Answer to Question #169785 in Statistics and Probability for Amedeka Israel Yao

Question

The continuous random variable x has the probability density function,

𝑓(𝑥) =

{

1/ 2(𝑥 - 2), 2 ≤ 𝑥 ≤ 3

𝑎, 3 < 𝑥 ≤ 5

2 - 𝑏𝑥, 5 < 𝑥 ≤ 6

0, 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒

i. Find the values of a and b, and sketch the graph of f(x).

ii. Find the cumulative distribution function, f(x) and sketch it.

Answer 1

Find the distribution function $F(x) = \int\limits_{ - \infty }^x {f(x)dx}$.

If $x < 2$ then $F(x) = \int\limits_{ - \infty }^x {0dx} = 0$

If $x \le 3$ then $F(x) = \int\limits_{ - \infty }^2 {0dx} + \int\limits_2^x {\frac{1}{2}(x - 2)dx} = \left. {\frac{{{x^2}}}{4}} \right|_2^x - \left. x \right|_2^x = \frac{{{x^2} - 4}}{4} - x + 2 = \frac{{{x^2}}}{4} - x + 1$

If $x \le 5$ then $F(x) = \int\limits_{ - \infty }^2 {0dx} + \int\limits_2^3 {\frac{1}{2}(x - 2)dx} + \int\limits_3^x {adx} = \left. {\frac{{{x^2}}}{4}} \right|_2^3 - \left. x \right|_2^3 + a\left. x \right|_3^x = \frac{{9 - 4}}{4} - 3 + 2 + a(x - 3) = \frac{5}{4} - 1 + ax - 3a = ax - 3a + \frac{1}{4}$

If $x \le 6$ then

$F(x) = \int\limits_{ - \infty }^2 {0dx} + \int\limits_2^3 {\frac{1}{2}(x - 2)dx} + \int\limits_3^5 {adx} + \int\limits_5^x {\left( {2 - bx} \right)} = \left. {\frac{{{x^2}}}{4}} \right|_2^3 - \left. x \right|_2^3 + a\left. x \right|_3^5 + 2\left. x \right|_5^x - \left. {\frac{{b{x^2}}}{2}} \right|_5^x = \frac{{9 - 4}}{4} - 3 + 2 + 2a + 2(x - 5) - \frac{b}{2}({x^2} - 25) = \frac{5}{4} - 1 + 2a + 2x - 10 - \frac{{b{x^2}}}{2} + \frac{{25b}}{2} = - \frac{{b{x^2}}}{2} + 2x + 2a + \frac{{25b}}{2} - \frac{{39}}{4}$

If $x > 6$ then $F(x) = \int\limits_{ - \infty }^2 {0dx} + \int\limits_2^3 {\frac{1}{2}(x - 2)dx} + \int\limits_3^5 {adx} + \int\limits_5^6 {\left( {2 - bx} \right)} = \left. {\frac{{{x^2}}}{4}} \right|_2^3 - \left. x \right|_2^3 + a\left. x \right|_3^5 + 2\left. x \right|_5^6 - \left. {\frac{{b{x^2}}}{2}} \right|_5^6 = \frac{{9 - 4}}{4} - 3 + 2 + 2a + 2(6 - 5) - \frac{b}{2}(36 - 25) = \frac{5}{4} - 1 + 2a + 2 - \frac{{11b}}{2} = \frac{9}{4} + 2a - \frac{{11b}}{2}$

Then $F(x) = \left\{ {\begin{matrix} {0,\,\,x < 2}\\ {\frac{{{x^2}}}{4} - x + 1,\,\,2 \le x \le 3}\\ {ax - 3a + \frac{1}{4},\,\,3 < x \le 5}\\ { - \frac{{b{x^2}}}{2} + 2x + 2a + \frac{{25b}}{2} - \frac{{39}}{4},\,\,5 < x \le 6}\\ {\frac{9}{4} + 2a - \frac{{11b}}{2},\,\,x > 6} \end{matrix}} \right.$

