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

MATH105 April 2015

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Question 04 (b)
A continuous random variable $X$ is given by the following probability density function $f(x)=\left\{{\begin{array}{ll}{\frac {1}{4}}+{\frac {1}{2}}|x|&{\mbox{if }}-1\leq x\leq 1\\0&{\mbox{otherwise.}}\end{array}}\right.$ Let $F(x)$ be the cumulative distribution function for the random variable $X.$ Find $F(x)$ for $0<x<1.$

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Hint
Recall that if $f(y)$ is the probability density function of a random variable $X$, its cumulative distribution function is defined as $F(x)=\int _{-\infty }^{x}f(y)\,dy$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 ${\color {Blue}F(x)={\frac {1}{2}}+{\frac {x}{4}}+{\frac {x^{2}}{4}}.}$