I have a question about a Markov's inequality, which states following.

Let $X : \Omega \rightarrow \mathbb{R}$ be a non-negative random variable on probability space $(\Omega, \mathscr{A}, P)$ and let $c > 0$. Then: $$\mathrm{P}[X > c] \leq \frac{\mathbb{E}(X)}{c}.$$

I know the proof where I actually use the fact that random variable is non-negative. But I found that this inequality holds even for (some) negative random variables. Can someone give me a proof or counterexample that $\mathbb{E}(X) > 0$ is (or is not) satisfactory condition?

As an example, I use the following random variable on probability space $(\Omega, \mathscr{A}, \lambda)$, where $\lambda$ is Lebesgue measure.

Example of random variable

Then the following is true: $\mathbb{E}(X) = 0.75$, and: $$P[X > c] = \begin{cases}0.75, &c\in ]0, 1], \\ 0, &c > 1.\end{cases}$$ And you can see that the inequality holds true.

up vote 1 down vote accepted

Say that $X$ is chosen uniformly at random to be either $1$ or $-1$. Then $\mathbb{E}[X]=0$; but, $$ P\left(X>\frac{1}{2}\right)=P(X=1)=\frac{1}{2}\nleq\frac{\ 0\ }{(\frac{1}{2})}. $$

  • I see, Nicholas , thank you, sometimes I just ask wrong questions :D But, for me, next questions arise... Obviously, there are random variables that can be negative and the inequality holds true. Is there a condition, where the set of random variables with this condition are all random variables for which Markov's inequality holds true? In other words, exists there extended version of Markov's inequality? Thank you one more time :) – Adam Feb 13 '14 at 13:04
  • @Adam I don't really know how to answer that question, but I'll think about it. However, I should point out that even your example doesn't work: the expectation of the random variable you describe is $.75\cdot1+.25\cdot(-1)=0.5$, not $0.75$. So, for instance, the inequality fails when $c=.7$. – Nick Peterson Feb 13 '14 at 13:18
  • ... I see, I've always been making the same mistake, thank you! :) – Adam Feb 13 '14 at 13:30

A nice concrete counterexample has already been given in the other answer. I'll just add a few notes that I think are related to your question.

The proof I read goes (more or less) like this: $$\int_{\{X>c\}}cdP \leq \int_{\{X>c\}}XdP\leq \int_{\Omega }XdP =\mathbb{E}\left[X\right].$$

The first inequality doesn't intrigue much ($c$ is already positive), but the second one does when it comes to non-negativity.

For given $A\subseteq B \subseteq \Omega$ and a non-negative variable $X$ we have: $$ \int_{A}XdP\leq \int_{B }XdP.$$

But we don't expect this to happen for a general $X$, that is, we don't expect: $$ \int_{A}X^+dP - \int_{A}X^-dP \leq \int_{B }X^+dP - \int_{B }X^-dP,$$ as it is equivalent to: $$ \int_{B }X^-dP - \int_{A}X^-dP \leq \int_{B }X^+dP - \int_{A}X^+dP,$$ and there is no general relationship between the non-negative variables $X^+$ and $X^-$ (other than the fact that they subtract up to $X$).

Finally, note that the "two-tailed" version of Markov inequality is: $$P\left(|X| > c \right) \leq c^{-1}\mathbb{E}\left[|X| \right]. $$ Aslo: $$P\left(X > c \right) \leq P\left(|X| > c \right), $$ so $c^{-1}\mathbb{E}\left[|X| \right]$ "safely" (true even if $X$ is not non-negative) dominates $P\left(X > c \right).$

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