Markov's Inequality
Theorem: Markov's Inequality
The inequality states that for any non-negative random variable X and any positive number a, the probability that X is greater than or equal to a is bounded by the expectation of X divided by a:
$$p(X \geq a) \leq \frac{E[X]}{a}$$
Proof of Markov's Inequality (Discrete Case):
$$E[X] = \sum_{x \in X} x \cdot p(x)$$
$$= \sum_{x < a} x \cdot p(x) + \sum_{x \geq a} x \cdot p(x)$$
As X is non-negative, we know that all values in the sum are positive and the following inequality holds:
$$\sum_{x < a} x \cdot p(x) + \sum_{x \geq a} x \cdot p(x) \geq \sum_{x \geq a} x \cdot p(x)$$
As x is always larger or equal to a, it holds:
$$\sum_{x \geq a} x \cdot p(x) \geq \sum_{x \geq a} a \cdot p(x)$$
$$ = a \cdot p(X \geq a)$$
So we got:
$$E[X] \geq a \cdot p(X \geq a)$$
$$ \Leftrightarrow p(X \geq a) \leq \frac{E[X]}{a} $$