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**Boysilver** If X and Y are two independent identically distributed real valued random variables, show that

$\displaystyle 5 \mathbb{P}(|X-Y| \leq 1) \geq \mathbb{P}(|X-Y| \leq 2)$.

Here's the hint: For $\displaystyle n,m \in \mathbb{Z}$ let $\displaystyle R_{n,m} = \{(x,y) \in \mathbb{R}^2 : n \leq x < n+1, m \leq y < m+1\} $ and let

$\displaystyle S = \{(x,y) \in \mathbb{R}^2 : |x-y| \leq 1\} $ and

$\displaystyle T = \{(x,y) \in \mathbb{R}^2 : |x-y| \leq 2\} $.

Draw and compare these sets and use independence.

Justify the following:

$\displaystyle P(T)\leq \sum_n \left(P(R_{n,n-2})+P(R_{n,n-1})+P(R_{n,n})+P(R_{n,n+1})+P(R_{n,n})\right) $ $\displaystyle \leq 5 \sum_n P(R_{n,n}) \leq 5 P(S)$

(you may need independence and Cauchy-Schwarz inequality for the middle inequality)