By the properties of the distribution function

$\mathop {\lim }\limits_{x \to 3 - 0} F(x) = \mathop {\lim }\limits_{x \to 3 + 0} F(x) \Rightarrow \frac{{{3^2}}}{4} - 3 + 1 = 3a - 3a + \frac{1}{4}$

$\mathop {\lim }\limits_{x \to 5 - 0} F(x) = \mathop {\lim }\limits_{x \to 5 + 0} F(x) \Rightarrow 5a - 3a + \frac{1}{4} = - \frac{{25b}}{2} + 10 + 2a + \frac{{25b}}{2} - \frac{{39}}{4}$

$\mathop {\lim }\limits_{x \to 6 - 0} F(x) = \mathop {\lim }\limits_{x \to 6 + 0} F(x) \Rightarrow - \frac{{36b}}{2} + 12 + 2a + \frac{{25b}}{2} - \frac{{39}}{4} = \frac{9}{4} + 2a - \frac{{11b}}{2}$

$F(\infty ) = 1 \Rightarrow \frac{9}{4} + 2a - \frac{{11b}}{2} = 1$

We have the system of equations:

$\left\{ \begin{array}{l} \frac{{{3^2}}}{4} - 3 + 1 = 3a - 3a + \frac{1}{4}\\ 5a - 3a + \frac{1}{4} = - \frac{{25b}}{2} + 10 + 2a + \frac{{25b}}{2} - \frac{{39}}{4}\\ - \frac{{36b}}{2} + 12 + 2a + \frac{{25b}}{2} - \frac{{39}}{4} = \frac{9}{4} + 2a - \frac{{11b}}{2}\\ \frac{9}{4} + 2a - \frac{{11b}}{2} = 1 \end{array} \right.$

$\left\{ \begin{array}{l} \frac{1}{4} = \frac{1}{4}\\ 2a + \frac{1}{4} = 2a + \frac{1}{4}\\ - \frac{{11b}}{2} + \frac{9}{4} + 2a = \frac{9}{4} + 2a - \frac{{11b}}{2}\\ 2a - \frac{{11b}}{2} = - \frac{5}{4} \end{array} \right.$

$2a - \frac{{11b}}{2} = - \frac{5}{4} \Rightarrow 8a - 22b = - 5 \Rightarrow a = \frac{{ - 5 + 22b}}{8}$

This equation has an infinite number of solutions. Let's find any particular solution.

Let $b = \frac{5}{{22}} \Rightarrow a = 0$

Then

$f(x) = \left\{ {\begin{matrix} {\frac{1}{2}(x - 2),\,\,2 \le x \le 3}\\ {0,\,\,3 < x \le 5}\\ {2 - \frac{5}{{22}}x,\,\,5 < x \le 6}\\ {0,\,\,otherwise} \end{matrix}} \right.$

$F(x) = \left\{ {\begin{matrix} {0,\,\,x < 2}\\ {\frac{{{x^2}}}{4} - x + 1,\,\,2 \le x \le 3}\\ {0 - 0 + \frac{1}{4},\,\,3 < x \le 5}\\ { - \frac{{5{x^2}}}{{44}} + 2x + \frac{{125}}{{44}} - \frac{{39}}{4},\,\,5 < x \le 6}\\ {\frac{9}{4} - \frac{{55}}{{44}},\,\,x > 6} \end{matrix}} \right.= \left\{ {\begin{matrix} {0,\,\,x < 2}\\ {\frac{{{x^2}}}{4} - x + 1,\,\,2 \le x \le 3}\\ {0 - 0 + \frac{1}{4},\,\,3 < x \le 5}\\ { - \frac{{5{x^2}}}{{44}} + 2x - \frac{{76}}{{11}},\,\,\,5 < x \le 6}\\ {1,\,\,x > 6} \end{matrix}} \right.$

sketch the graph of f(x):

sketch the graph of F(x):