Random variables: multiple medians

Dependencies:

1. Median of a random variable

Let $m_1$ and $m_2$ be medians of a random variable $X$ such that $m_1 < m_2$. Then $\Pr(X \le m_1) = 1/2$, $\Pr(m_1 < X < m_2) = 0$, and $\Pr(X \ge m_2) = 1/2$.

Proof

\begin{align} & 1 = \Pr(X \le m_1) + \Pr(m_1 < X < m_2) + \Pr(X \ge m_2) \\ &\implies 0 = (\Pr(X \le m_1) - 1/2) + \Pr(m_1 < X < m_2) + (\Pr(X \ge m_2) - 1/2) \end{align} Since $m_1$ and $m_2$ are medians, all these terms are non-negative and they sum to 0. Hence, each term must be 0.

Dependency for:

1. Minimizer of f(z) = E(|X-z|) is median

Info:

• Depth: 7
• Number of transitive dependencies: 13

Transitive dependencies:

1. /analysis/topological-space
2. /sets-and-relations/countable-set
3. /sets-and-relations/de-morgan-laws
4. σ-algebra
5. Generated σ-algebra
6. Borel algebra
7. Measurable function
8. Generators of the real Borel algebra (incomplete)
9. Measure
10. σ-algebra is closed under countable intersections
11. Probability
12. Random variable
13. Median of a random variable