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

a) What is the value of k?

$\displaystyle k\intop_{0}^{\infty}(1+\frac{x}{2.5})^{-7}dx$=$\displaystyle \frac{-5k}{12}(1+\frac{x}{2.5})^{-6}\mid_{0}^{\infty}=\frac{5k}{12}$ 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)= $\displaystyle (12/5)\intop_{0}^{\infty}x(1+\frac{x}{2.5})^{-7}dx$ Let $\displaystyle u=(1+\frac{x}{2.5})$ then $\displaystyle 2.5du=dx$ and $\displaystyle x=2.5(u -1)$. We now have $\displaystyle E(X)= (12/5)\intop_{1}^{\infty}2.5(u-1)(u)^{-7}(2.5)du$

$\displaystyle = (2.5)^{2}(12/5)[\intop_{1}^{\infty}(u)^{-6}+(u)^{-7}du]$$\displaystyle =(2.5)^{2}(12/5)[\frac{-1}{5}u^{-5}-\frac{-1}{6}u^{-6}]\mid_{1}^{\infty} \displaystyle =(2.5)^{2}(12/5)[\frac{(1+\frac{x}{2.5})^{-5}}{-5}-\frac{(1+\frac{x}{2.5})^{-6}}{-6}]_{0}^{\infty}$$\displaystyle =(2.5)^{2}(12/5)[\frac{0}{-5}-\frac{0}{-6}-(\frac{1}{-5}-\frac{1}{-6})]$$\displaystyle =(2.5)^{2}(12/5)(\frac{1}{5}-\frac{1}{6})=.5 \displaystyle V(X)=(12/5)\intop_{0}^{\infty}x^{2}(1+\frac{x}{2.5})^{-7}dx$$\displaystyle =(12/5)\intop_{1}^{\infty}(2.5)^{2}(u-1)^{2}(u)^{-7}(2.5)du$$\displaystyle =(12/5)(2.5)^{3}\intop_{1}^{\infty}(u-1)^{2}(u)^{-7}du \displaystyle =(12/5)(2.5)^{3}\intop_{1}^{\infty}(u^{2}-2u+1)(u)^{-7}du$$\displaystyle =(12/5)(2.5)^{3}\intop_{1}^{\infty}(u^{-5}-2u^{-6}+u^{-7})du$$\displaystyle =(12/5)(2.5)^{3}[\frac{u^{-4}}{-4}-\frac{2u^{-5}}{-5}+\frac{u^{-6}}{-6}]_{1}^{\infty} \displaystyle =(12/5)(2.5)^{3}[\frac{(1+\frac{x}{2.5})^{-4}}{-4}-\frac{2(1+\frac{x}{2.5})^{-5}}{-5}+\frac{(1+\frac{x}{2.5})^{-6}}{-6}]_{0}^{\infty}$$\displaystyle =(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$

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

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 $2500. Then write an expression for Y as a function of X (which involves several different pieces) and calculate the expected value of this function.]$\displaystyle X=x; Y=h(x)= 0\: when\: 0\leq x\leq500$;$\displaystyle 0.8x\: when\;500<x\leq2500$;$\displaystyle and\; x\; when\;2500<x$.$\displaystyle E(Y)=0\intop_{0}^{500}xdx+\intop_{500}^{2500}.8x^{ 2}dx+\intop_{2500}^{\infty}x^{2}dx\displaystyle =0[\frac{x^{2}}{2}]_{0}^{500}+.8[\frac{x^{3}}{3}]_{500}^{2500}+[\frac{x^{3}}{3}]_{2500}^{\infty}\$