.
(In fact, I only need that $\mathbb Ee^{rX/Y} < \infty$ for any one positive value of $r = r_0$.  In that case, we may choose $r_0 = c/2$, and take $\delta = 0$ in all the above arguments.  Then the integral $\int_0^\infty e^{r_0 u} e^{-cu} ~du$ still converges.)
 A: Let $D$ denote a set of probability distributions of positive random variables and $r$ a positive real number. Consider the following properties:


*
* For every random variables $X$ and $T$ with probability distributions in $D$, $E(\mathrm{e}^{rXT})$ is finite.

* For every random variable $X$ with probability distribution in $D$, $E(\mathrm{e}^{rX^2})$ is finite.


Property 1 implies Property 2 (choose $X=T$). Conversely, if Property 2 holds, note that $$
\mathrm{e}^{ab}\le\mathrm{e}^{a^2}+\mathrm{e}^{b^2}-1$$ for every nonnegative $a$ and $b$ since the inequality is symmetric and, if $a\le b$, then the LHS is $\le\mathrm{e}^{b^2}$. For $a=\sqrt{r}X$ and $b=\sqrt{r}T$, one gets 
$$
E(\mathrm{e}^{rXT})\le E(\mathrm{e}^{rX^2})+E(\mathrm{e}^{rT^2})-1,
$$
which proves that Property 1 holds. Hence Properties 1 and 2 are equivalent.
Coming back to your context, you ask for Property 1 to hold with $T=1/Y$ and with no further constraint on the joint distribution of $(X,T)$ that $XT\ge1$ almost surely, a condition which holds as soon as $X\ge1$ and $T\ge1$ almost surely, for example. Thus, it seems that you are left with the condition that the set of probability distributions $D$ to which the distributions of $X$ and $T=1/Y$ belong, fulfills Property 2.
Basically, this means that one can replace your control $C\mathrm{e}^{-2ru^{2}}$ of the tails $P(X\ge u)$ and $P(1/Y\ge u)$ by $$Cu^{-(2+\delta)}\mathrm{e}^{-ru^2},
$$
as soon as $\delta>0$ and that, without further information on the relationship between $X$ and $Y$, one cannot expect a much better sufficient condition.
