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Math Help - Very nice (but quite tough) puzzle

  1. #1
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    Very nice (but quite tough) puzzle

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

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

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

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

    Draw and compare these sets and use independence.
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  2. #2
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    Quote Originally Posted by Boysilver View Post
    If X and Y are two independent identically distributed real valued random variables, show that

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

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

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

    Draw and compare these sets and use independence.
    Justify the following:

    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) \leq 5 \sum_n P(R_{n,n}) \leq 5 P(S)

    (you may need independence and Cauchy-Schwarz inequality for the middle inequality)
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  3. #3
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    I'd tried this briefly but hadn't considered using Cauchy Schwarz. This seems to give the result immediately - thanks!
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