$(1-F(y))y \rightarrow 1$ as $y \rightarrow \infty$ I am trying to understand the proof given in Appendix A of https://arxiv.org/pdf/2103.00083.pdf.
At the final part, the authors are proving that $(1-F(y))y \rightarrow 1$ as $y \rightarrow \infty$, where $F(y)$ is a cumulative probability distribution.
They claim that since
\begin{align}
1-F(y) \lessapprox y^{-p} \text{ for } p>1.
\end{align}
Then
\begin{align}
(1-F(y))y \lessapprox y^{-p+1} \rightarrow 1 \text{ as } y\rightarrow\infty.
\end{align}
Where $p>1$.
First of all, why does $1-F(y) \lessapprox y^{-p}$ apply? It has to do with $\int(1-F(y)) \lessapprox \int y^{-p}$?
And can someone please illuminate me on this concept of inequality convergence?
 A: First, the authors made an error. From statements before, it is clear they need $(1-F(y))y\to0,$ not $(1-F(y))y\to 1.$
And, indeed, if $p>1,$ then $y^{1-p}=y^{-p+1}\to 0.$
The second question is: How do we know $p>1$ exists?
There, I think the authors have fudged the argument. It is true that $1-F(y)$ has to converge to zero faster than any constant multiple of $y^{-1},$ since otherwise the random variable would not have an expected value.
But that doesn't mean $p$ exists. For example, $1-F(y)\sim\frac C{y\log^2 y}$ would have an expected value, but converges slower than any $y^{-p}$ for $p>1.$
So you are going to need to use something different to prove $(1-F(y))y\to0$ a little more rigorously.
Luckily, there is a more direct way to prove this limit.
If $E$ is the expected value,
$$E=\int_{-\infty}^{\infty} tF'(t)\,dt \geq \int_{-\infty}^y tF'(t)\,dt+y(1-F(y))$$
That's because:
$$\int_y^{\infty} tF'(t)\,dt \geq \int_{y}^\infty yF'(t)\,dt=y(1-F(y)).$$
So we have, for $y>0,$ $$0\leq y(1-F(y))\leq E-\int_{-\infty}^ytF'(t)\,dt$$
But the limit as $y\to\infty$ of the integral on the right is just $E$ again, so, by the squeeze theorem, $y(1-F(y))\to0.$

The authors have left out a related needed result, $yF(y)\to 0$ as $y\to -\infty.$ That's the same, only using the left hand side. You get, for $y<0:$
$$0\geq yF(y)\geq \int_{-\infty}^y tF'(t)\,dt\to 0$$ as $y\to-\infty.$
This is really just the same argument, for $G(y)=1-F(-y).$ Then $G(y)$ is the CDF for $-Y$ if $F$ is the CDF for $Y.$
(Technically, $G(y)$ might disagree with the CDF of $-Y$ at any point of discontinuity of $G,$ but there are at most countably many discontinuities, so they don't matter for integrals.)
