Variance of Continuous Random Variables

Recall that for a discrete random variable $X$ with mean $\mu = E[X]$, the variance is defined as $\text{Var}(X) = E[(X-\mu)^2]$. The same definition applies when $X$ is continuous with probability density function $f(x)$; expectations are simply computed by integration:

$$\text{Var}(X) = E[(X-\mu)^2] = \int_{-\infty}^{\infty} (x-\mu)^2\, f(x)\,dx.$$

Expanding $(X-\mu)^2$ and using linearity of expectation yields the computational formula

$$\text{Var}(X) = E[X^2] - \mu^2,$$

which is almost always easier to evaluate than the definition directly. The two ingredients are the first and second moments

$$\mu = E[X] = \int_{-\infty}^{\infty} x\, f(x)\,dx, \qquad E[X^2] = \int_{-\infty}^{\infty} x^2\, f(x)\,dx.$$

Example. Let $X$ have density $f(x) = \dfrac{1}{3}$ on $[0,3]$ (and $0$ elsewhere). Then

$$\begin{align*} \mu &= \int_0^3 x \cdot \tfrac{1}{3}\,dx = \tfrac{1}{3} \cdot \tfrac{9}{2} = \tfrac{3}{2}, \\[3pt] E[X^2] &= \int_0^3 x^2 \cdot \tfrac{1}{3}\,dx = \tfrac{1}{3} \cdot \tfrac{27}{3} = 3. \end{align*}$$

Therefore

$$\text{Var}(X) = E[X^2] - \mu^2 = 3 - \tfrac{9}{4} = \tfrac{3}{4}.$$