The total claim amount for a health insurance policy follows a distribution with density function:
$f(x)= \frac{1}{1000}e^{\frac{-x}{1000}}$ for $x\gt 1000$
The premium for the policy is set at the total expected claim amount plus 100. If 100 policies are sold calculate the approximate probability that the insurance claim will have claims exceeding the premiums collected.
Let $Pr$ denote the premium. Then $Pr$= $E[X]+100$
We want $P(x \gt Pr)$ which can be found by $\int_{100}^{Pr}f(x)dx$
$Pr= \int_0^{\infty} \frac{x}{1000}e^{\frac{-x}{1000}}dx+100 = 100+\frac{1}{1,000,000,000}$ by integration by parts. It is clear that at this point my answer is wrong I don't think it was integration. What is wrong with my logic?
Best Answer
First I assume you mean for $f(x)$ to be for $x>0$ rather than $x>1000$ (you seem to be assuming $x>0$ below and $x>1000$ wouldn't make any sense).
You have misplaced some factors of $1000$: $$ \int_0^\infty x \frac{1}{1000}e^{-x/1000}dx = 1000\int_0^\infty \frac{x}{1000}e^{-x/1000} \frac{dx}{1000} = 1000\int_0^\infty u e^{-u}du.$$ Now that the annoying factor of $1000$ is handled, you should be able to get more easily by IBP that $\int_0^\infty u e^{-u}du =1.$
So the expected claim is $1000$ and we set the premium at $1100.$
To get the probability that you lose money on $100$ claims, we can invoke the central limit theorem. The mean claim is $1000,$ as seen above. Similarly the standard deviation can be computed to be $1000$ as well. The central limit theorem says that the average of the $100$ claims is approximately Gaussian with mean $1000$ and standard deviation $\frac{1000}{\sqrt{100}} = 100.$
We will lose money if the average claim is larger than the premium. Since the premium is $\$100$ larger than the mean, and our standard deviation is also $100,$ this is the probability of a normal exceeding the mean by more than one standard deviation, which is $1-\Phi(1) \approx 16\%.$
(The CLT approach is probably more useful, but we can also express the answer exactly without the central limit theorem by using the fact that the sum of exponentials is Gamma distributed. We get $$ \int_{1100\times 100}^\infty \frac{1}{\Gamma(100)}\frac{x^{99}}{(1000)^{100}}e^{-x/1000}dx = \frac{1}{\Gamma(100)}\int_{110}^\infty u^{99} e^{-u} du = \frac{\Gamma(100,110)}{\Gamma(100)} \approx 16\%$$ where $\Gamma(z)$ and $\Gamma(z,a)$ are the gamma function and incomplete gamma function.)