[proofplan] Part (i) follows from non-negativity of expectation applied to $(X - \mathbb{E}[X])^2$, with the equality case handled by the fact that a non-negative random variable with zero expectation is almost surely zero. Part (ii) is a direct computation using linearity. Part (iii) uses the expansion of $(X - \mathbb{E}[X])^2$ and linearity. [/proofplan] [step:Prove $\operatorname{Var}(X) \ge 0$ with equality iff $X$ is almost surely constant] By definition, $\operatorname{Var}(X) = \mathbb{E}[(X - \mathbb{E}[X])^2]$. The random variable $(X - \mathbb{E}[X])^2 \ge 0$, so by non-negativity of expectation, $\operatorname{Var}(X) \ge 0$. For the equality case: $\operatorname{Var}(X) = 0$ means $\mathbb{E}[(X - \mu)^2] = 0$ where $\mu = \mathbb{E}[X]$. Since $(X - \mu)^2 \ge 0$, we have a non-negative random variable with zero expectation, so $\mathbb{P}((X - \mu)^2 = 0) = 1$. Since $(X - \mu)^2 = 0$ if and only if $X = \mu$, this gives $\mathbb{P}(X = \mu) = 1$. Conversely, if $\mathbb{P}(X = \mu) = 1$, then $(X - \mu)^2 = 0$ almost surely, so $\operatorname{Var}(X) = \mathbb{E}[0] = 0$. [/step] [step:Prove $\operatorname{Var}(a + bX) = b^2 \operatorname{Var}(X)$] Let $Y = a + bX$. By linearity, $\mathbb{E}[Y] = a + b\,\mathbb{E}[X]$. Then \begin{align*} Y - \mathbb{E}[Y] = (a + bX) - (a + b\,\mathbb{E}[X]) = b(X - \mathbb{E}[X]). \end{align*} Squaring and taking expectations: \begin{align*} \operatorname{Var}(a + bX) = \mathbb{E}[(Y - \mathbb{E}[Y])^2] = \mathbb{E}[b^2(X - \mathbb{E}[X])^2] = b^2\,\mathbb{E}[(X - \mathbb{E}[X])^2] = b^2 \operatorname{Var}(X), \end{align*} where we used linearity to extract the constant $b^2$ from the expectation. [/step] [step:Prove the computational formula $\operatorname{Var}(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2$] Let $\mu = \mathbb{E}[X]$. Expanding the square in the definition: \begin{align*} \operatorname{Var}(X) &= \mathbb{E}[(X - \mu)^2] \\ &= \mathbb{E}[X^2 - 2\mu X + \mu^2] \\ &= \mathbb{E}[X^2] - 2\mu\,\mathbb{E}[X] + \mu^2 \\ &= \mathbb{E}[X^2] - 2\mu^2 + \mu^2 \\ &= \mathbb{E}[X^2] - \mu^2, \end{align*} where the third line uses [linearity of expectation](/theorems/1117), and the fourth line uses $\mathbb{E}[X] = \mu$. [/step]