I'm reading a paper in which the authors want to prove something like this:
\begin{align}\tag{1} E(X^{-p})< \infty, \end{align} where $p \geq 2$ and $X$ is a nonnegative random variable.
They say it's enough to show the following:
Fix $a \in (1/2,1)$. Then, for all $p \geq 2$, there exists a $c_0$ (depending on $p$ and $a$) such that if $c \leq c_0$, then \begin{align}\tag{2} P(X\leq c^a) \leq c^{(1-a)p}. \end{align}
I am trying to prove that this is indeed sufficient for $X$ to have negative moments (i.e $(2)\implies (1)$). My attempt:
- Use integral representation of expectation, split the integral at $c_0^{-ap}$, and bound the first term by a constant.
\begin{align} E(X^{-p}) &= \int_0^{\infty} P(X^{-p}>y) dy \\ &\leq c_0^{-ap} + \int_{c_0^{-ap}}^{\infty} P(X<y^{-1/p})dy \end{align}
Change variables: $ y= c^{-ap} $, so $dy = -apc^{-ap-1}dc$, and $$ \int_{c_0^{-ap}}^{\infty} P(X<y^{-1/p})dy = ap \int_0^{c_0} P(X < c^a)c^{-ap-1}dc. $$
Use the assumed (2) to get $$ \int_0^{c_0} P(X < c^a)c^{-ap-1}dc \leq \int_0^{c_0} c^{(1-a)p}c^{-ap-1}dc = \int_0^{c_0} c^{p(1-2a)-1}dc. $$
The issue here is that the exponent of $c$ is in the interval $(-2,-1)$, which makes the integral diverge. The authors say that $(2)\implies (1)$ is "clearly" true, so there's probably something "simple" that is not occurring to me. Any ideas?