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.)