Moments of Continuous Random Variables

For a discrete random variable, the $k$-th moment is $E[X^k] = \sum\limits_{x} x^k \, p(x).$ For a continuous random variable, we replace the sum by an integral against the probability density function (pdf).

The $k$-th moment of a continuous random variable $X$ with pdf $f$ is

$$\mu_k = E[X^k] = \int_{-\infty}^{\infty} x^k \, f(x) \, dx.$$

The first moment $\mu_1 = E[X]$ is the mean. The second moment $\mu_2 = E[X^2]$ will turn out to be the key ingredient in computing the variance.

To illustrate, let $X$ have pdf $f(x) = \dfrac{1}{4}$ on $[0,4]$ (and zero elsewhere). Then

$$E[X^2] = \int_0^4 x^2 \cdot \dfrac{1}{4} \, dx = \dfrac{1}{4} \left[ \dfrac{x^3}{3} \right]_0^4 = \dfrac{1}{4} \cdot \dfrac{64}{3} = \dfrac{16}{3}.$$