# Bivariate One-Sided Chebyshev Inequality (Symmetric Case)

Let $X$ and $Y$ be random variables with finite means $\mu_X$ and $\mu_Y,$ finite variances $\sigma_X^2$ and $\sigma_Y^2,$ and correlation $\rho.$ Let $A$ be the event that $X \leq \mu_X + k\sigma_X$ and $Y \leq \mu_Y + k\sigma_Y,$ where $k \gt 1.$ Using the one-sided univariate Chebyshev inequality, I can find this lower bound on the probability of $A$ occurring: $$P[A] \geq {{k^2-1} \over {k^2+1}}$$ But this does not use the correlation. If the correlation is known, what is the tightest bound possible?

• Are you sure you don't want $A$ to be the event $|X-\mu_X|\le k\ \sigma_X$ and $|Y-\mu_Y|\le k\ \sigma_Y$? That would be closer to the univariate Chebyshev situation.
– whuber
Feb 12, 2014 at 19:35
• Formulas for that case are available from Wikipedia's article on Chebyshev's inequality. I couldn't find anything on the one-sided scenario. Feb 12, 2014 at 20:10

Using the results from wikipedia page of Chebyshev's inequality (first inequality in section "two correlated variables"), I dont see anything for your case with one-sided inequalities. The more usual case with double inequalities, one obtains, with the event $A$ defined as $A=|\frac{X-\mu_X}{\sigma_X}|\le k ~~\text{and}~~ |\frac{Y-\mu_Y}{\sigma_Y}| \le k$ we get the inequality $$Pr(A) \ge 1 - \frac{2k^2+\sqrt{4k^4-4k^4\rho}}{2k^4} = 1- \frac{1+\sqrt{1-\rho}}{k^2}$$ where $\rho$ is the correlation between $X$ and $Y$. Observe that negative correlation reduces the bound, while positive correlation increases it.