# Probability of a random variable greater than its expected value

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)

• 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. – Brendan McKay Apr 22 '20 at 3:31
• 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 ? – exteral Apr 22 '20 at 3:33
• My examples are bounded and have all moments bounded. – Brendan McKay Apr 22 '20 at 3:40
• your statements are quite unclear "certain constraint"? – kodlu Apr 22 '20 at 7:35
• 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. – Brendan McKay Apr 22 '20 at 8:19

## 3 Answers

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 $$P(Y>0)$$.

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

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