Question
Let $X$ be a random variable with finite expectation, i.e. $\mathbb E|X|<\infty$.
I know the tail probability must then be $\mathbb P(X > x) = o(1/x)$ as $x \to \infty$. (See here for proof.)
But is there any (known) bound that's better than $1/x$ ?
Thoughts
Using the same proof idea as in the post above, we can show that if $\mathbb E|f(X)| < \infty$ for some measurable function $f(x) \to \infty$, then $\mathbb P(X > x) = o \big( 1/f(x) \big)$.
For example, $\mathbb E|X|^k < \infty$ implies $\mathbb P(X > x) = o(1/x^k)$ as $x \to \infty$.
However, I'd only like to assume $\mathbb E|X| < \infty$.