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arrow_back Prove Markov’s inequality.

by Platinum
(106,844 points)
in Data Science & Statistics
45 views
Prove

(Markov's inequality). Suppose \(X\) is a nonnegative random variable and \(a \in\) is a positive constant. Then
\[
P(X \geq a) \leq \frac{E X}{a}
\]

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
(106,844 points)

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