Chebyshev's inequality
Dependencies:
$\newcommand{\E}{\operatorname{E}}$ $\newcommand{\Var}{\operatorname{Var}}$ Let $X$ be a real-valued random variable and let $a > 0$. Then \[ \Pr(|X - \E(X)| \ge a) \le \frac{\Var(X)}{a^2} \]
Proof
\begin{align} \Pr(|X - \E(X)| \ge a) &= \Pr((X - \E(X))^2 \ge a^2) \\ &\le \frac{\E((X - \E(X))^2)}{a^2} \tag{by Markov's bound} \\ &= \frac{\Var(X)}{a^2} \end{align}
Dependency for: None
Info:
- Depth: 9
- Number of transitive dependencies: 22
Transitive dependencies:
- /analysis/topological-space
- /sets-and-relations/countable-set
- /sets-and-relations/de-morgan-laws
- /measure-theory/linearity-of-lebesgue-integral
- /measure-theory/lebesgue-integral
- σ-algebra
- Generated σ-algebra
- Borel algebra
- Measurable function
- Generators of the real Borel algebra (incomplete)
- Measure
- σ-algebra is closed under countable intersections
- Group
- Ring
- Field
- Vector Space
- Probability
- Random variable
- Expected value of a random variable
- Linearity of expectation
- Variance of a random variable
- Markov's bound