Question

Find the probability density function p.d.f and the cumulative distribution function $F\left( x \right)$ associated with the following p.d.f function for $f\left( x \right)$
$f\left( x \right)=\left\{ \begin{matrix}
   &3\left( 1-2{{x}^{2}} \right),\text{ }0< x< 1 \\
   &0, \text{ Otherwise} \\
\end{matrix} \right.$ Also find $p\left( \dfrac{1}{4}< x< \dfrac{1}{3} \right)$

Answer 1

Hint: To solve this question, we should know the concepts related to continuous random variables in probability. We know that the total sum of the probability is 1. We will check if the given $f\left( x \right)$ is the p.d.f of a continuous random variable. Any p.d.f of a continuous random variable in x should satisfy $\int\limits_{-\infty }^{\infty }{f\left( x \right)dx=1}$. In our question, we need the integral between 0 and 1 to be equal to 1. After checking it, the cumulative distribution function is given by the integral $F\left( y \right)=\int\limits_{\text{Lower Limit}}^{y}{f\left( x \right)dx}$. In our question, the lower limit is 0.

Complete step-by-step solution
We are given a probability density function p.d.f as
$f\left( x \right)=\left\{ \begin{matrix}
   3\left( 1-2{{x}^{2}} \right),\text{ }0< x< 1 \\
   0, \text{ Otherwise} \\
\end{matrix} \right.$
We know that the total probability should be equal to 1. We can apply it to the continuous random variable as
Any p.d.f of a continuous random variable in x should satisfy $\int\limits_{-\infty }^{\infty }{f\left( x \right)dx=1}$.
In our question, the value of the p.d.f is zero everywhere except the region $\left( 0,1 \right)$. So, we can apply the total probability condition between the interval $\left( 0,1 \right)$. By applying, we get
\[\begin{align}
 & \int\limits_{0}^{1}{3\left( 1-2{{x}^{2}} \right)dx}=3\int\limits_{0}^{1}{\left( 1-2{{x}^{2}} \right)dx}=3\left[ x-2\dfrac{{{x}^{3}}}{3} \right]_{0}^{1} \\
&\Rightarrow 3\left[ \left( 1-2\dfrac{{{1}^{3}}}{3} \right)-\left( 0-2\dfrac{{{0}^{3}}}{3} \right) \right]=3\left[ \dfrac{1}{3} \right]=1 \\
\end{align}\]
So, we got $\int\limits_{0}^{1}{f\left( x \right)dx=1}$. SO, the given $f\left( x \right)$ is a p.d.f.
We know that the cumulative distribution function until a value y is given by the integral $F\left( y \right)=\int\limits_{\text{Lower Limit}}^{y}{f\left( x \right)dx}$.
In our problem, the lower limit is 0.
So, we get the cumulative distribution as
$\begin{align}
  & F\left( y \right)=\int\limits_{0}^{y}{3\left( 1-2{{x}^{2}} \right)dx}=3\int\limits_{0}^{y}{\left( 1-2{{x}^{2}} \right)dx}=3\left[ x-2\dfrac{{{x}^{3}}}{3} \right]_{0}^{y} \\
 &\Rightarrow F\left( y \right)=3\left( y-2\dfrac{{{y}^{3}}}{3} \right)=3y-2{{y}^{3}} \\
\end{align}$
SO, the cumulative distribution is $F\left( y \right)=3y-2{{y}^{3}}$. The meaning of the cumulative distribution is that the probability of the value of x being in the range of $\left( 0,y \right)$ is given by$F\left( y \right)=3y-2{{y}^{3}}$. We can write it as
$P\left( 0< x< y \right)=3y-2{{y}^{3}}$
We are asked to find the value of $p\left( \dfrac{1}{4}< x< \dfrac{1}{3} \right)$. We can find this value by subtracting the value of cumulative distribution until $x=\dfrac{1}{4}$ from the value of cumulative distribution until $x=\dfrac{1}{3}$. We can write this as
$p\left( \dfrac{1}{4}< x< \dfrac{1}{3} \right)=p\left( 0< x< \dfrac{1}{3} \right)-p\left( 0< x< \dfrac{1}{4} \right)$
From the definition of the cumulative distribution that we have got, we can write the above values by substituting $y=\dfrac{1}{3},\dfrac{1}{4}$
$\begin{align}
  & p\left( \dfrac{1}{4}< x< \dfrac{1}{3} \right)=\left( 3\dfrac{1}{3}-2\dfrac{{{1}^{3}}}{{{3}^{3}}} \right)-\left( 3\dfrac{1}{4}-2\dfrac{{{1}^{3}}}{{{4}^{3}}} \right) \\
&\Rightarrow p\left( \dfrac{1}{4}< x< \dfrac{1}{3} \right)=1-\dfrac{2}{27}-\dfrac{3}{4}+\dfrac{1}{32}=\dfrac{25}{27}-\dfrac{23}{32} \\
\end{align}$
By simplifying, we get
$p\left( \dfrac{1}{4}< x< \dfrac{1}{3} \right)=\dfrac{25}{27}-\dfrac{23}{32}=\dfrac{25\times 32-23\times 27}{27\times 32}=\dfrac{800-621}{864}=\dfrac{179}{864}$
$\therefore $ The required value is $p\left( \dfrac{1}{4}< x< \dfrac{1}{3} \right)=\dfrac{179}{864}$

Note: While checking the p.d.f of a function, if we don’t get the total probability as 1, we should divide by the constant which is equal to the integral value to the given p.d.f to make the total probability as 1.Generally, we will be given a p.d.f which has a total probability as 1. In our question, it is equal to 1. Another way to get the required expression is by integrating the given p.d.f within the limits $\dfrac{1}{4},\dfrac{1}{3}$. That is
$p\left( \dfrac{1}{4}< x< \dfrac{1}{3} \right)=\int\limits_{\dfrac{1}{4}}^{\dfrac{1}{3}}{3\left( 1-2{{x}^{2}} \right)dx}$. By solving, we can get the same answer.