Chapter 4.2, Problem 26E

Let X be the total medical expenses (in 1000s of dollars) incurred by a particular individual during a given year. Although X is a discrete random variable, suppose its distribution is quite well approximated by a continuous distribution with pdf f (x) = k(1 + x /2.5)⁻⁷ for x ≥ 0.

a. What is the value of k?

b. Graph the pdf of X.

c. What are the expected value and standard deviation of total medical expenses?

d. This individual is covered by an insurance plan that entails a $500 deductible provision (so the first $500 worth of expenses are paid by the individual). Then the plan will pay 80% of any additional expenses exceeding $500, and the maximum payment by the individual (including the deductible amount) is $2500. Let Y denote the amount of this individual’s medical expenses paid by the insurance company. What is the expected value of Y?

[Hint: First figure out what value of X corresponds to the maximum out-of-pocket expense of $2500. Then write an expression for Y as a function of X (which involves several different pieces) and calculate the expected value of this function.]

To determine

Find the value of k.

Answer to Problem 26E

Thevalue of k is 2.4.

Explanation of Solution

Given info:

The probability density function (pdf) of total medical expenses (in 1,000s of dollars), X, is:

$f\left(x\right)=\{\begin{array}{l}k{\left(1+\frac{x}{2.5}\right)}^{-7}x\ge 0\\ 0,\text{Otherwise}\end{array}$

Calculation:

Properties of a legitimate pdf:

For a pdf to be legitimate,

• $f\left(x\right)\ge 0$, for all possible values of x.
• ${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f\left(x\right)dx}=1$, when $-\infty <x<\infty$ is the full range of x.

The constant k must be a non-negative quantity. Moreover,

$\begin{array}{c}1={\displaystyle \underset{0}{\overset{\infty }{\int }}f\left(x\right)dx}\\ ={\displaystyle \underset{0}{\overset{\infty }{\int }}k{\left(1+\frac{x}{2.5}\right)}^{-7}dx}\end{array}$

By using substitution rule,

$\begin{array}{c}u=1+\frac{x}{2.5}\\ du=\frac{dx}{2.5}\\ 2.5du=dx\end{array}$

The interval is ,

x	0	$\infty$
u	$1\left(=1+\frac{0}{2.5}\right)$	$\infty \left(=1+\frac{\infty }{2.5}\right)$

On simplifying,

$\begin{array}{c}1={\displaystyle \underset{1}{\overset{\infty }{\int }}2.5k{u}^{-7}du}\\ 1=2.5k{\frac{u^{-6}}{-6}]}_1^{\infty }\\ 1=2.5k\left(\frac{{\infty }^{-6}}{-6}-\frac{1^{-6}}{-6}\right)\\ 1=2.5k\left(\frac{1}{6}\right)\\ 1=\frac{k}{2.4}\\ k=2.4\end{array}$

Hence, thevalue of k is 2.4.

To determine

Sketch thegraph of probability density functionX.

Answer to Problem 26E

The graph of $f\left(x\right)$ is:

Explanation of Solution

Calculation:

The graph of the $f\left(x\right)$ obtained as follows:

For x at 0:

Substitute 0 for x in the equation $f\left(x\right)=k{\left(1+\frac{x}{2.5}\right)}^{-7}$

$\begin{array}{c}f\left(x\right)=2.4{\left(1+\frac{0}{2.5}\right)}^{-7}\\ =2.4\end{array}$

Similarly the remaining points are obtained as follows:

x	$f\left(x\right)$
0	2.4
2	0.04
4	0.003
6	0.00046
8	0.0001
10	0.000031

Table 1

Procedure for drawing the density curve of the variable $f\left(x\right)=k{\left(1+\frac{x}{2.5}\right)}^{-7}$ for $x\ge 0$ is as follows:

• Let horizontal axis take x values and vertical axis take $f\left(x\right)$ values
• Plot the points obtained from Table 1.

The graph of $f\left(x\right)$ is given below:

Figure 1

In the above graph, x takes the values from the interval of $\left[0,\infty \right]$ and $f\left(x\right)$ represents the probability density function.

Thus, the graph for the $f\left(x\right)$ has been obtained.

To determine

Find the expected value and standard deviation of medical expenses.

Answer to Problem 26E

Theexpected value of total medical expenses is 0.5.

The standard deviation of total medical expenses is 0.612.

