Chebychev's inequality for two random variables

  • Thread starter rayge
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  • #1
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(I wasn't sure how to title this, it's just that the statement resembles Chebychev's but with two RV's.)

Homework Statement




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]

Homework Equations


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


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:

Answers and Replies

  • #2
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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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