# an upper bound of mean absolute difference?

Let $$X$$ be an integrable random variable with CDF $$F$$ and inverse CDF $$F^*$$. $$Y$$ is iid with $$X$$. Prove $$E|X-Y| \leq \frac{2}{\sqrt{3}}\sigma,$$ where $$\sigma=\sqrt{Var(X)} = \sqrt{E[(X-\mu)^2]}$$.

I am looking for some hint for this proof.

What I've got is $$E|X-Y|=2\int_{0}^{1}(2u-1)F^*(u)du$$. But I am not sure if this is correct direction.

I also noticed that $$\frac{2}{\sqrt{3}}$$ may be related to the variance of the uniform distribution.

• Hint: $$\mathbb E[|X-Y|]=2\int F(y)(1-F(y))\,\text{d}y$$ Oct 23, 2020 at 5:51
• Thanks, @Xi'an. If there any further steps? I know how to derive this but not sure how to use it. I feel that it is even further than the hint in my question..
– Tan
Oct 23, 2020 at 6:12
• You got an expression containing $\mu$. That can't be right, since the distribution of $| X-Y|$ do not depend on $\mu$. Oct 24, 2020 at 14:38
• Thank you! I corrected it.
– Tan
Oct 24, 2020 at 15:36
• @Kjetil Because $\mu$ was a bound integration variable, there was no reason to change it to "$u$" and there was no ambiguity in its use (but I admit it was initially confusing to see it in the integral!)
– whuber
Oct 24, 2020 at 15:55

## 1 Answer

Theorem 3.3 from p. 86 of "Cerone, Pietro, and Sever S. Dragomir. "A survey on bounds for the Gini Mean Difference." Advances in Inequalities from Probability Theory and Statistics (2008)" states that
$$R_G(f) \le \frac{2}{(q+1)^{1/q}}\left[M_{E,p}(f)\right]^{1/p}$$
where $$R_G(f)=\frac{1}2 E|X-Y|$$, $$p>1$$, $$1/p+1/q=1$$, and $$M_{E,p}(f)=E\left[|X-\mu|^{p}\right]$$.
The proof is short and uses Holder's inequality. Now, Remark 3.2 says to take $$p=q=2$$ in the inequality to find
$$R_G(f) \le \frac{2}{\sqrt{3}}\sigma$$
The reference says this inequality is known and refers to
https://galton.uchicago.edu/~wichura/stat304/handouts/L09.means3.pdf
But, I could not access that website. It also states the upper bound is obtained for the Unif(0,1) distribution. It seems like there is a misprint in the reference because I think the inequality should be $$R_G(f) \le \frac{1}{\sqrt{3}}\sigma$$. There is a $$\frac{1}2$$ included as part of the definition of the Gini mean difference $$R_G(f)$$.

• Is this the resource you're talking about? core.ac.uk/download/pdf/82733855.pdf (written by 2 of the same authors). They do prove the $1/\sqrt{3} \cdot \sigma$ bound that you're talking about (Thm. 6, p. 605) after the above theorem you posted (Thm. 5, p. 604) -- then that inequality is equivalent to what OP wants to show. Dec 28, 2020 at 22:45
• @tchainzzz yes, that is the reference and that has the correct bound. Dec 29, 2020 at 1:19