Cantelli's Inequality states that for a random variable $X$ with mean $μ$ and variance $\sigma^2$:

$$ P(X-μ\geq \alpha)\leq\frac{\sigma^2}{\sigma^2 + \alpha^2} $$

Now, I read that if I consider a Bernoulli random variable with parameter $p$, then it can be shown that there is no better bound. I took this line of thinking but the farthest I got was that:

$$ P(X-p \geq \alpha) \leq \frac{p(1-p)}{p(1-p) + \alpha^2} $$

I am not sure even if I simplify here and get an equality what it could possibly show in terms of the efficiency of the bound. Does anyone have any insights? Thanks!

up vote 2 down vote accepted

Hint: Take $\alpha = 1-p$.

In detail:

$$\mathbb{P}\{X - p \geq \alpha\} = \mathbb{P}\{X \geq 1\} = p$$


$$\frac{\sigma^2}{\sigma^2+\alpha^2} = \frac{p(1-p)}{p(1-p)+(1-p)^2}= \frac{p}{p+(1-p)} = p.$$

(Place your move over the gray areas to reveal them.)

  • A hint that was given was to consider a Bernoulli random variable. Is your approach exactly implicitly considering a Bernoulli random variable? – user136503 Jan 29 '16 at 16:17
  • @user136503 Yes: in the above, I assume as in your question that $X\sim\operatorname{Bernoulli}(p)$ for some $p\in (0,1)$. – Clement C. Jan 29 '16 at 16:18
  • I see, so if we take $\alpha$ to be a fixed number then that means that we are only showing that Cantelli's inequality can have no better result in some cases, but not all cases? In other words, are we just showing the existence of an $\alpha$ that makes it so that Cantelli's inequality is sharp? – user136503 Jan 29 '16 at 16:22
  • 1
    Yes, but that's exactly what you are asked to prove: there can be no general improvement on the inequality, since there is at least one particular setting of the parameters for which the inequality is tight. – Clement C. Jan 29 '16 at 16:23

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