Question

The following is a p.d.f. (Probability density function) of a continuous random variable $X$:
$f\left( x \right) = \dfrac{x}{{32}}{\rm{ }}0 < x < 8; = 0$ otherwise
Find the value of the c.d.f. at $x = 0.5$ and $x = 9$.

Answer 1

Here, we will use to find the formula of c.d.f. of a probability density function with the limits 0 and $0.5$ to find the value of the c.d.f. at $x = 0.5$. Then, we will use to find the formula of c.d.f. of a probability density function with the limits 0 and 9. We will then use the property of definite integrals to split the integral into a sum of two integrals (with limits 0 to 8, and 8 to 9 respectively). Finally, we will integrate the expressions to find the value of the c.d.f. at $x = 9$.
Formula Used: We will use the following formulas:
1.The c.d.f. of a p.d.f. $f\left( x \right)$ is given as $F\left( x \right) = \int\limits_{ - \infty }^x {f\left( x \right)} dx$.
2.The integral of a function of the form $af\left( x \right)$ can be written as $\int {af\left( x \right)} dx = a\int {f\left( x \right)} dx$.
3.The integral of a function of the form ${x^n}$ can be written as $\int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$.
4.The definite integral $\int\limits_a^b {f\left( x \right)} dx$ can be written as the sum of the definite integrals $\int\limits_a^c {f\left( x \right)} dx$ and $\int\limits_c^b {f\left( x \right)} dx$, where $a < c < b$.

