Question

Let be a random variable with probability density function

$f\left(x\right)=\left\{\begin{array}{ll}c\left(1-x^2\right)& -1<x<1\\ 0& otherwise\end{array}\right.$

(a) What is the value of?

(b) What is the cumulative distribution function of?

Short answer

(a) The value of c is$\frac{3}{4}$

(b) The cumulative distribution function of X is

$F_x\left(x\right)=\left\{\begin{array}{ll}0& x<-1\\ \frac{3}{4}x-\frac{1}{4}{x}^3+\frac{1}{2}& -1\le x<1\\ 1& x\ge 1\end{array}\right.$

Step-by-step solution

Step 1

Given: Density function of a random variable X as f(x)=$\left\{\begin{array}{l}c(1-x^2)-1<x<1\\ 0otherwise\end{array}\right.$

Step 2

We know that any random variable integrates to 1. Therefore, we have

$$\begin{gathered} {\int }_{-\infty }^{\infty }f\left(x\right)dx=1 \\ {\int }_{\{-\infty \}}^{-1}0dx+{\int }_{\{-1\}}^{1}c(1-x^2)dx+{\int }_1^{\infty }0dx=1 \\ {\int }_{\{-1\}}^{1}c(1-x^2)dx=1 \\ c{{\left[x-\frac{x^3}{3}\right]}^1}_{\left\{-1\right\}}=1 \\ c\left[1-\frac{1}{3}-(-1)+\frac{-1}{3}\right]=1 \\ \frac{4}{3}c=1 \\ c=\frac{3}{4} \end{gathered}$$

Step 3

Thus,the density function is$f\left(x\right)=\left\{\begin{array}{ll}\frac{3}{4}(1-x^2)& -1<x<1\\ 0& otherwise\end{array}\right.$

Step 4

Cummulative distribution function of X is given as $F_X\left(x\right)$as

$$\begin{gathered} F_x\left(x\right)=P\left[X\le x\right]{\int }_{\left\{-\infty \right\}}^{\mathrm{\pi }}0dt+{\int }_{\left\{-1\right\}}^{\mathrm{\pi }}\frac{3}{4}(1-t^2)dt \\ =\frac{3}{4}{{\left[t-\frac{t^3}{3}\right]}^x}_{\left\{-1\right\}} \\ =\frac{3}{4}\left[x-\frac{x^3}{3}-\left(-1\right)+\frac{-1}{3}\right] \\ =\frac{3}{4}\left[x-\frac{x^3}{3}+\frac{2}{3}\right] \\ =\frac{3}{4}x-\frac{1}{4}{x}^3+\frac{1}{2} \end{gathered}$$

Step 5

Thus, CDF of X is

$F_X\left(x\right)=\left\{\begin{array}{cc}0& x\le -1\\ \frac{3}{4}x-\frac{1}{4}{x}^3+\frac{1}{2}& -1\le x<1\\ 1& x\ge 1\end{array}\begin{array}{}\end{array}\right.$