# Very nice (but quite tough) puzzle

• Nov 15th 2009, 08:20 AM
Boysilver
Very nice (but quite tough) puzzle
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.
• Nov 15th 2009, 10:45 AM
Laurent
Quote:

Originally Posted by 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)
• Nov 15th 2009, 12:25 PM
Boysilver
I'd tried this briefly but hadn't considered using Cauchy Schwarz. This seems to give the result immediately - thanks!