# proof of Markov’s inequality

Define

$$Y=\{\begin{array}{cc}d\hfill & X\ge d\hfill \\ 0\hfill & \text{otherwise}\hfill \end{array}.$$ |

Then $0\le Y\le X$. Additionally, it follows immediately from the definition that $Y$ is a random variable^{} (i.e., that it is measurable). Computing the expected value^{} of $Y$, we have that

$$\mathbb{E}[X]\ge \mathbb{E}[Y]=d\cdot \mathbb{P}\{X\ge d\},$$ |

and the inequality^{} follows.

Title | proof of Markov’s inequality |
---|---|

Canonical name | ProofOfMarkovsInequality |

Date of creation | 2013-03-22 12:47:42 |

Last modified on | 2013-03-22 12:47:42 |

Owner | Andrea Ambrosio (7332) |

Last modified by | Andrea Ambrosio (7332) |

Numerical id | 7 |

Author | Andrea Ambrosio (7332) |

Entry type | Proof |

Classification | msc 60A99 |