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I'm assuming that the random variable $X$ has mean $0$ and finite variance ${\sigma}^2$. It is immediate from Chebyshev's inequality that $$P(X\geq x)\leq \frac{{\sigma}^2}{x^2},$$ but I'm still trying to show that $$P(X\geq x)\leq \frac{{\sigma}^2}{{\sigma}^2+x^2}$$ for $x>0$. Any hints would be appreciated.

(I've shown that $$P(X\geq x)\leq \frac{E[(X+a)^2]}{(a+x)^2}$$ for $a\geq 0$.)

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  • $\begingroup$ this is known as the Cantelli inequality. $\endgroup$
    – mookid
    Feb 9, 2015 at 20:53
  • $\begingroup$ @mookid Thanks! $\endgroup$
    – Aubrey
    Feb 9, 2015 at 21:02
  • $\begingroup$ @mookid Your comment helped me look up a hint and find the answer. I guess I could now delete the question since I don't need an answer anymore? $\endgroup$
    – Aubrey
    Feb 9, 2015 at 21:08
  • $\begingroup$ It could help someone. Maybe try to answer yourself! $\endgroup$
    – mookid
    Feb 9, 2015 at 21:09

3 Answers 3

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After showing that $$P(X\geq x)\leq \frac{E[(X+a)^2]}{(a+x)^2}$$ for $a\geq 0$, we only need to minimize over $a$ and we obtain the result.

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Note that $t=E(t-X) \leq\sum_{i:x_i<t}(t-x_i)p_i$

Now, $t^2\leq (\sum_{i:x_i<t}(t-x_i)p_i)^2=(\sum_{i:x_i<t}(t-x_i)\sqrt{p_i} \sqrt{p_i})^2\\\le E(t-X)^2 P[X<t],$

having used the Cauchy-Schwarz inequality.

So, $P[X<t]\ge \frac{t^2}{t^2+\sigma^2}$ as $E(t-X)^2=t^2+\sigma^2$

So, $P[X\ge t] \le \frac{\sigma^2}{t^2+\sigma^2}$

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    $\begingroup$ How would this argument work if the probability space is not discrete? Aubrey's hint is one without any assumption on the space. $\endgroup$ Jul 24, 2015 at 20:42
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Use Markov Inequality:

$P(|X - \mu| > \alpha)$

= $P((X-\mu)^2 > \alpha^2)$

= $P((X-\mu)^2 + \sigma^2 > \alpha^2 + \sigma^2)$

Note that $E((X-\mu)^2 + \sigma^2) = 2\sigma^2$

And thus use Markov to get the following inequality:

$P(|X - \mu| > \alpha)$ <= $\frac{2\sigma^2}{\sigma^2 + \alpha^2}$

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