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# proof of Markov’s inequality

Define

$Y=\begin{cases}d&X\geq d\\ 0&\text{otherwise}\\ \end{cases}.$ |

Then $0\leq Y\leq X$. 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

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

and the inequality follows.

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