Explanation of Solution

Calculation:

The mean or expectation of total medical expenses is:

$\begin{array}{c}\mu =E\left(X\right)\\ ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}xf\left(x\right)dx}\\ ={\displaystyle \underset{0}{\overset{\infty }{\int }}2.4x{\left(1+\frac{x}{2.5}\right)}^{-7}dx}\\ =2.4{\displaystyle \underset{0}{\overset{\infty }{\int }}x{\left(1+\frac{x}{2.5}\right)}^{-7}dx}\end{array}$

By using substitution rule,

$\begin{array}{c}u=1+\frac{x}{2.5}\\ x=2.5\left(u-1\right)\\ du=\frac{dx}{2.5}\\ 2.5du=dx\end{array}$

The interval is ,

x	0	$\infty$
u	$1\left(=1+\frac{0}{2.5}\right)$	$\infty \left(=1+\frac{\infty }{2.5}\right)$

On simplifying,

$\begin{array}{c}=2.4{\displaystyle \underset{1}{\overset{\infty }{\int }}2.5\left(u-1\right)u^{-7}2.5du}\\ =15{\displaystyle \underset{1}{\overset{\infty }{\int }}\left(u^{-6}-u^{-7}\right)du}\\ =15{\left(\frac{u^{-5}}{-5}-\frac{u^{-6}}{-6}\right)]}_1^{\infty }\\ =15\left\{\left[\frac{{\infty }^{-5}}{-5}-\frac{{\infty }^{-6}}{-6}\right]-\left[\frac{1^{-5}}{-5}-\frac{1^{-6}}{-6}\right]\right\}\end{array}$

$\begin{array}{c}=15\left(0+\frac{1}{5}-\frac{1}{6}\right)\\ =15\left(0.0333\right)\\ =0.5\end{array}$

Hence, the mean of total expense is 0.5 or $500.

The variance of a continuous random variable is given as ${\sigma }^2=V\left(X\right)=E\left(X^2\right)-{\left(E\left(X\right)\right)}^2$.

Here,

$\begin{array}{c}E\left(X^2\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{x}^{2}f\left(x\right)dx}\\ ={\displaystyle \underset{0}{\overset{\infty }{\int }}2.4x^2{\left(1+\frac{x}{2.5}\right)}^{-7}dx}\\ =2.4{\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^2{\left(1+\frac{x}{2.5}\right)}^{-7}dx}\end{array}$

By using substitution rule,

$\begin{array}{c}u=1+\frac{x}{2.5}\\ x=2.5\left(u-1\right)\\ du=\frac{dx}{2.5}\\ 2.5du=dx\end{array}$

The interval is ,

x	0	$\infty$
u	$1\left(=1+\frac{0}{2.5}\right)$	$\infty \left(=1+\frac{\infty }{2.5}\right)$

On simplifying,

$\begin{array}{c}E\left(X^2\right)=2.4{\displaystyle \underset{1}{\overset{\infty }{\int }}{2.5}^2{\left(u-1\right)}^2{u}^{-7}2.5du}\\ =37.5{\displaystyle \underset{1}{\overset{\infty }{\int }}\left(u^2+1-2u\right)u^{-7}du}\\ =37.5{\displaystyle \underset{1}{\overset{\infty }{\int }}\left(u^{-5}+u^{-7}-2u^{-6}\right)du}\\ =37.5{\left(-\frac{u^{-4}}{4}-\frac{u^{-6}}{6}+\frac{2u^{-5}}{5}\right)]}_1^{\infty }\\ =37.5\left\{\left[\frac{{\infty }^{-4}}{-4}+\frac{{\infty }^{-6}}{-6}-\frac{2{\left(\infty \right)}^{-5}}{-5}\right]-\left[-\frac{1^{-4}}{4}-\frac{1^{-6}}{6}+\frac{2{\left(1\right)}^{-5}}{5}\right]\right\}\end{array}$

$\begin{array}{c}=37.5\left\{0-\left[-\frac{1}{4}-\frac{1}{6}+\frac{2}{5}\right]\right\}\\ =37.5\left\{\frac{1}{4}+\frac{1}{6}+\frac{2}{5}\right\}\\ =37.5\left(0.25+0.1667-0.4\right)\\ =0.625\end{array}$

Thus, the variance is:

$\begin{array}{c}V\left(X\right)=E\left(X^2\right)-{\left(E\left(X\right)\right)}^2\\ =0.625-{\left(0.5\right)}^2\\ =0.375\end{array}$

Hence thestandard deviation of total expenses is:

$\begin{array}{c}SD\left(X\right)=\sqrt{V\left(X\right)}\\ =\sqrt{0.375}\\ =\underset{\_}{0.612}.\end{array}$

To determine

Find the expected value of Y.

Answer to Problem 26E

Theexpected value of Y is 0.16037.

Explanation of Solution

Given info:

The insurance plan entails $500 deduction provision, then the 80% of payment will be done by the insurance company for any additional medical expenses greater than $500.

Calculation:

Here, the maximum amount paid by individual is $2,500.

The company will pay when the medical expenses is more than $500, that is $\$500+20\%\left(X-\$500\right)$.

