Science:Math Exam Resources/Courses/MATH105/April 2015/Question 04 (a)/Solution 1

To compute $E(X)$, we will compute the integral $E(X)=\int _{-\infty }^{\infty }xf(x)\,dx$, where $f(x)=\left\{{\begin{array}{ll}{\frac {1}{4}}+{\frac {1}{2}}|x|&{\mbox{if }}-1\leq x\leq 1\\0&{\mbox{otherwise}}\end{array}}\right.$ is the probability density function of $X$.

Since $f(x)=0$ for all $x\leq -1$ and $x\geq 1,E(X)=\int _{-1}^{1}xf(x)\,dx$.

Since $f(x)={\frac {1}{4}}-{\frac {x}{2}}$ for $-1\leq x\leq 0$ and $f(x)={\frac {1}{4}}+{\frac {x}{2}}$ for $0\leq x\leq 1$,

${\begin{aligned}E(X)&=\int _{-1}^{0}\left({\frac {x}{4}}-{\frac {x^{2}}{2}}\right)\,dx+\int _{0}^{1}\left({\frac {x}{4}}+{\frac {x^{2}}{2}}\right)\,dx\\&=\left.\left({\frac {x^{2}}{8}}-{\frac {x^{3}}{6}}\right)\right|_{-1}^{0}+\left.\left({\frac {x^{2}}{8}}+{\frac {x^{3}}{6}}\right)\right|_{0}^{1}\\&=-\left({\frac {1}{8}}+{\frac {1}{6}}\right)+\left({\frac {1}{8}}+{\frac {1}{6}}\right)\\&=0.\end{aligned}}$

Hence ${\color {blue}E(X)=0}$.

Note that we could have also obtained this result by simply noting that the integrand $xf(x)$ is an odd function, and if $g(x)$ is an odd function, $\int _{-a}^{a}g(x)=0$ for any $a\in \mathbb {R}$.