# Cantelli's inequality and Chebyshev's inequality in comparison

There is an exercise in my probability book

Problem. Let $$X$$ be a random variable with finite variance $$\sigma^{2}$$.

Prove that for non-negative $$\lambda \in \mathbb{R}$$ a one-sided Chebyshev inequality holds such that

$$P(X - E(X) \geq \lambda) \leq \frac{\sigma^{2}}{\lambda^{2} + \sigma^{2}}.$$

I managed to do this via Markov's inequality and minimizing the value.

When is Cantelli's inequality better than Chebyshev's inequality. I am not quite sure I understand what "better" stands for in this context. Any suggestions are welcomed.

And the final thing the problem asks for is:

Find $$X$$ assuming two values for which equality holds.

"Better" stands for a sharper bound, that is the bound for Cantelli's inequality is smaller than the bound in Chebyshev's inequality. More on this after I answer the second question.

To get equality for Cantelli's inequality: Let the distribution of $$X$$ be a Bernoulli distribution with parameter $$p$$, $$X \in Be(p)$$. Take $$a = 1 - p$$. Then the Cantelli inequality gives \begin{align} P(X - EX \geq \alpha) &\leq \frac{\sigma^2}{\sigma^2 + \alpha^2)} \\\\ &= \frac{p(1-p)}{p(1-p) + (1-p)^2} \\\\ &= p \end{align}

But the true probability \begin{align} P(X - EX \geq \alpha) &= P(X - p \geq 1-p) \\\\ &= P(X \geq 1) \\\\ &= p \end{align} is the same as the bound from Cantelli's inequality. This means that the Cantelli inequality cannot be improved without further assumptions.

The question now is when is Cantelli's inequality better than Chebyshev's inequality.

Cantelli's Inequalities : $$P(X-EX \geq \alpha) \leq \frac{\sigma^2}{\sigma^2 + \alpha^2}$$ $$P(|X-EX| \geq \alpha) \leq \frac{2\sigma^2}{\sigma^2 + \alpha^2}$$

Chebyshev's inequality: $$P(|X-EX| \geq x) \leq \frac{\sigma^2}{\alpha^2}$$

A sharper bound means $$\frac{2\sigma^2}{\sigma^2 + \alpha^2} \leq \frac{\sigma^2}{\alpha^2}$$ which becomes $$\alpha^2 \leq \sigma^2$$. This makes both bounds more than 1, hence not very useful. However the one sided Cantelli inequality is indeed useful as we have shown that it can produce sharp bounds. The one-sided version is also useful when Chebyshev's inequality isn't good enough.