0
$\begingroup$

We have a lot of probabilities lower bounding as (e.g. chernoff bound, reverse markov inequality, Paley–Zygmund inequality) $$ P( X-E(X) > a) \geq c, a > 0 \quad and \quad P(X > (1-\theta)E[X]) \geq c, 0<\theta < 1 $$

However, It would be great to know if there is any inequality bounding exactly
$$ P(X > E[X]) \geq c $$ i.e., the probability that a r.v greater than its exact expected value ? (e.g., Suppose X is bounded and with bounded first and second moments)

$\endgroup$
  • 5
    $\begingroup$ Put $1-2p$ at 0 and $p$ at $\pm 1$. Then $P(X\gt E[X])=p$ which can be as small as you like. The only value it can't have is 1. $\endgroup$ – Brendan McKay Apr 22 at 3:31
  • $\begingroup$ Sure if without any constraint, do you know if with constraint, e.g. bounded X in [a,b], bounded moments in the previous related literature ? $\endgroup$ – exteral Apr 22 at 3:33
  • 2
    $\begingroup$ My examples are bounded and have all moments bounded. $\endgroup$ – Brendan McKay Apr 22 at 3:40
  • 1
    $\begingroup$ your statements are quite unclear "certain constraint"? $\endgroup$ – kodlu Apr 22 at 7:35
  • 3
    $\begingroup$ Just add 1 to $X$ to make it positive. You won't get a useful answer unless you specify the conditions more precisely, as kodlu wrote. $\endgroup$ – Brendan McKay Apr 22 at 8:19
4
$\begingroup$

Let $Y:=X-EX$. We need to obtain a lower bound on $P(Y>0)$.

Suppose that $-a\le Y\le b$ for some real $a>0$ and $b>0$, and that $EY^2\ge s^2$ for some real $s$. Then $$1_{Y>0}\ge\frac{aY+Y^2}{ab+b^2}.$$ Taking expectations of both sides of this inequality, we get $$P(Y>0)\ge\frac{s^2}{ab+b^2}. \tag{1}$$

In terms of $X$, (1) can be rewritten as $$P(X>EX)\ge\frac{Var\,X}{ab+b^2},$$ provided that $-a\le X-EX\le b$.


The condition $-a\le Y\le b$ implies that $$Y^2\le\frac{Y+a}{a+b}\,b^2+\frac{b-Y}{a+b}\,a^2.$$ Taking expectations of both sides of this inequality, we get $$s^2\le EY^2\le\frac{a}{a+b}\,b^2+\frac{b}{a+b}\,a^2=ab.$$ So, letting now $$p:=\frac{s^2}{(a+b)a}\quad\text{and}\quad r:=\frac{s^2}{(a+b)b}, $$ we see that $$p+r=\frac{s^2}{ab}\le1.$$ Letting then $Y$ be a random variable taking values $-a,0,b$ with probabilities $p,1-p-r,r$ respectively, we see that $-a\le Y\le b$, $EY=0$, $EY^2=s^2$, and $$P(Y>0)=\frac{s^2}{ab+b^2}.$$ So, the lower bound on $P(Y>0)$ in (1) is attained.


Without the condition $EY^2\ge s^2$, no nonzero lower bound on $P(Y>0)$ exists even if we still assume that $-a\le Y\le b$ for some real $a\ge0$ and $b\ge0$ -- just let $Y$ be the constant $0$.

Also, obviously, the exact lower bound $\frac{s^2}{ab+b^2}$ on $P(Y>0)$ goes to $0$ if either $a\to\infty$ or $b\to\infty$. It follows that no nonzero lower bound on $P(Y>0)$ exists if we replace $a$ or $b$ by $\infty$.

Thus, none of the conditions imposed on $Y$ can be removed if one wants to have a nonzero lower bound on $P(Y>0)$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Can we obtain non-zero bound if we have higher order moment information about $Y$? For example, bounded fourth moment? Also, how about bounds for $\mathbb{P}(Y_1+...Y_n\geq0)$ for iid copies of $Y$? $\endgroup$ – neverevernever Jun 23 at 15:30
  • $\begingroup$ @neverevernever : I suggest you ask these questions in separate posts. $\endgroup$ – Iosif Pinelis Jun 23 at 20:25
  • $\begingroup$ Thanks for the suggestion, I will. Does it mean the answer is not obvious? $\endgroup$ – neverevernever Jun 23 at 23:26

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.