State Markov's inequality.

medium

Answer

$For a non-negative random variable X and a > 0: P(X \geq a) \leq E[X] / a.$
$Even weaker than Chebyshev but requires only non-negativity and a finite mean.$
$(\mathrm{apply}\;\mathrm{Markov}\;\mathrm{to}\;(X - {\mu})^{2})$
$Building block of concentration inequalities.$

Check yourself — multiple choice

Requires normality
P(X ≥ a) ≤ E[X]/a for non-negative X with finite mean — building block of Chebyshev and Chernoff
P = E[X]
Doesn't exist

Markov: P(X ≥ a) ≤ E[X]/a; requires only non-negativity + finite mean.

#probability#theory

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