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Example For $f(x) = \left\{\begin{array}{ll} 1 - \vert x\vert & (-1 \leq x \leq 1)\\ 0 & (\mbox{else}) \end{array}\right. $,
find the expectation and the variance.

Answer We find the expectation

$\displaystyle E(X)$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty} x f(x) dx$  
  $\displaystyle =$ $\displaystyle \int_{-\infty}^{-1} 0dx + \int_{-1}^{0}x(1+x)dx + \int_{0}^{1}x(1-x)dx + \int_{1}^{\infty} 0dx$  
  $\displaystyle =$ $\displaystyle \left[\frac{x^2}{2} + \frac{x^3}{3}\right]_{-1}^{0} + \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_{0}^{1} = 0$  


$\displaystyle E(X^2)$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty} x^2 f(x) dx$  
  $\displaystyle =$ $\displaystyle \int_{-\infty}^{-1} 0dx + \int_{-1}^{0}x^2(1+x)dx + \int_{0}^{1}x^2(1-x)dx + \int_{1}^{\infty} 0dx$  
  $\displaystyle =$ $\displaystyle \left[\frac{x^3}{3} + \frac{x^4}{4}\right]_{-1}^{0} + \left[\frac{x^3}{3} - \frac{x^4}{4}\right]_{0}^{1}$  
  $\displaystyle =$ $\displaystyle -(-\frac{1}{3} + \frac{1}{4}) + \frac{1}{3} - \frac{1}{4} = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}$  

Thus, the variance is give by

$\displaystyle V(X) = E(X^2) - E(X)^2 = \frac{1}{6} $