Chebychev's inequality for two random variables

[rayge]

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

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

Homework Equations

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

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 E[(X_1-\mu_1)(X_2-\mu_2)]/\sigma_1\sigma_2, but I really don't know how to evaluate that. I was wondering about using Markov's inequality and substituting u(X_1,X_2) for u(X_1), but of course there's no equation linking X_1 and X_2. Feeling stumped. Any suggestions welcome!

 

[mfb]

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 $\rho$.