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Assume that $ f: \mathbb{R} \to \mathbb{R}_+ $ is a concave, non-decreasing and positive function. Let $\mathbb{X}$ be a finite set consisting of $ 0\leq x_1 \leq x_2 \leq x_3 \leq \ldots \leq x_n$. Further, let $ X $ be a random variable supported on $\mathbb{X}$.

I am interested in the following inequality:

$$ \frac{ f(x_n) - \mathbb{E}[f(X)] }{ \sqrt{ \mathbb{E}\left[ \left(f(X) - \mathbb{E}[f(X)]\right)^2 \right] } } \leq \frac{ x_n - \mathbb{E}[X] }{ \sqrt{ \mathbb{E}\left[ \left(X - \mathbb{E}[X]\right)^2 \right] } } $$

Is this inequality always true under the given conditions? If so, can you provide a proof or some intuition behind why this inequality holds? If not, can you provide a counterexample or conditions under which it might fail?

Alireza Bakhtiari
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1 Answers1

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It is true, fairly simple, and reasonably well-known (to people who know it well) in a slightly different form.

Shifting $x_n$ and $f(x_n)$ to $0$, flipping both axes and the fraction, squaring, writing the variance as $E[Y^2]-E[Y]^2$ and adding $1$, we arrive at the following (classical?) version of this inequality:

Let $X\ge 0$ be a random variable, and let $f:[0,+\infty)\to [0,+\infty)$ be a convex increasing function with $f(0)=0$. Then $$ \frac{E[f(X)^2]}{E[f(X)]^2}\ge \frac{E[X^2]}{E[X]^2}\,. $$

The proof is a three-liner. First, note that the LHS is invariant under multiplication of $f$ by a positive constant, so we can normalize so that $E[f(X)]=E[X]$. Then $E[(f(X)-X)_+]=E[(X-f(X))_+]$ (here $a_+=\max(0,a)$). Note that due to the convexity of $f$ and the condition $f(0)=0$, we have $f(x)\le x$ for $x\le x_0$ and $f(x)\ge x$ for $x\ge x_0$ with some $x_0>0$.

Hence $$ E[(f(X)^2-X^2)_+]\ge E[2x_0(f(X)-X)_+] \\ =E[2x_0(X-f(X))_+]\ge E[(X^2-f(X)^2)_+]\,, $$ whence $E[f(X)^2]\ge E[X^2]$ and we are done.

fedja
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  • There is another bounty with some hours left (there is an answer but not sure if it really solves the question) that I feel it could be solved with a version of the inequality you used because what wants to be proved $\sum (p-\frac{1}{n})^2\ge\sum(\frac{p\log 1/p}{\sum p\log 1/p} -\frac{1}{n})^2$ reduces to $\sum p^2\ge\sum(\frac{p\log 1/p}{\sum p\log 1/p})^2$ or $\frac{\sum p^2}{(\sum p)^2}\ge\frac{\sum f(p)^2}{(\sum f(p))^2}$, if you want to look at it. – Dabed Aug 05 '24 at 05:15
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    @Dabed $f(p)=-p \log p$ is concave but not increasing, which makes it challenging (see the existing answer for more details). – Amir Aug 05 '24 at 06:54
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    @Amir Ah right did follow the proof but put my attention was drawn just to the similitude of the inequalities and didn't think this much trough, thanks. – Dabed Aug 05 '24 at 07:28
  • @Amir Of course, $f(p)=-p\log p$ is not increasing on $(0,1)$ but $-cp\log p+p$ (where $c\in(0,1)$ is the normalizing factor such that $\sum cf(p)=1$) is increasing on the set of $p$'s involved (though it is not at all obvious), and that suffices. Can you finish it from here or should I post the full solution? – fedja Aug 05 '24 at 11:53
  • @fedja Thank you! If you are not busy now, you may post it as a full solution so that I can award the bounty (it expires shortly). – Amir Aug 05 '24 at 12:20
  • @Amir Right now I have some other fish to fry, but I'll try to do it later in the day :-) – fedja Aug 05 '24 at 12:26
  • @fedja As it expires in 2-3 hours, you may provide a mini answer and then complete it later. Otherwise, I will complete it later. Grilling fish is healthier and faster than frying it :D – Amir Aug 05 '24 at 12:38
  • @fedja thanks a lot for the answer! Btw, do you recommend a source for inequalities of this type? I read most parts of the "Cauchy-Schwarz Master Class" but didn't find it useful here. – Alireza Bakhtiari Aug 08 '24 at 12:25
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    @AlirezaBakhtiari You are welcome. I'm not sure if I know any good textbook that would go over such trickery (maybe Amir can recommend something?), but the idea of the general setup with "one crossing point and balanced sums/integrals" is simple enough to understand just from a couple of examples and it is surprisingly useful. – fedja Aug 09 '24 at 00:24
  • @fedja I have a new question that is similar to this one, so I think you may be able to help me! I really appreciate that! – Alireza Bakhtiari Sep 10 '24 at 08:52