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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?

Gary
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user1125373
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  • Without searching for the proof in that article, if $p>1$ then $y^{1-p}=y^{-p+1}\to 0$ as $y\to\infty.$ So unless $p$ is some function of $y,$ rather than a constant, it seems like there is a problem there. – Thomas Andrews Nov 29 '22 at 21:56
  • Okay, the article is really proving convergence to $0,$ not $1,$ so that $\to1$ statement is an error in the article - the author means $\to0.$ – Thomas Andrews Nov 29 '22 at 22:04
  • The expected value existing for $F$ means $\int_{\mathbb R} tF'(t),dt$ exists, which, the author is claiming implies $p$ exists. I don't know the result that lets the author make this claim. It's definitely true that $1-F(y)$ has to go to $0$ faster than $y^{-1}.$ Something like $1-F(y)\sim \frac1{y\log^2y}$ seems to work. But you still have $(1-F(y))y\to 0.$ – Thomas Andrews Nov 29 '22 at 22:09
  • The reality is, if $X$ is a random variable with$F$ as the CDF, then $$E(X)\leq F(y)E(X\mid X< y)+y(1-F(y))$$ so if $y(1-F(y))$ does not converge to $0,$ then $E(X\mid X<y)$ does not converge to $E(X)$ as $y\to\infty.$ But that seems to conflict with the condition that $E(X)$ exists. – Thomas Andrews Nov 29 '22 at 22:19

1 Answers1

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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.)

Thomas Andrews
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  • Thanks a lot!! I can't believe I have missed that mistake. I can understand most of your proof, one quick question though: is there any formal proof for $\int_y^{\infty}tF'(t)dt \ge \int_y^{\infty}yF'(t)dt$? I definitely get the idea though, we are multiplying by numbers greater than $y$ itself so the integral on the left has to be bigger than the one on the right. – user1125373 Nov 29 '22 at 23:10
  • For $t\in [y,+\infty),$ $t\geq y,$ so, since $F'(t)\geq 0,$ $$tF'(t)\geq yF'(t)$$ so the integrals from $y$ to infinity of the left side is $\geq$ the integral from $y$ to infinity of the right side. @user1125373 – Thomas Andrews Nov 29 '22 at 23:18
  • Of course. Thanks a lot!! – user1125373 Nov 29 '22 at 23:32