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)

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    $\begingroup$ 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 '20 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 '20 at 3:33
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    $\begingroup$ My examples are bounded and have all moments bounded. $\endgroup$ – Brendan McKay Apr 22 '20 at 3:40
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    $\begingroup$ your statements are quite unclear "certain constraint"? $\endgroup$ – kodlu Apr 22 '20 at 7:35
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    $\begingroup$ 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 '20 at 8:19

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 $P(Y>0)$.

  • $\begingroup$ 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 '20 at 15:30
  • $\begingroup$ @neverevernever : I suggest you ask these questions in separate posts. $\endgroup$ – Iosif Pinelis Jun 23 '20 at 20:25
  • $\begingroup$ Thanks for the suggestion, I will. Does it mean the answer is not obvious? $\endgroup$ – neverevernever Jun 23 '20 at 23:26

Not sure how interesting it is, given that computing $\mathbb{E}[|X-\mathbb{E}[X]|]$ may be unwiedly, but Iosif Pinelis' argument can be adapted to give the following statement, which does not require existence of a finite second moment nor a lower bound on the support.

Suppose $Y := X - \mathbb{E}[X]$ satisfies $Y \leq a$ a.s., for some $a>0$. Then $$ \mathbb{P}\{ Y > 0\} \geq \frac{\mathbb{E}[|Y|]}{2a}\,. $$ Note that this is achieved for, e.g., $Y$ Rademacher; and that it improves on the variance-base bound from Iosif Pinelis' answer in some cases. (For instance, $Y$ uniform on $[-1,1]$, where we get $1/2$ instead of $1/6$ as a lower bound.)

The proof is just adapting Iosif's, by writing $$ \mathbf{1}_{Y>0} \geq \frac{Y+|Y|}{2a} $$ and taking expectations.


The Cantelli inequality asserts that

$$ \Pr(X-\mathbb{E}[X]\ge\lambda)\quad\begin{cases} \le \frac{\sigma^2}{\sigma^2 + \lambda^2} & \text{if } \lambda > 0, \\[8pt] \ge 1 - \frac{\sigma^2}{\sigma^2 + \lambda^2} & \text{if }\lambda < 0 \end{cases} $$ for square integrable $X$ with $\sigma^2$ its variance.

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    $\begingroup$ But this tells us nothing about the case $\lambda = 0$ considered in the question, unless I misunderstood something. $\endgroup$ – Mateusz Kwaśnicki May 5 at 15:45
  • $\begingroup$ Indeed I misunderstood the question. $\endgroup$ – coudy May 5 at 15:47

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