# 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

$$R_G(f) \le \frac{2}{(q+1)^{1/q}}\left[M_{E,p}(f)\right]^{1/p}$$
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
$$R_G(f) \le \frac{2}{\sqrt{3}}\sigma$$
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