6
$\begingroup$

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?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
10
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • $\begingroup$ 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
    Commented Jul 25, 2014 at 4:19
  • 2
    $\begingroup$ 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
    Commented Jul 25, 2014 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
    Commented Jul 26, 2014 at 17:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .