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Chebychev's inequality for two random variables

  1. Oct 22, 2013 #1
    (I wasn't sure how to title this, it's just that the statement resembles Chebychev's but with two RV's.)

    1. The problem statement, all variables and given/known data


    Let [itex]\sigma_1^1 = \sigma_2^2 = \sigma^2[/itex] be the common variance of [itex]X_1[/itex] and [itex]X_2[/itex] and let [roh] (can't find the encoding for roh) be the correlation coefficient of [itex]X_1[/itex] and [itex]X_2[/itex]. Show for [itex]k>0[/itex] that

    [itex]P[|(X_1-\mu_1) + (X_2-\mu_2)|\geq k\sigma]\leq2(1+[roh])/k^2[/itex]

    2. Relevant equations
    Chebychev's inequality:
    [itex]P(|X-\mu|\geq k\sigma) \leq 1/k^2[/itex]


    3. The attempt at a solution

    I'm really only looking for a place to start. I can try working backwords, and expanding [roh] into its definition, which is [itex]E[(X_1-\mu_1)(X_2-\mu_2)]/\sigma_1\sigma_2[/itex], but I really don't know how to evaluate that. I was wondering about using Markov's inequality and substituting [itex]u(X_1,X_2)[/itex] for [itex]u(X_1)[/itex], but of course there's no equation linking [itex]X_1[/itex] and [itex]X_2[/itex]. Feeling stumped. Any suggestions welcome!
     
    Last edited: Oct 22, 2013
  2. jcsd
  3. Oct 22, 2013 #2

    mfb

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    Staff: Mentor

    I would try to find the variance and mean (okay, mean is obvious) for X1+X2. I think this converts Chebychev's inequality to the one you have to show.

    The greek letter is rho, and \rho gives ##\rho##.
     
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