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Let $\sigma^2$ be the common variance of the random variables $X$ and $Y$, with their correlation coefficient being $\rho$.

Show that $\forall k>0$, $P\{|(X-\mu_X)+(Y-\mu_Y)| \ge k\sigma\} \le (2(1+\rho))/k^2$.

I know this looks similar to the Chebyshev Inequality: $P\{|X-\mu_x| \ge k\sigma \} \le 1/k^2$. However, I am struggling to find a way to apply Chebyshev's Inequality to show the above inequality. Would I need to use a different inequality, like the covariance inequality or Minkowski Inequality? How many I go about doing this?

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    $\begingroup$ $2(1+\rho)\sigma^2$ is the variance of $Z=X+Y.$ Apply Chebyshev to $Z.$ $\endgroup$
    – whuber
    Nov 3 '19 at 22:10
  • $\begingroup$ This was a great comment. Can I accept it as an answer? $\endgroup$
    – Jen Snow
    Nov 5 '19 at 3:12
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Just to expand on whuber's comment (and give you an official answer), suppose you take $Z=X+Y$ and then find the mean and variance of this random variable. You have mean:

$$\mathbb{E}(Z) = \mathbb{E}(X+Y) = \mathbb{E}(X) + \mathbb{E}(Y) = \mu_X + \mu_Y,$$

and variance:

$$\mathbb{V}(Z) = \mathbb{V}(X+Y) = \mathbb{V}(X) + 2 \cdot \mathbb{C}(X,Y) + \mathbb{V}(Y) = \sigma^2 + 2 \rho \sigma^2 + \sigma^2 = 2(1+\rho) \sigma^2.$$

Applying Chebychev's inequality to $Z$ yields the desired inequality.

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  • $\begingroup$ Wonderful. Thank you! $\endgroup$
    – Jen Snow
    Nov 6 '19 at 1:39

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