Markov inequality

For any random variable X and and a>0, the following inequality holds;

$$P(|X|\geq a) \leq \frac{E\{|X|\}}{a}$$

explanation
For any continuous distribution Y with a density function f_Y(y) with support in the $$\mathbb{R}+$$ set, the the E(Y) is;

$$E(Y)=\int_0^\infty y f_Y(y) d y $$

but as a>0 and the integral is +ve for all 00, the following inequality holds;

$$P(|X|\geq a) \leq \frac{E(X^2)}{a^2}$$

@todo

Probability and expectation
link between probabilities and expectations

the inequalities impose limits on the tail probabilities that a distribution can take. Useful for demonstrating convergence.