0
$\begingroup$

I am having a trouble understanding a subtle difference in the answer of two problems.

Problem one:

An insurance policy reimburses a loss up to a benefit limit of 10. The policyholder's loss, Y. follows a distribution with density function:

$ f(x) = \begin{cases} \frac{2}{x^3}, & \text{if $x \gt 1$} \\ 0, & \text{otherwise} \end{cases}$

What is the expected value of the benefit paid under the insurance policy?

Problem two:

An auto insurance company insures an automobile woth 15,000 for one year under a policy with a 1000 deductible. During the policy year there is a .04 chance of partial damage to the car and .02 chance of total loss of the car. If there is partial damage to the car, The amount $X$ of damage in the thousands follows a distribution with density function:

$f(x) = \begin{cases} .5003e^{\frac{-x}{2}}, & \text{for $0 \lt x \lt 15$} \\ 0, & \text{otherwise} \end{cases}$

In problem two, Inherently if the policy covers a $15000$ car and the deductible is $1000$ then the maximum loss or payout would be $14,000$ so why when we calculate the answer is it defined as $(0)(.94)+(.04)(.5003)\int_1^{15} e^{\frac{-x}{2}}+(.02)(14)$. The part that I am having the biggest issue with is the last part $(.02)(14)$ Which is contradictory to the first problem where:

$\int_1^{10} y\frac{2}{y^2}dy + \int_{10}^{\infty} 10 \frac{2}{y^2}dy$

The probability of the maximum payout is multiplied by the pdf (which is the likely hood of a value y occurring).

So why is 14 not multiplied by the pdf and integrated in problem two?

$\endgroup$
0
$\begingroup$

So why is 14 not multiplied by the pdf and integrated in problem two?

It is.   The integral of the pdf over the entire support equals $1$, by definition, and so that is by what it is multiplied.   Also recall the pdf is zero outside the support.

$$\int_0^{15} 14\cdot(0.20)\cdot 0.5003 \mathsf e^{-x/2}~\mathsf d x +\int_{15}^\infty 14\cdot(0.20)\cdot 0~\mathsf d x ~=~ 14\cdot(0.20)$$


PS: also the answer should be : $(0)(.94)+(.04)(.5003)\int_0^{15}x\mathsf e^{-x/2}\mathsf d x+(14)(0.20)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.