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.)
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