Complete step-by-step answer:
We need to find the value of the cumulative density function at $x = 0.5$.
The value $x = 0.5$ lies between 0 and 8.
Thus, the lower limit becomes 0 and the upper limit becomes $0.5$.
Therefore, substituting $x = 0.5$ and $f\left( x \right) = \dfrac{x}{{32}}$ in the c.d.f. $F\left( x \right) = \int\limits_{ - \infty }^x {f\left( x \right)} dx$, we get
$\Rightarrow F\left( {0.5} \right) = \int\limits_0^{0.5} {\dfrac{x}{{32}}} dx$
Now, we will integrate the expression.
The integral of a function of the form $af\left( x \right)$ can be written as $\int {af\left( x \right)} dx = a\int {f\left( x \right)} dx$.
Therefore, we get
$\Rightarrow F\left( {0.5} \right) = \dfrac{1}{{32}}\int\limits_0^{0.5} x dx$
Rewriting the expression, we get
$\Rightarrow F\left( {0.5} \right) = \dfrac{1}{{32}}\int\limits_0^{0.5} {{x^1}} dx$
The integral of a function of the form ${x^n}$ can be written as $\int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$.
Therefore, we get
$\Rightarrow F\left( {0.5} \right) = \dfrac{1}{{32}}\left. {\left( {\dfrac{{{x^{1 + 1}}}}{{1 + 1}}} \right)} \right|_0^{0.5}$
Simplifying the expression, we get
$\Rightarrow F\left( {0.5} \right) = \dfrac{1}{{32}}\left. {\left( {\dfrac{{{x^2}}}{2}} \right)} \right|_0^{0.5}$
Substituting the limits into the expression, we get
$\Rightarrow F\left( {0.5} \right) = \dfrac{1}{{32}}\left[ {\dfrac{{{{\left( {0.5} \right)}^2}}}{2} - \dfrac{{{0^2}}}{2}} \right]$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow F\left( {0.5} \right) = \dfrac{1}{{32}}\left[ {\dfrac{{0.25}}{2} - \dfrac{0}{2}} \right]\\\ \Rightarrow F\left( {0.5} \right) = \dfrac{1}{{32}}\left[ {\dfrac{{\dfrac{1}{4}}}{2} - 0} \right]\\\ \Rightarrow F\left( {0.5} \right) = \dfrac{1}{{32}}\left[ {\dfrac{1}{8}} \right]\end{array}$
Multiplying the terms of the expression, we get
$\Rightarrow F\left( {0.5} \right) = \dfrac{1}{{256}}$
Therefore, we get the value of the c.d.f. at $x = 0.5$ as $\dfrac{1}{{256}}$.
Now, we need to find the value of the cumulative density function at $x = 9$.
Substituting $x = 9$ in the c.d.f. $F\left( x \right) = \int\limits_{ - \infty }^x {f\left( x \right)} dx$, we get
$\Rightarrow F\left( 9 \right) = \int\limits_0^9 {f\left( x \right)} dx$
We can observe that 9 does not lie between 0 and 8.
We will split the limits using the property of definite integrals.
The definite integral $\int\limits_a^b {f\left( x \right)} dx$ can be written as the sum of the definite integrals $\int\limits_a^c {f\left( x \right)} dx$ and $\int\limits_c^b {f\left( x \right)} dx$, where $a < c < b$.
Therefore, we can rewrite the equation as
$\Rightarrow F\left( 9 \right) = \int\limits_0^8 {f\left( x \right)} dx + \int\limits_8^9 {f\left( x \right)} dx$
The function in the first integral lies between 0 and 8. Therefore, $f\left( x \right) = \dfrac{x}{{32}}$.
The function in the second integral does not lie between 0 and 8. Therefore, $f\left( x \right) = 0$.
Therefore, the equation becomes
$\Rightarrow F\left( 9 \right) = \int\limits_0^8 {\dfrac{x}{{32}}} dx + \int\limits_8^9 {\left( 0 \right)} dx$
Rewriting the equation, we get
$\Rightarrow F\left( 9 \right) = \int\limits_0^8 {\dfrac{x}{{32}}} dx + \int\limits_8^9 {\left( {0x} \right)} dx$
The integral of a function of the form $af\left( x \right)$ can be written as $\int {af\left( x \right)} dx = a\int {f\left( x \right)} dx$.
Therefore, we get
$\begin{array}{l} \Rightarrow F\left( 9 \right) = \dfrac{1}{{32}}\int\limits_0^8 x dx + 0 \times \int\limits_8^9 x dx\\\ \Rightarrow F\left( 9 \right) = \dfrac{1}{{32}}\int\limits_0^8 x dx + 0\\\ \Rightarrow F\left( 9 \right) = \dfrac{1}{{32}}\int\limits_0^8 x dx\end{array}$
Rewriting the expression, we get
$\Rightarrow F\left( 9 \right) = \dfrac{1}{{32}}\int\limits_0^8 {{x^1}} dx$
Therefore, we get
$\Rightarrow F\left( 9 \right) = \dfrac{1}{{32}}\left. {\left( {\dfrac{{{x^{1 + 1}}}}{{1 + 1}}} \right)} \right|_0^8$
Simplifying the expression, we get
$\Rightarrow F\left( 9 \right) = \dfrac{1}{{32}}\left. {\left( {\dfrac{{{x^2}}}{2}} \right)} \right|_0^8$
Substituting the limits into the expression, we get
$\Rightarrow F\left( 9 \right) = \dfrac{1}{{32}}\left[ {\dfrac{{{8^2}}}{2} - \dfrac{{{0^2}}}{2}} \right]$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow F\left( 9 \right) = \dfrac{1}{{32}}\left[ {\dfrac{{64}}{2} - \dfrac{0}{2}} \right]\\\ \Rightarrow F\left( 9 \right) = \dfrac{1}{{32}}\left[ {32 - 0} \right]\\\ \Rightarrow F\left( 9 \right) = \dfrac{1}{{32}}\left[ {32} \right]\end{array}$
Multiplying the terms of the expression, we get
$\Rightarrow F\left( 9 \right) = 1$
Therefore, we get the value of the c.d.f. at $x = 9$ as 1.

Note: Here we found out the cumulative density function. The cumulative density function gives the probability that a continuous random variable $X$ takes a value less than or equal to $x$. A common mistake is to answer that the c.d.f. is 0 at $x = 9$, and $\dfrac{{0.5}}{{32}} = \dfrac{1}{{64}}$ at $x = 0.5$. This is incorrect because these are the values of the p.d.f. at the given values of $x$.