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

The cumulative distribution function $F(x)$ of a random variable $X$ is defined as $F(x)=\int _{-\infty }^{x}f(y)\,dy$, where $f(y)$ is the probability density function of $X$. Suppose $0<x<1$. Then

$F(x)=\int _{-\infty }^{x}f(y)\,dy=\int _{-\infty }^{-1}f(y)\,dy+\int _{-1}^{0}f(y)\,dy+\int _{0}^{x}f(y)\,dy$

Since $f(x)=0$ for all $x<-1$,

$\int _{-\infty }^{-1}f(y)\,dy=\int _{-\infty }^{-1}0\,dy=0.$

For $-1\leq x\leq 0,f(x)={\frac {1}{4}}-{\frac {x}{2}}.$ Therefore,

$\int _{-1}^{0}f(y)\,dy=\int _{-1}^{0}\left({\frac {1}{4}}-{\frac {y}{2}}\right)\,dy=\left.\left({\frac {y}{4}}-{\frac {y^{2}}{4}}\right)\right|_{-1}^{0}=-\left({\frac {-1}{4}}-{\frac {1}{4}}\right)={\frac {1}{2}}.$

For $0\leq x\leq 1$, $f(x)={\frac {1}{4}}+{\frac {x}{2}}.$ Therefore,

$\int _{0}^{x}f(y)\,dy=\int _{0}^{x}\left({\frac {1}{4}}+{\frac {y}{2}}\right)\,dy={\frac {x}{4}}+{\frac {x^{2}}{4}}.$

Putting all these results together means that the cumulative distribution function for the random variable $X$ in the range $0<x<1$ is

${\color {Blue}F(x)={\frac {1}{2}}+{\frac {x}{4}}+{\frac {x^{2}}{4}}.}$