Question

Given the probability density function (p.d.f.) of a continuous random variable \[x\] as
\[f\left( x \right)=\left\{ \begin{matrix}
   \dfrac{{{x}^{2}}}{3} & -1< x< 2 \\
   0 & \text{otherwise} \\
\end{matrix} \right\}\]
Determine the cumulative distribution function (c.d.f.) of $X$ and hence find $P\left( x< 1 \right),P\left( x\le -2 \right),P\left( x> 0 \right),P\left( 1< x< 2 \right)$

Answer 1

Hint: We will first write the definition of the probability density function and cumulative density function. Then, for the given PDF, we will apply the formula for CDF that is \[{{F}_{X}}\left( x \right)=\int\limits_{-1}^{x}{{{f}_{X}}\left( t \right)dt}\] and we will get the CDF. After that, we will put the values if required probabilities to get the answer.

Complete step by step answer:
Let’s define what a probability density function is. So, the Probability Density Function (P.D.F.) is the probability function that is represented for the density of a continuous random variable lying between a certain range of values. The probability density function is defined in the form of an integral of the density of the variable density over a given range. It is denoted by \[f\left( x \right)\] . This function is positive or non-negative at any point of the graph and the integral of PDF over the entire space is always equal to one.
Now, we will see the meaning of the cumulative distribution function. The Cumulative Distribution Function (CDF), of a real-valued random variable $X$ , evaluated at $x$ , is the probability function that $X$ will take a value less than or equal to $x$. It is used to describe the probability distribution of random variables in a table.
The CDF defined for a discrete random variable is given as ${{F}_{X}}\left( x \right)=P\left( X\le x \right),\text{ for all }x\in \mathbb{R}$ Where $X$ is the probability that takes a value less than or equal to $x$ and that lies in the semi-closed interval $\left( a,b \right]$ , where $a< b$ . Therefore, the probability within the interval is written as: $P\left( a< X \le b \right)={{F}_{X}}\left( a \right)-{{F}_{X}}\left( b \right)$ .
 The CDF defined for a continuous random variable is given as: ${{F}_{X}}\left( x \right)=\int\limits_{-\infty }^{x}{{{f}_{X}}\left( t \right)dt}$ . Here, $X$ is expressed in terms of integration of its probability density function $f\left( x \right)$ .

Now, we are given the probability density function (p.d.f.) of a continuous random variable \[x\] as \[f\left( x \right)=\left\{ \begin{matrix}
   \dfrac{{{x}^{2}}}{3} & -1< x< 2 \\
   0 & \text{otherwise} \\
\end{matrix} \right\}\]

Now, we will find CDF , as we saw above in the definition of CDF, here the CDF defined for a continuous random variable is given as:
\[{{F}_{X}}\left( x \right)=\int\limits_{-1}^{x}{{{f}_{X}}\left( t \right)dt}\Rightarrow {{F}_{X}}\left( x \right)=\int\limits_{-1}^{x}{\dfrac{{{t}^{2}}}{3}dt\text{ }}\text{for }x\in \left( -1,2 \right)\] ,
Now we will integrate it using power rule: ${{F}_{X}}\left( x \right)=\left( \dfrac{{{t}^{3}}}{9} \right)_{-1}^{x}\Rightarrow \dfrac{{{x}^{3}}}{9}-\left( \dfrac{{{\left( -1 \right)}^{3}}}{9} \right)\Rightarrow \dfrac{{{x}^{3}}+1}{9}\text{ , }x\in \left( -1,2 \right)$ and \[{{F}_{X}}\left( x \right)=0\text{ , otherwise}\]
Therefore, the cumulative distribution function will be: ${{F}_{X}}\left( x \right)=\dfrac{{{x}^{3}}+1}{9}\text{ , }x\in \left( -1,2 \right)\text{ and }{{F}_{X}}\left( x \right)=0\text{ , otherwise}$
Now, consider $P\left( x< 1 \right)={{F}_{X}}\left( 1 \right)=\dfrac{{{\left( 1 \right)}^{3}}+1}{9}=\dfrac{2}{9}$
$P\left( x\le -2 \right)={{F}_{X}}\left( -2 \right)=0$
$\begin{align}
  & P\left( x> 0 \right)=1-P\left( x\le 0 \right)=1-{{F}_{X}}\left( 0 \right) \\
 & \Rightarrow P\left( x> 0 \right)=1-\left( \dfrac{1}{9} \right)=\dfrac{8}{9} \\
 & \Rightarrow P\left( x> 0 \right)=\dfrac{8}{9} \\
\end{align}$
And finally:
$\begin{align}
  & P\left( 1< x< 2 \right)={{F}_{X}}\left( 2 \right)-{{F}_{X}}\left( 1 \right)=1-\left( \dfrac{2}{9} \right) \\
 & \Rightarrow P\left( 1< x< 2 \right)=\dfrac{7}{9} \\
\end{align}$
Hence, the value of
$\begin{align}
  & P\left( x< 1 \right)=\dfrac{2}{9}, \\
 & P\left( x\le -2 \right)=0, \\
 & P\left( x> 0 \right)=\dfrac{8}{9}, \\
 & P\left( 1< x< 2 \right)=\dfrac{7}{9} \\
\end{align}$

Note:
 Student can make the mistake while applying the formula for CDF that is ${{F}_{X}}\left( x \right)=\int\limits_{-\infty }^{x}{{{f}_{X}}\left( t \right)dt}$ we have taken the lower limit value as $-1$ as our function is defined from$-1$. Note that, due to the property of continuous random variable, the density function curve is continuous for all over the given range which defines itself over a range of continuous values or the domain of the variable.