One-sided Chebyshev-type inequalities for bounded probability distributions

Created by Olli Niemitalo on 2007-12-14, last modified 2013-05-30

Chebyshev's inequality states that, for any probability distribution, at most 1/k^2 of the area of the probability density function lies more than k standard deviations away from the mean. We can do better, if we know that the distribution is bounded and we know the bounds.

Let X be a random variable bounded by 0 \le X \le M, where M > 0. Given the first two moments, E(X) and E(X^2), of its probability distribution, a sharp lower bound to P(X <L), where L > 0, is given by:

P(X < L)\ge \left\{ \begin{array}{ll} 0 & \mbox{if $E(X) > L$ and $E(X^2) < L E(X) + M E(X) - L M$,}\\\\ 1 - \frac{L E(X) + M E(X) - E(X^2)}{L M} & \mbox{if $\big (E(X) > L$ and $E(X^2) \ge L E(X) + M E(X) - L M \big )$}\\ & \mbox{or $\big(E(X) \le L$ and $E(X^2) \ge L E(X)\big)$,}\\\\ \frac{E(X)^2 - 2 L E(X) + L^2}{E(X^2) - 2 L E(X) + L^2} & \mbox{if $E(X) \le L$ and $E(X^2) < L E(X)$.} \end{array} \right.

That's it. On a related note...

Let X be a random variable and L a constant both bounded by 0 \le X \le M and 0 \le L \le M, where M > 0.

Let Y be a random variable otherwise equal to X, but collecting the tail of X exceeding L to L:

Y = \left\{\begin{array}{ll} X & \mbox{if $X \le L$,} \\ L & \mbox{if $X > L$.}\\ \end{array}\right.

E(Y) and E(Y^2) cannot be determined knowing only E(X), E(X^2), M and L. However, sharp upper bounds to E(Y) and E(Y^2) are given by:

 \begin{array}{c} E(Y) \le \left\{ \begin{array}{ll} L & \mbox{if $E(X) > L$ and $E(X^2) < L E(X) + M E(X) - L M$,} \\\\ \frac{LE(X)+ME(X)-E(X^2)}{M} & \mbox{if $\big(E(X) > L$ and $E(X^2) \ge L E(X) + M E(X) - L M\big)$ or} \\ & \mbox{$\big(E(X) \le L$ and $E(X^2) \ge L E(X)\big)$,} \\\\ E(X) & \mbox{if $E(X) \le L$ and $E(X^2) < L E(X)$,} \end{array}\right.\\\\ E(Y^2) \le \left\{ \begin{array}{ll} L^2 & \mbox{if $E(X) > L$ and $E(X^2) < L E(X) + M E(X) - L M$,} \\\\ \frac{L^2E(X)+LME(X)-LE(X^2)}{M} & \mbox{if $\big(E(X) > L$ and $E(X^2) \ge L E(X) + M E(X) - L M\big)$ or} \\ & \mbox{$\big(E(X) \le L$ and $E(X^2) \ge L E(X)\big)$,} \\\\ E(X^2) & \mbox{if $E(X) \le L$ and $E(X^2) < L E(X)$.} \end{array}\right. \end{array}

Beats me how to prove these formulas, but I tried with tens of thousands of randomly generated distributions and they always worked. The way I got the formulas was by trying to think of the "worst-case" distributions (that's why these are sharp bounds). There were a few of these, corresponding to the different conditions. So I think I know the worst cases, but I don't know how to show that these truly are the worst cases.

Henry Bottomley has written more about Chebyshev type inequalities.

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