Suppose $X$ is a non negative r.v. with mean $\mu$ and variance $\sigma^2$and $c$ is a constant such that $0<c<\mu$.

For any real number $a(\neq -c)$, $ P(X\geq c)=P(X+a\geq c+a)=P([X+a]^{2}\geq [c+a]^{2})\leq \frac{E(X+a)^{2}}{(c+a)^{2}}=\frac{\sigma ^{2}+(\mu +a)^{2}}{(c+a)^{2}}$ by Markov's Inequality.

$\frac{\sigma ^{2}+(\mu +a)^{2}}{(c+a)^{2}}$ has minimum value $\frac{\sigma ^2}{\sigma ^2+(c-\mu)^2}$ when $a=\frac{\sigma ^{2}}{c-\mu}-\mu$.

Hence, $P(X\geq c)\leq \frac{\sigma ^2}{\sigma ^2+(c-\mu)^2}$.

Now from by Cantelli's inequality, $P(X\leq \mu -a) \leq \frac{\sigma ^2}{\sigma ^2+a^2}$ for $a>0$. Putting $a=\mu -c$ we get,$P(X\leq c)\leq \frac{\sigma ^2}{\sigma ^2+(\mu-c)^2}$.

So is it possible that the maximum value of both $P(X\geq c)$ and $P(X\leq c)$ be same for all values of $c$ or am I missing something here?


Sure, let $X$ follow a Bernoulli distribution: $P(X = 0) = P(X = 1) = \frac{1}{2}$.

  • $\begingroup$ This should hold for all distribution. It seems strange for any other distribution. Is there any way to improve the bounds for either $P(X\geq c)$ or $P(X\leq c)$? $\endgroup$ – Dip Dec 19 '12 at 8:55
  • 1
    $\begingroup$ Since there exists a distribution that attains both bounds, obviously the bounds cannot be improved without putting some restriction on the space of distributions you are interested in. $\endgroup$ – Jonathan Christensen Dec 19 '12 at 14:30

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