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Prove

(Markov's inequality). Suppose \(X\) is a nonnegative random variable and \(a \in\) is a positive constant. Then
\[
P(X \geq a) \leq \frac{E X}{a}
\]
in Data Science & Statistics by Platinum (132,156 points) | 134 views

1 Answer

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Best answer
By definition of expectation, we have
\[
\begin{aligned}
E X &=\sum_{k \in X(\Omega)} k P(X=k) \\
&=\sum_{k \in X(\Omega) \text { st. } k \geq a} k P(X=k)+\sum_{k \in X(\Omega) \text { s.t. } k<a} k P(X=k) \\
& \geq \sum_{k \in X(\Omega) \text { s.t. } k \geq a} k P(X=k) \\
& \geq \sum_{k \in X(\Omega) \text { s.t. } k \geq a} a P(X=k) \\
&=a \sum_{k \in X(\Omega) \text { s.t. } k \geq a} P(X=k) \\
&=a P(X \geq a)
\end{aligned}
\]

where the first inequality follows from the fact that \(X\) is nonnegative and probabilities are nonnegative, and the second inequality follows from the fact that \(k \geq a\) over the set \(\{k \in X(\Omega)\) s.t. \(k \geq a\}\).
Notation: "s.t." stands for "such that".
Dividing both sides by \(a\), we recover
\[
P(X \geq a) \leq \frac{E X}{a}
\]
by Platinum (132,156 points)

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