Maximum mean absolute difference of two iid random variables : a random variable proof?

Question

The post is a follow-up to this question :

Maximum mean absolute difference of two iid random variables

The question is to show that for two iid random variables $X$ and $Y$ on the unit interval, one has:

$$\mathbb E[|Y-X|] \le 1/2 $$

(The maximizer then is the $1/2 (\delta_0+\delta_1)$ distribution.)

The proposed proof (there is only one), by Sergei Golovan, and the tricks it uses, is quite striking. Still I don't see a way to convert that proof in term of random variables only, which leaves me unsatisfied (in particular, it is difficult to interpret probabilistically the integration by parts step).

Also, the upper bound, 1/2, lets me wondering if some symmetry argument could be used here.

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So I'm asking if there would be a proof that sticks with the random variables only, in the sense, say that it does not use integration by parts.

It may well be there is no such proof, and that one ultimately has to resort to integration by parts, I have no idea...

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Some equalities one may write to start with but that seem not to help much :

\begin{align*} \mathbb E[|Y-X|] & = \mathbb E[(Y-X) 1_{Y>X}]+ \mathbb E[(X-Y) 1_{X>Y}] \\ & = 2 \mathbb E[(Y-X) 1_{Y>X}] \\ & = 2 (\mathbb E[Y 1_{Y>X}] - \mathbb E[X 1_{Y>X}]) \\ & = 2 (\mathbb E[Y 1_{Y>X}] - \mathbb E[Y 1_{X>Y}]) \\ & = 2 \mathbb E[Y (1_{Y>X}-1_{X>Y}] ]\end{align*}

Answer 1

Per the answer I have just posted at the reference question, there is a simple (essentially one line) proof using only operations on random variables, together with the famous identity expressing the expectation as the integral of the tail probabilities (which also has a one line proof using Tonelli's Theorem).

The idea is to write the expectation as an integral over $[0,1]$ of a quantity that is bounded by $\tfrac 12$. The precise identity is $$ \frac{\mathbb E|X-Y|}{2}=\int_0^1 \mathbb P(X>t)\cdot \mathbb P(X\leq t)\ dt, $$ and you can bound the integrand by $\tfrac 14$ using the AM-GM inequality.

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I realize you asked for a proof that avoids integration by parts, and this is technically true for the proof I have given. However, it is important to be aware that while the tail integral formula I have used has a simple Tonelli proof that does not use integration by parts, it is often referred to as an "integration-by-parts" formula since that is the simplest way to prove it in elementary probability theory when one does not have access to Fubini / Tonelli theorems.

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I realized I can make the proof of the identity more explicitly probabilistic as follows. Observe the almost sure identity $$ (X-Y)\cdot 1[X>Y]=\int_0^1 1[X>t\geq Y]\ dt, $$ and similarly with $X$ and $Y$ reversed. Since $$|X-Y|=(X-Y)\cdot 1[X>Y]+(Y-X)\cdot 1[Y>X],$$ it follows that $$ |X-Y|=\int_0^1 1[X>t\geq Y]+1[Y>t\geq X]\ dt. $$ Now by Tonelli's Theorem we can put in the expectations on both sides to obtain $$ \frac{\mathbb E|X-Y|}{2}=\int_0^1 \mathbb P(X>t\geq Y)\ dt=\int_0^1 \mathbb P(X>t)\cdot \mathbb P(t\geq Y)\ dt, $$ using independence of $X$ and $Y$. Since $X$ and $Y$ have the same distribution, the identity at the top of my post follows.