Chapter 5, Problem 5.1P

Let X be a random variable with probability density function $f\left(x\right)=\{\begin{array}{cc}c\left(1-x^2\right)& -1<x<0\\ 0& \text{otherwise}\end{array}$

a. What is the value of c?

b. What is the cumulative distribution function of X?

To determine

(a) To compute: To find the value of c

Answer to Problem 5.1P

The value of c is 0.75.

Explanation of Solution

Given information:

The probability density function of X

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

Calculation:

As the probability density function integrates to 1

  $\underset{-1}{\overset{1}{{\displaystyle \int }}}c \left(1-x^2\right) dx$ = 1

= ${\left[c \times \left(x-\frac{x^3}{3}\right)\right]}_{-1}^1$ = 1

= $c \times \left(1-\frac{1}{3}-\left(-1-\frac{-1}{3}\right)\right)$ = 1

= $c \times \left(2-\frac{2}{3}\right)=1$

= $c \times \left(\frac{4}{3}\right)=1$

  $c=\frac{3}{4}=0.75$

To determine

To find: Cumulative distribution of X

Answer to Problem 5.1P

The cumulative distribution of X is $0.75 \times \left(x-\frac{x^3}{3}+\frac{2}{3}\right)$ .

Explanation of Solution

Formula used:

F (x) = $\underset{-1}{\overset{x}{{\displaystyle \int }}}f\left(t\right)dt$

Calculation:

The cumulative distribution of X is

  $= \underset{-1}{\overset{x}{{\displaystyle \int }}}0.75 \times \left(1-t^2\right) dt$

  $= {\left[0.75 \times \left(t-\frac{t^3}{3}\right)\right]}_{-1}^x$

  $=0.75 \times \left[x-\frac{x^3}{3}- \left(-1-\frac{-1}{3}\right)\right]$

  $=0.75 \times \left(x-\frac{x^3}{3}+\frac{2}{3}\right)$