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

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MATH105 April 2015

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Question 04 (a)
A continuous random variable $X$ is given by the following probability density function $f(x)=\left\{{\begin{array}{ll}{\frac {1}{4}}+{\frac {1}{2}}\|x\|&{\mbox{if }}-1\leq x\leq 1\\0&{\mbox{otherwise.}}\end{array}}\right.$ Find the expected value $E(X)$ of the random variable $X$ .

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Hint
Recall the definition of expected value: if $f(x)$ is the probability density function of a random variable $X$ , its expected value is defined as the integral $E(X)=\int _{-\infty }^{\infty }xf(x)\,dx$ .

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Solution
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 $a\in \mathbb {R}$ .