proof of Markov’s inequality

Define

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

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

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