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Company XYZ provides a warranty on a product that it produces. Each year, the number of warranty claims follows a Poisson distribution with mean c. The probability that no warranty claims are received in any given year is 0.60. Company XYZ purchases an insurance policy that will reduce its overall warranty claim payment costs. The insurance policy will pay nothing for the first warranty claim received and 5000 for each claim thereafter until the end of the year. Calculate the expected amount of annual insurance policy payments to Company XYZ.

My solution y= payment by insurance company $$y=\begin{cases} {0} & \text{if } x=0\\ 5000(x-1) & \text{if } x\geq1,\\ \end{cases}$$

c=0.5018 $p(x=0)=\frac{(e^-c)*(c^0)}{0!}=0.60$ so $c=0.5108$ so the mean of x is 0.5108 so $E[x]=0.5108$ $E[y]=\sum 0*p(x=0)+5000(x-1)*p(x\geq1)=0*06+5000(x-1)*0.4=2000E[x]-2000=-976$ Correct answer is 554,I would like to know if my function for payment is correct and what would be my next step.

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  • $\begingroup$ I don't understand your method. The expected payout is $E=\sum_{n=2}^{\infty}5000\times (n-1)\times p_n$ where $p_n$ is the probability of getting exactly $n$ claims. Fairly easy computation from there. By the way, I get $c\sim .51082$ which is slightly different than what you say. $\endgroup$
    – lulu
    Commented Aug 30, 2016 at 17:43
  • $\begingroup$ Where in your calculation are you using the fact that this is a Poisson process? In my formula (in my comment) I use it in evaluating $p_n$. Indeed, $p_n=\frac {c^ne^{-c}}{n!}$. I then advise computing the sum numerically...you only need to sum the first dozen or so terms (probably fewer, really). $\endgroup$
    – lulu
    Commented Aug 30, 2016 at 17:49
  • $\begingroup$ yes I'm using the Poisson probability mass function,and c=0.5108 I rounded wrong.So when do I know when to stop? $\endgroup$
    – Theo Robinson
    Commented Aug 30, 2016 at 18:01
  • $\begingroup$ show me where in your formula you use the Poisson formula. As it stands, you have written $E$ as a function of a variable $x$, which makes no sense. $E$ is a number, not a function. $\endgroup$
    – lulu
    Commented Aug 30, 2016 at 18:03
  • $\begingroup$ $E$ is expected value in this case $\endgroup$
    – Theo Robinson
    Commented Aug 30, 2016 at 18:17

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We know: $X\sim\mathcal {Pois}(c)$ and $\mathsf P(X=0)=0.60$.

Since the first fact means that $\mathsf P(X=x)~=~\dfrac{c^x\mathsf e^{-cx}}{x!}\mathbf 1_{x\in\Bbb N}$ , we can easily calculate $c$ knowing the second fact.

We know $Y := 5000(X-1)^+~$ which is $~Y=5000\max(X-1,0)$

Then $\mathsf E(Y) = 5000~\mathsf E(X-1\mid X\geq 1)~\mathsf P(X\geq 1) \color{silver}{+ \require{cancel}\cancel{0~\mathsf P(X=0)}}$

If only the Poisson distribution had some convenient property that allowed us to easily find this conditional expectation without messy summation.   Hmm...

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  • $\begingroup$ would it be possible for you to please take a look at this one: math.stackexchange.com/questions/2524383/… it's similar and I believe that you might be able to help. Thank you. $\endgroup$
    – user100463
    Commented Nov 17, 2017 at 11:07

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