Markov's Inequality, only non-negative random variables

Question

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.

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.

Answer 1

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

Answer 2

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 $P\left(X > c \right).$