Question

Let $X$be a random variable with probability density function

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

(a) What is the value of $c$?

(b) What is the cumulative distribution function of $X$?

Short answer

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

(b) The cumulative distribution function of $X$is

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

Step-by-step solution

Part (a) Step 1. Given Information.

Here, the density function of a random variable X is given as

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

Part (a)  Step 2. Integration of probability density function.

We know that the probability density function of any random variable integrates to $1$. Therefore, we have

${\int }_{-\infty }^{\infty }f\left(x\right)dx=1$

Part (a)  Step 3. Solve the integration of probability density function. 

${\int }_{\left\{-\infty \right\}}^{1}0dx+{\int }_{\left\{-1\right\}}^{1}c(1-x^2)dx+{\int }_1^{\infty }0dx=1$

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

role="math" localid="1646480390300" $c{\left[x-\frac{x^3}{3}\right]}_{\left\{-1\right\}}^1=1$

$$\begin{gathered} c\left[1-\frac{1}{3}-\left(-1\right)+\frac{-1}{3}\right]=1 \\ c\left[\frac{3-1+3-1}{3}\right]=1 \\ \frac{4}{3}c=1 \end{gathered}$$

Part (a)  Step 4. Determine the value of c.

$$\begin{gathered} c=1\times \frac{3}{4} \\ \therefore c=\frac{3}{4} \end{gathered}$$

Part (b)  Step 1. Given Information.

Here, the density function of a random variable $X$is given as

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

Part (b)  Step 2.  Define cumulative distribution function of X. 

Cumulative distribution function of $X$is given as

$F_X\left(x\right)=P\left|X\le x\right|$

Part (b)  Step 3.  Solve cumulative distribution function of X. 

$f_X\left(x\right)={\int }_{\left\{-\infty \right\}}^{\mathrm{\pi }}0dt+{\int }_{\left\{-1\right\}}^{\mathrm{\pi }}\frac{3}{4}(1-t^2)dt$

as $c=\frac{3}{4}$

$$\begin{gathered} =\frac{3}{4}{\left[t-\frac{t^3}{3}\right]}_{\left\{-1\right\}}^x \\ =\frac{3}{4}\left[x-\frac{x^3}{3}-\left(-1\right)+\frac{-1}{3}\right] \\ =\frac{3}{4}\left[x-\frac{x^3}{3}+\frac{3-1}{3}\right] \\ =\frac{3}{4}\left[x-\frac{x^3}{3}+\frac{2}{3}\right] \\ =\frac{3}{4}x-\frac{x^3}{4}+\frac{1}{2} \end{gathered}$$

Part (b)  Step 4. Write the cumulative distribution function of X.

Therefore, the cumulative distribution function of $X$is

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