ThmDex – An index of mathematical definitions, results, and conjectures.
Signed expectation finite iff absolutely integrable random basic number
Formulation 0
Let $X \in \text{Random}(\mathbb{R})$ be a D3161: Random real number.
Then \begin{equation} \mathbb{E} |X| < \infty \quad \iff \quad - \infty < \mathbb{E} X < \infty \end{equation}
Formulation 1
Let $X \in \text{Random}(\mathbb{R})$ be a D3161: Random real number.
Then \begin{equation} \mathbb{E} |X| < \infty \quad \iff \quad \mathbb{E} X \in (-\infty, \infty) \end{equation}
Proofs
Proof 0
Let $X \in \text{Random}(\mathbb{R})$ be a D3161: Random real number.
If $\mathbb{E} |X| < \infty$, then by applying R2100: Triangle inequality for signed basic expectation, one has \begin{equation} \mathbb{E} X \leq |\mathbb{E} X| \leq \mathbb{E} |X| < \infty \end{equation} On the other hand, suppose that $\mathbb{E} X \in (-\infty, \infty)$. By the definition of a D5102: Basic expectation, we have $\mathbb{E} X = \mathbb{E} X^+ - \mathbb{E} X^-$ and thus \begin{equation} - \infty < \mathbb{E} X^+ - X^- < \infty \end{equation} Since $\infty - \infty$ (or $- \infty + \infty$) is not an allowed operation, then this implies that $\mathbb{E} X^+, \mathbb{E} X^- \in (-\infty, \infty)$ so that, in particular, $\mathbb{E} X^+ + \mathbb{E} X^- < \infty$. Result R3583: Random basic number absolutisation equals sum of positive and negative parts provides the partition $|X| = X^+ + X^-$ and, taking expectations on both sides, we conclude \begin{equation} \mathbb{E} |X| = \mathbb{E} X^+ + \mathbb{E} X^- < \infty \end{equation} $\square$