ThmDex – An index of mathematical definitions, results, and conjectures.
Real variance partition into first and second moments

Let $X \in \text{Random}(\mathbb{R})$ be a D3161: Random real number such that
 (i) $$\mathbb{E} |X|^2 < \infty$$
Then $$\text{Var} X = \mathbb{E}(X^2) - (\mathbb{E} X)^2$$

Let $X \in \text{Random}(\mathbb{R})$ be a D3161: Random real number such that
 (i) $$\mathbb{E} |X|^2 < \infty$$
Then $$\mathbb{E}[(X - \mathbb{E} X)^2] = \mathbb{E}(X^2) - (\mathbb{E} X)^2$$

Let $X \in \text{Random}(\mathbb{R})$ be a D3161: Random real number such that
 (i) $$\mathbb{E} |X|^2 < \infty$$
Then $$\text{Var} X = \mathbb{E}(X X) - \mathbb{E} X \mathbb{E} X$$

Let $P = (\Omega, \mathcal{F}, \mathbb{P})$ be a D1159: Probability space such that
 (i) $X : \Omega \to \mathbb{R}$ is a D3161: Random real number on $P$ (ii) $$\mathbb{E} |X|^2 < \infty$$
Then $$\text{Var}(X) = \mathbb{E} (X^2) - (\mathbb{E} X)^2$$

Let $P = (\Omega, \mathcal{F}, \mathbb{P})$ be a D1159: Probability space such that
 (i) $X : \Omega \to \mathbb{R}$ is a D3161: Random real number on $P$ (ii) $$\mathbb{E} |X|^2 < \infty$$
Then $$\mathbb{E}[(X - \mathbb{E} X)^2] = \mathbb{E} (X X) - \mathbb{E} X \mathbb{E} X$$
Proofs
Proof 0
Let $X \in \text{Random}(\mathbb{R})$ be a D3161: Random real number such that
 (i) $$\mathbb{E} |X|^2 < \infty$$
Since $\mathbb{E} |X|^2 < \infty$, result R3581: Absolute moment inherits finiteness from greater exponents for random complex number guarantees that also $\mathbb{E} |X| < \infty$. Additionally, result R4514: Even integer absolute moments coincide with moments for random basic real number shows that $\mathbb{E} |X|^2 = \mathbb{E} X^2$, whence $\mathbb{E} X^2 < \infty$ as well. That is, all the relevant quantities exist and are finite.

Then, by direct computation, we have $$\begin{split} \mathsf{Var}(X) & = \mathbb{E}[(X - \mathbb{E} X)^2] \\ & = \mathbb{E}(X^2) - \mathbb{E}(2 X \mathbb{E} X) + (\mathbb{E} X)^2 \\ & = \mathbb{E}(X^2) - 2 \mathbb{E} (X) \mathbb{E}(X) + (\mathbb{E} X)^2 \\ & = \mathbb{E}(X^2) - 2 \mathbb{E} (X)^2 + (\mathbb{E} X)^2 \\ & = \mathbb{E}(X^2) - (\mathbb{E} X)^2 \end{split}$$ $\square$