Let $X$ be a real-valued random variable with $X \geq 0$ and $\mathbb E X >0$. I would like to bound $\mathbb P(X >0)$ from below using information about the first few moments of the variable.

From Cauchy-Schwarz inequality we know $\mathbb P(X>0) \geq \frac{(\mathbb E X)^2}{\mathbb E X^2 }$. Unfortunately, this is not a good enough bound (for instance, if you apply it to $X$ being the absolute value of a standard Gaussian you get $\mathbb P(X>0) \geq \frac{2}{\pi}=0.636..$).

Is there a refined lower bound?


Let $f(x) = \sum_{j=0}^d a_j x^j$ be any polynomial of degree $d$ such that $f(0) = 0$ and $f(x) \le 1$ for all $x \ge 0$. Then $$P(X > 0) \ge E[f(X)] = \sum_{j=0}^d a_j E[X^j]$$ Your Cauchy-Schwarz bound is the case $f(x) = 1 - (x - c)^2/c^2 = 2 x/c - x^2/c^2$ where $c = E[X^2]/E[X]$.

If you want a bound that scales properly (so the bound for $X$ is the same as the bound for $kX$ for any $k > 0$), you can take $a_j = b_j/E[X]^j$.

These bounds are best possible in the sense that they are exact for any $X$ whose distribution is concentrated on $0$ and the points where $f(x) = 1$.

If $X$ is a random variable supported on $[0, L]$, you can take a polynomial $f$ such that $f(0) = 0$, $0 \le f(x) \le 1$ for $0 \le x \le 1$, and $f(x) \ge 1-\epsilon$ for $\delta \le x \le L$, and then $E[f(X)] \ge (1-\epsilon) P(X \ge \delta)$. So you can tailor the estimate to be as close to $P(X > 0)$ as you want for this particular distribution.

  • $\begingroup$ Thank you very much. Using this method I was able to check that the Cauchy-Schwarz bound is the best you can get using degree two polynomials. I will try to see if it is possible to improve the $2/\pi$ bound for the half-normal distribution using polynomials of degree three or four. $\endgroup$ – user16436 Jul 25 '14 at 4:19
  • 1
    $\begingroup$ For the half-normal distribution, with the degree-4 polynomial $f(x) = 1 - (x - p_1)^2 (x - p_2)^2/(p1^2 p_2^2)$ with $p_1 = .773065746933609$, $p2 = 2.14702617243363$ I get $P(X>0) \ge 0.801242498847562$. So some improvement, but not huge. $\endgroup$ – Robert Israel Jul 25 '14 at 7:24
  • $\begingroup$ I agree with you. That's the polynomial one should use. Thank you again, your suggestion is very helpful to me. $\endgroup$ – user16436 Jul 26 '14 at 17:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.