To find the total medical expenses for an individual,

Solve the expression:

$\begin{array}{c}\$2,500=\$500+20\%\left(X-\$500\right)\\ \$2,500=\$500+\frac{20}{100}\left(X-\$500\right)\\ \$2,000=\frac{20}{100}\left(X-\$500\right)\\ \$200,000=20\left(X-\$500\right)\\ \$10,000=X-\$500\\ X=\$10,500\end{array}$

Hence the total medical expense for an individual is $10,500.

From that, $8,000 will be paid by the insurance company and $2,500 will be paid by an individual.

Let Y represents the medical expenses paid by the insurance company.

If the medical expense of an individual is less than $500, the insurance company will not pay any of it.

If the medical expense of an individual is between $500 and $10,500, the insurance company will 80% of the total medical expenses, that is, $Y=0.8\left(X-0.5\right)$, for $0.5\le X\le 10.5$.

If the medical expense of an individual is greater than $10,500, the insurance company will 80% of the total medical expenses, that is, $Y=\left(X-10.5\right)+8$, for $X\ge 10.5$.

The expenses paid by the company (in 1,000s of dollars), Y, is:

$Y=\{\begin{array}{l}0,x\le 0.5\\ 0.8\left(X-0.5\right)0.5\le x\le 10.5\\ \left(X-10.5\right)+8,x\ge 10.5\end{array}$

The expected value of Y obtained as given below:

$\begin{array}{c}E\left(Y\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}xf\left(x\right)dx}\\ ={\displaystyle \underset{0}{\overset{\infty }{\int }}2.4y{\left(1+\frac{x}{2.5}\right)}^{-7}dx}\\ =\left\{\begin{array}{c}2.4{\displaystyle \underset{0}{\overset{0.5}{\int }}y{\left(1+\frac{x}{2.5}\right)}^{-7}dx}+2.4{\displaystyle \underset{0.5}{\overset{10.5}{\int }}y{\left(1+\frac{x}{2.5}\right)}^{-7}dx}\\ +2.4{\displaystyle \underset{10.5}{\overset{\infty }{\int }}y{\left(1+\frac{x}{2.5}\right)}^{-7}dx}\end{array}\right\}\\ =\left\{\begin{array}{c}0+2.4{\displaystyle \underset{0.5}{\overset{10.5}{\int }}0.8\left(x-0.5\right){\left(1+\frac{x}{2.5}\right)}^{-7}dx}\\ +2.4{\displaystyle \underset{10.5}{\overset{\infty }{\int }}\left(x-10.5\right)+8{\left(1+\frac{x}{2.5}\right)}^{-7}dx}\end{array}\right\}\end{array}$

By using substitution rule,

$\begin{array}{c}u=1+\frac{x}{2.5}\\ x=2.5\left(u-1\right)\\ du=\frac{dx}{2.5}\\ 2.5du=dx\end{array}$

The interval is ,

x	0	$\infty$
u	$1\left(=1+\frac{0}{2.5}\right)$	$\infty \left(=1+\frac{\infty }{2.5}\right)$

On simplifying,

$\begin{array}{c}E\left(Y\right)=\left\{\begin{array}{c}0+2.4{\displaystyle \underset{0.5}{\overset{10.5}{\int }}\left[0.8\left(2.5\left(u-1\right)-0.5\right)\right]u^{-7}2.5du}\\ +2.4{\displaystyle \underset{10.5}{\overset{\infty }{\int }}\left[\left(2.5\left(u-1\right)-10.5\right)+8\right]u^{-7}2.5du}\end{array}\right\}\\ =\left\{\begin{array}{l}2.4{\displaystyle \underset{0.5}{\overset{10.5}{\int }}\left[0.8\left(2.5\left(u-1\right)-0.5\right)\right]u^{-7}2.5du}+\\ 2.4{\displaystyle \underset{10.5}{\overset{\infty }{\int }}\left[\left(2.5\left(u-1\right)-10.5\right)+8\right]u^{-7}2.5du}\end{array}\right\}\\ =4.8{\displaystyle \underset{0.5}{\overset{10.5}{\int }}\left[\left(2.5\left(u-1\right)-0.5\right)\right]u^{-7}du}+6{\displaystyle \underset{10.5}{\overset{\infty }{\int }}\left(2.5\left(u-1\right)-2.5\right)u^{-7}du}\\ =4.8{\displaystyle \underset{0.5}{\overset{10.5}{\int }}\left[\left(2.5u-3\right)\right]u^{-7}du}+6{\displaystyle \underset{10.5}{\overset{\infty }{\int }}\left(2.5u-5\right)u^{-7}du}\end{array}$

$\begin{array}{l}=4.8{\left(2.5\frac{u^{-5}}{-5}-3\frac{u^{-6}}{-6}\right)]}_{0.5}^{10.5}+6{\left(2.5\frac{u^{-5}}{-5}-5\frac{u^{-6}}{-6}\right)]}_{10.5}^{\infty }\\ =0.16024+0.00013\\ =0.16037\end{array}$

Hence, theexpected value of Y is 0.16037 or $167.37.