ThmDex – An index of mathematical definitions, results, and conjectures.
Result R4132 on D1719: Expectation
Strict version of probabilistic Markov inequality
Formulation 0
Let $X \in \text{Random}[0, \infty]$ be a D5101: Random unsigned basic number.
Let $\lambda > 0$ be a D993: Real number.
Then \begin{equation} \mathbb{P}(X > \lambda) \leq \frac{1}{\lambda} \mathbb{E} (X) \end{equation}
Proofs
Proof 0
Let $X \in \text{Random}[0, \infty]$ be a D5101: Random unsigned basic number.
Let $\lambda > 0$ be a D993: Real number.
According to result R4145: Binary union is an upper bound to both sets in the union, we have the inclusion \begin{equation} \{ X > \lambda \} \subseteq \{ X > \lambda \} \cup \{ X = \lambda \} = \{ X \geq \lambda \} \end{equation} Now, applying results
(i) R2090: Isotonicity of probability measure
(ii) R2016: Probabilistic Markov's inequality

we conclude with \begin{equation} \mathbb{P}(X > \lambda) \leq \mathbb{P}(X \geq \lambda) \leq \frac{1}{\lambda} \mathbb{E}(X) \end{equation} $\square$