Let $X$ be a positive (i.e. nonzero) random variable which is bounded in probability as follows: there ex. constants $A, C > 0$ s.t. for every $\varepsilon > 0$
$$ \mathbb{P}(X \geq \varepsilon) \leq A \exp (-C \varepsilon^2) $$
Knowing this, is there anything we can say about the expression $ \mathbb{E} [ \exp (-1/X) ]$? I feel like intuitively this should mean we can somehow bound this expression as well, at least in probability, but I can't figure out how to go about it. Any help would be appreciated, I don't even really know where to start on this one.
Edit: the sub-gaussian tail bound on $X$ does lead to equivalent properties for $X$, but since I have the reciprocal here, I don't really know much about it… it isn't sub-gaussian itself as far as I know, so I can't apply any results for the moment-generating function here…
Best Answer
Recall that, for non-negative random variables $Y$, we have $$\mathbb{E}[Y] = \int_0^\infty \mathbb{P}(Y > \lambda) \, d\lambda$$ (this is a consequence of Fubini's theorem). Applying this to $Y = \exp(-1/X)$, which is supported on $[0,1]$, \begin{align} \mathbb{E}[\exp(-1/X)] &= \int_0^1 \mathbb{P}(\exp(-1/X) > \lambda )\, d\lambda \\ &= \int_0^1 \mathbb{P}(X > -1/\ln(\lambda)) \, d \lambda\\ &\leq \int_0^1 A \exp(-C/(\ln(\lambda))^2) \, d\lambda. \end{align}