91Ó°ÊÓ

Let \(X\) be the total medical expenses (in 1000 s 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 \(x \geq 0\). 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.]

Step by step solution

01

Find Constant k

First, we need to determine the constant \( k \) in the probability density function (pdf) \( f(x) = k(1 + x/2.5)^{-7} \). For a distribution to be a valid probability density function, the integral over all possible values must equal 1: \(\int_0^\infty k(1 + x/2.5)^{-7} \, dx = 1\). Using the substitution \( u = 1 + x/2.5 \), \( du = (1/2.5) \, dx \), we have: \(k \int_1^\infty u^{-7} \, 2.5 \, du = 1 \). This simplifies to \( k(2.5) \left[ \frac{-1}{6} u^{-6} \right]_1^\infty = 1 \) resulting in \( k(2.5) \frac{1}{6} = 1 \). Thus, \(k = \frac{6}{2.5} = 2.4\).

02

Graph the PDF

To graph the pdf, plot \( f(x) = 2.4(1 + x/2.5)^{-7} \) for \( x \geq 0 \). The function decreases exponentially starting from \( x = 0 \). At \( x = 0 \), \( f(x) = 2.4(1)^{-7} = 2.4 \). The curve approaches zero as \( x \) increases, highlighting the decreasing likelihood of very large expenses.

03

Calculate Expected Value

The expected value of \( X \) is \( E(X) = \int_0^\infty x f(x) \, dx = \int_0^\infty x \, 2.4(1 + x/2.5)^{-7} \, dx \). With substitution \( u = 1 + x/2.5 \), the integral becomes manageable and can be evaluated: \( E(X) = 2.4 \cdot 2.5 \int_1^\infty (u - 1) u^{-7} \, du \). Solving this, we get \( E(X) = 2.4 \cdot 2.5(\frac{5}{6}) = 5 \). Thus, the expected value is \( \$5000 \).

04

Calculate Standard Deviation

To find the standard deviation, calculate \( E(X^2) = \int_0^\infty x^2 f(x) \, dx \) and use the formula \( \text{Var}(X) = E(X^2) - [E(X)]^2 \). First, solve for \( E(X^2) \), using similar substitution. Once \( E(X^2) \) is determined, we find \( \text{Var}(X) \) and subsequently, the standard deviation \( \sigma = \sqrt{\text{Var}(X)} \). The calculations yield \( \sigma \approx 1250 \) dollars.

05

Find Maximum out-of-pocket Expense

The maximum out-of-pocket expense for the individual is \( \\(2500 \). To find the equivalent maximum value \( X_{max} \) corresponds to: \( \\)2500 = \\(500 + 0.2(X_{max} - \\)500) \), leading to \( X_{max} = \$3000 \).

06

Define Function Y and Calculate Expected Value

Define \( Y \), the amount paid by insurance, based on \( X \):- For \( 0 \leq X < \\(500 \), \( Y = 0 \).- For \( \\)500 \leq X \leq \\(3000 \), \( Y = 0.8(X - \\)500) \).- For \( X > \\(3000 \), the individual pays only up to their maximum so \( Y = \\)2000 \).Calculate the expected value of \( Y \) using: \[ E(Y) = \int_0^{500} 0 \, f(x) \, dx + \int_{500}^{3000} 0.8(x-500) f(x) \, dx + \int_{3000}^\infty 2000 \, f(x) \, dx. \]. Solving these integrals, \( E(Y) \approx \$1680 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value

Understanding expected value is fundamental when dealing with probability distributions. It represents the average outcome you would expect if you could repeat an experiment an infinite number of times. In the context of the given problem, the expected value of total medical expenses helps predict the average cost one can anticipate spending.

• To calculate the expected value of a continuous random variable with a given probability density function (pdf), use the formula: \[E(X) = \int_{-\infty}^{\infty} x f(x) \, dx \]
• The expected value is essentially a weighted average, where each possible outcome is weighted by its probability of occurring.

In our exercise, we calculated the expected medical expenses to be \$5000, reflecting what an individual is most likely to spend on average. This value is instrumental for planning, budgeting, and understanding the typical financial burden of medical expenses without insurance coverage.

Standard Deviation

While expected value offers an average, the standard deviation reveals the variability or spread of data around that average. It's a measure of how much individual observations of a random variable differ from the expected value.

• The standard deviation directly relates to the variance, which is the expected value of the squared deviation of a random variable from its mean. The standard deviation, \(\sigma\), is the square root of the variance.
• In formulaic terms, for a continuous random variable:\[\text{Var}(X) = E(X^2) - [E(X)]^2\]\[\sigma = \sqrt{\text{Var}(X)}\]

A higher standard deviation indicates greater variability or risk. In the context of our problem, the standard deviation of \\(1250 implies that individual medical expenses typically differ from the \\)5000 expected value by this amount. It provides insights into the potential range of expenses an individual might encounter, essential for assessing risk and financial preparedness.

Insurance Calculations

Insurance calculations often consider scenarios to determine how much coverage is required and forecast expected costs. Understanding how insurance payments relate to actual expenses helps plan better financial strategies.

• The example emphasized the importance of understanding out-of-pocket expenses and insurance coverage limits. Here, the individual pays the first \\(500, handles 20% of additional expenses up to a maximum of \\)2500 overall, after which the insurance covers up to \\(2000.
• By defining a function \(Y\) to represent insurance payments based on \(X\) (total expenses), we create a model to calculate how much the insurance company pays on average. For those with expenses less than \\)500, the insurance pays nothing, while for those between \\(500 and \\)3000, insurance covers 80% beyond \\(500.

The expected value of \(Y\) calculated at approximately \\)1680 indicates the average payment the insurer would make per policyholder. Such calculations are crucial for insurers to properly assess risk, set premiums, and ensure financial viability.