# Health insurance coverage

• Jun 30th 2010, 04:49 AM
oldguynewstudent
Health insurance coverage
Could someone check this very complicated problem?

I believe I have the correct answer for part a, the graph is omitted, part c the Variance and Standard Deviation seem a little high, and finally I'm not sure at all on part d, because the expected value would go to infinity. I especially need help with part d. Thanks!

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 $f(x)=k(1 + x/2.5)^{-7}for x\geq0$.

a) What is the value of k?

$k\intop_{0}^{\infty}(1+\frac{x}{2.5})^{-7}dx$= $\frac{-5k}{12}(1+\frac{x}{2.5})^{-6}\mid_{0}^{\infty}=\frac{5k}{12}$ This integral must equal 1 so k=12/5.

b) Graph the pdf of X.

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

E(X)= $(12/5)\intop_{0}^{\infty}x(1+\frac{x}{2.5})^{-7}dx$ Let $u=(1+\frac{x}{2.5})$ then $2.5du=dx$ and $x=2.5(u -1)$. We now have $E(X)= (12/5)\intop_{1}^{\infty}2.5(u-1)(u)^{-7}(2.5)du$

$= (2.5)^{2}(12/5)[\intop_{1}^{\infty}(u)^{-6}+(u)^{-7}du]$ $=(2.5)^{2}(12/5)[\frac{-1}{5}u^{-5}-\frac{-1}{6}u^{-6}]\mid_{1}^{\infty}$

$=(2.5)^{2}(12/5)[\frac{(1+\frac{x}{2.5})^{-5}}{-5}-\frac{(1+\frac{x}{2.5})^{-6}}{-6}]_{0}^{\infty}$ $=(2.5)^{2}(12/5)[\frac{0}{-5}-\frac{0}{-6}-(\frac{1}{-5}-\frac{1}{-6})]$ $=(2.5)^{2}(12/5)(\frac{1}{5}-\frac{1}{6})=.5$

$V(X)=(12/5)\intop_{0}^{\infty}x^{2}(1+\frac{x}{2.5})^{-7}dx$ $=(12/5)\intop_{1}^{\infty}(2.5)^{2}(u-1)^{2}(u)^{-7}(2.5)du$ $=(12/5)(2.5)^{3}\intop_{1}^{\infty}(u-1)^{2}(u)^{-7}du$

$=(12/5)(2.5)^{3}\intop_{1}^{\infty}(u^{2}-2u+1)(u)^{-7}du$ $=(12/5)(2.5)^{3}\intop_{1}^{\infty}(u^{-5}-2u^{-6}+u^{-7})du$ $=(12/5)(2.5)^{3}[\frac{u^{-4}}{-4}-\frac{2u^{-5}}{-5}+\frac{u^{-6}}{-6}]_{1}^{\infty}$

$=(12/5)(2.5)^{3}[\frac{(1+\frac{x}{2.5})^{-4}}{-4}-\frac{2(1+\frac{x}{2.5})^{-5}}{-5}+\frac{(1+\frac{x}{2.5})^{-6}}{-6}]_{0}^{\infty}$ $=(12/5)(2.5)^{3}[\frac{0}{-4}-\frac{0}{-5}+\frac{0}{-6}-(\frac{1}{-4}-\frac{1}{-5}+\frac{1}{-6})]=8.125$

$\sigma=\sqrt{8.125}= 2.85$

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.]

$X=x; Y=h(x)= 0\: when\: 0\leq x\leq500$; $0.8x\: when\;500; $and\; x\; when\;2500.

$E(Y)=0\intop_{0}^{500}xdx+\intop_{500}^{2500}.8x^{ 2}dx+\intop_{2500}^{\infty}x^{2}dx$ $=0[\frac{x^{2}}{2}]_{0}^{500}+.8[\frac{x^{3}}{3}]_{500}^{2500}+[\frac{x^{3}}{3}]_{2500}^{\infty}$