Class of c.d.f such that $\lim_{\infty}x(1-F(x))=0$

Question

Say random variable $x$ has c.d.f $F()$. I am looking for simple sufficient conditions such that the $\lim_{x\to \infty}x(1-F(x))=0$? Bounded support is clearly sufficient. I could not prove that finite mean (Ex finite) was sufficient (even if $x$ has positive support).

Answer 1

The simplest thing to assume here is to assume that there exists $a_0,\epsilon$ so that for all $x > a_0$

$$ 1-F(x) < \frac{1}{x^{1+\epsilon}} $$

But you have a good approach, use partial integration and assume $X>0,EX<\infty$, then $$ \int_0^a x F(dx)dx = -a(1 - F(a)) + \int_0^a(1-F(x))\,dx $$

But we also have the identity $$ EX = \int_0^{\infty}(1-F(x))\,dx $$ This beacause you calculate the same area by this as using $'xf_X'$

This then should indicate the limit.