Cantelli's inequality

If $X$ and $Y$ are two independent random variables with means and variances $\mu_X,\mu_Y,\sigma^2_X,\sigma_Y^2$ respectively. If $\mu_X+\mu_Y>0$, by Cantelli's inequality then $P(X+Y\leq0)\leq\dfrac{\sigma_X^2+\sigma_Y^2}{\sigma_X^2+\sigma_Y^2+(\mu_X+\mu_Y)^2}$. Is this bound sharp? If yes, an example?

Thanks

• Are you assuming $X$ and $Y$ are independent? – Robert Israel May 19 '17 at 1:10

Let $Z = X + Y$. $Z$ has mean $\mu=\mu_1 + \mu_2$ and, assuming $X$ and $Y$ are independent, variance $\sigma^2 = \sigma_1^2 + \sigma_2^2$. The inequality is then Cantelli's inequality for $Z$. Now Cantelli's is equivalent to
$$I_{z > 0} \ge 1 - (z - b)^2/b^2$$ with $$b = \mu + \sigma^2/\mu$$ Thus the inequality is sharp if and only if almost surely $Z$ is one of the two values of $z$ that make this an equality, namely $0$ and $b$. But the only way the sum of two independent random variables can have only two possible values is that one of $X$ and $Y$ is almost surely constant.
• Thanks Robert. I was wondering if one can get an optimal bound (assuming independency between $X$ and $Y$) with the limited information of the first two moments of these random variables, of course without one being almost surely constant. – Quema May 19 '17 at 12:18