Chapter 8.5, Problem 1PT

To determine

Whether the statement “The function whose graph is given at the right is a probability density function” is true or false.

Answer to Problem 1PT

The given statement is $\underset{\_}{\text{false}}$.

Explanation of Solution

Definition used:

Probability density function:

(1) Every continuous random variable X has a probability density function f. That is, $P\left(a\le X\le b\right)={\displaystyle {\int }_a^{b}f\left(x\right)}dx$

(2) The probability density function f of a random variable X satisfies the condition $f\left(x\right)\ge 0$ for all x.

(3) The probabilities are measured on a scale from 0 to 1, it follows that ${\displaystyle {\int }_{-\infty }^{\infty }f\left(x\right)dx}=1$.

Formula used:

The slope of the tangent line is, $m=\frac{y_2-y_1}{x_2-x_1}$.

Calculation:

From the Figure, it is observed that the function $f\left(x\right)$ is discontinuous at $x=0\text{ and }x=3$.

Therefore, the given graph is not continuous and it is not a probability density function.

This can be proved with the help of the formula (3) mentioned above.

Here, $f\left(x\right)=0\text{for }-2\le x\le 0$.

Compute the equation of the line passing through the points $\left(0,0.6\right)$ and $\left(2,0\right)$ in the interval $0<x\le 2$ as follows.

Compute the slope of the tangent line by substituting $x_1=0$, $x_2=2$, $y_1=0.6$ and $y_2=0$ in the formula used above.

$\begin{array}{c}\text{slope}m=\frac{0-\left(0.6\right)}{2-\left(0\right)}\\ =\frac{-0.6}{2}\\ =-0.3\end{array}$

Substitute $m=-0.3$ in the linear function $f\left(x\right)=mx+b$ mentioned above.

Therefore, the linear equation is, $f\left(x\right)=-0.3x+b$.

Compute the value of b by substituting the point $\left(2,0\right)$ in $f\left(x\right)=-0.3x+b$.

$\begin{array}{c}0=-0.3\left(2\right)+b\\ 0=-0.6+b\\ b=0.6\end{array}$

Therefore, the linear function is $f\left(x\right)=-0.3x+0.6$ when $0<x\le 2$.

Compute the equation of the line passing through the points $\left(2,0\right)$ and $\left(3,0.4\right)$ in the interval $2\le x\le 3$.

Compute the slope of the tangent line by substituting $x_1=2$, $x_2=3$, $y_1=0$ and $y_2=0.4$ in the formula used above.

$\begin{array}{c}\text{slope}m=\frac{0.4-0}{3-2}\\ =\frac{0.4}{1}\\ =0.4\end{array}$

Substitute $m=0.4$ in the linear function $f\left(x\right)=mx+b$ mentioned above.

Therefore, the linear equation is, $f\left(x\right)=0.4x+b$.

Compute the value of b by substituting the point $\left(2,0\right)$ in $f\left(x\right)=0.4x+b$.

$\begin{array}{c}0=0.4\left(2\right)+b\\ 0=0.8+b\\ b=-0.8\end{array}$

Therefore, the linear function is $f\left(x\right)=0.4x-0.8$ when $2\le x\le 3$.

Thus the function is $f\left(x\right)=\{\begin{array}{l}0\text{for }-2\le x\le 0\\ -0.3x+0.6\text{for }0\le x\le 2\\ 0.4x-0.8\text{for }2\le x\le 3\\ 0\text{for }3<x\le 5\end{array}$.

Compute ${\displaystyle {\int }_{-\infty }^{\infty }f\left(x\right)dx}$ as follows.

$\begin{array}{c}{\displaystyle {\int }_{-\infty }^{\infty }f\left(x\right)dx}={\displaystyle {\int }_{-2}^{5}f\left(x\right)dx}\\ ={\displaystyle {\int }_{-2}^{0}f\left(x\right)dx}+{\displaystyle {\int }_0^{2}f\left(x\right)dx}+{\displaystyle {\int }_2^{3}f\left(x\right)dx}+{\displaystyle {\int }_3^{5}f\left(x\right)dx}\\ =0+{\displaystyle {\int }_0^2\left(-0.3x+0.6\right)dx}+{\displaystyle {\int }_2^3\left(0.4x-0.8\right)dx}+0\\ =-0.3{\left[\frac{x^2}{2}\right]}_0^2+0.4{\left[\frac{x^2}{2}\right]}_2^3\end{array}$

Further simplified as,

$\begin{array}{c}{\displaystyle {\int }_{-\infty }^{\infty }f\left(x\right)dx}=-0.3\left[2\right]+0.4\left[\frac{9}{2}\right]\\ =-0.6+1.8\\ =1.2\\ \ne 1\end{array}$

Hence, the given graph is not a probability density function by the definition mentioned above.

Therefore, the given statement is $\underset{\_}{\text{false}}$.