TITLE: Estimate probabilities from its moments
QUESTION [15 upvotes]: I want to estimate probability $Pr(X \leq a)$, where $X$ is a continuous random variable and $a$ is given, only based on some moments of $X$ (e.g., the first four moments, but without knowing its distribution type).

REPLY [6 votes]: Let the first four moments be $E(X^j) = m_j$, $j=1\ldots 4$.  Suppose $g(x)$ is a polynomial of degree $d$ so that $g(x) \ge I_{x \le a}$, i.e. $g(x) \ge 1$ for $x \le a$ and $g(x) \ge 0$ for all real $x$.  Then for any random variable $X$ such that $E[X^d]$ exists, $P(X \le a) = E[I_{X \le a}] \le E[g(X)]$, and $E[g(X)]$ can be calculated using the first $d$ moments of $X$, i.e. if $g(x) = \sum_{j=0}^d c_j x^j$, $E[g(X)] = c_0 + \sum_{j=1}^d c_j E[X^j]$.  Moreover, suppose $g(x) = I_{x\le a}$ at some points.  Then this upper bound is optimal in the sense that it gives the exact value of $P(X \le a)$ for any probability distribution concentrated on those points.  In the case $d=4$, for any $b_1$ and $b_2$ with $b_1 < a < b_2$, there is a unique polynomial $g(x)$ of degree 4 with $g(b_1) = g(a) = 1$, $g'(b_1) = g(b_2) = g'(b_2) = 0$; this will satisfy $g(x) \ge I_{x \le a}$, and the corresponding estimate is tight for distributions concentrated on $\{b_1, a, b_2\}$.  For example, with $b_1 = -1$, $a=0$ and $b_2 = 1$, $g(x) =  \frac{x^4}{2} + \frac{x^3}{4} - x^2 - \frac{3x}{4} + 1$,
leading to the estimate $P(X \le 0) \le \frac{1}{2} E[X^4] + \frac{1}{3} E[X^3] - E[X^2] - \frac{3}{4} E[X] + 1$.