To show that $P(|X-Y| \leq 2) \leq 3P(|X-Y| \leq 1)$ [duplicate]

probability inequality

I found this question while browsing through "The Probabilistic Method", by Noga Elon.

Let X and Y be 2 independent and identically distributed real valued random variables. Prove that: $$P(|X-Y| \leq 2) \leq 3P(|X-Y| \leq 1)$$

So I tried the following:

$$P\{|X-Y| \leq 2\} = P\{|X-Y| \leq 1\} + P\{X-Y \in (1,2]\cup[-2,-1)\}$$ $$= P\{|X-Y| \leq 1\} + P\{X-Y \in (1,2]\} + P\{X-Y \in [-2,-1)\}$$ $$= P\{|X-Y| \leq 1\} + 2P\{X-Y \in (1,2]\}$$ where the last step follows because $X-Y$ has a symmetric distribution. NB: A random variable Z has symmetric distribution if $$P(Z \leq z) = P(Z \geq -z) \quad \forall z \in \mathbb{R}$$

Thus the problem boils down to showing $$P\{X-Y \in (1,2]\} \leq P(|X-Y| \leq 1)$$ and I would be done. Unfortunately, I don't know how to proceed from here. I appreciate any help, hints, useful comments etc. I receive.


Solution 1:

A proof is given in "The 123 Theorem and its extensions" by Noga Alon, Raphael Yuster. (See also this question.)

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