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.