Computing the Expectation of the Square of a Random Variable: $ \text{E}[X^{2}] $. What is the rule for computing $ \text{E}[X^{2}] $, where $ \text{E} $ is the expectation operator and $ X $ is a random variable?
Let $ S $ be a sample space, and let $ p(x) $ denote the probability mass function of $ X $.
Is
$$
\text{E}[X^{2}] = \sum_{x \in S} x^{2} \cdot p(x),
$$
or do I also need to square the $ x $ appearing in $ p(x) $?
 A: In general, if $ (\Omega,\Sigma,P) $ is a probability space and $ X: (\Omega,\Sigma) \to (\mathbb{R},\mathcal{B}(\mathbb{R})) $ is a real-valued random variable, then
$$
\text{E}[X^{2}] = \int_{\Omega} X^{2} ~ d{P}.
$$
Although this formula works for all cases, it is rarely used, especially when $ X $ is known to have certain nice properties.
Examples:


*

*If $ X $ is a discrete random variable (i.e., its cumulative distribution function (cdf) is a step-function) and $ p $ is its probability mass function (pmf), then we can use the formula
$$
\text{E}[X^{2}] = \sum_{x \in \text{Range}(X)} x^{2} \cdot p(x).
$$

*If $ X $ is an absolutely continuous random variable (i.e., its cdf is an absolutely continuous function), then it possesses a probability density function (pdf) $ f $. We thus have the formula
$$
\text{E}[X^{2}] = \int_{\mathbb{R}} x^{2} f(x) ~ d{\mu(x)},
$$
where $ \mu $ is the standard Borel measure on $ \mathbb{R} $. Of course, if $ f $ is continuous, then we can simply compute the improper Riemann integral
$$
\text{E}[X^{2}] = \int_{- \infty}^{\infty} x^{2} f(x) ~ d{x}.
$$
