# How can I show that $P\{|(X-\mu_X)+(Y-\mu_Y)| \ge k\sigma\} \le (2(1+\rho))/k^2$?

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?

• $2(1+\rho)\sigma^2$ is the variance of $Z=X+Y.$ Apply Chebyshev to $Z.$
– whuber
Nov 3 '19 at 22:10
• This was a great comment. Can I accept it as an answer? Nov 5 '19 at 3:12

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:
$$\mathbb{E}(Z) = \mathbb{E}(X+Y) = \mathbb{E}(X) + \mathbb{E}(Y) = \mu_X + \mu_Y,$$
$$\mathbb{V}(Z) = \mathbb{V}(X+Y) = \mathbb{V}(X) + 2 \cdot \mathbb{C}(X,Y) + \mathbb{V}(Y) = \sigma^2 + 2 \rho \sigma^2 + \sigma^2 = 2(1+\rho) \sigma^2.$$
Applying Chebychev's inequality to $$Z$$ yields the desired inequality.