Let $X$ be a random variable with characteristic function $\varphi_X(y)$ such that $\varphi_X(y) = 1 - o(|y|^{\alpha})$ for $y \rightarrow 0$. Prove that $P(|X| \geq t) = o(t^{-\alpha})$ for $t \rightarrow \infty$.
I think I need to use Markov's inequity, but I don't know how...
$$P(|X| \geq t) \leq \frac {E|X|} {t} \, \text{for} \, E|X| < \infty$$ But we don't know $E|X| < \infty$ or $E|X| = \infty$. And even if $<$, we don't know what $E|X|$ equals to. Or know? Can we find characteristic function for modulus X if we have it for just X?
Please, explain me this task.
This follows easily from the following well known inequality which you can find on page171 of Breiman's book: $P(|X|> t) \leq \alpha t \int_0^{1/t} [1-\Re \phi_X(s)] ds$. [I have replaced $u$ in that inequality by $\frac 1 t$]. Here $\alpha$ is a universal constant belonging to $(0,\infty)$.