# Poisson distribution question: Company XYZ provides a warranty on a product that it produces...

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 attempt:

Let $$Y$$ be the RV denoting the amount of annual insurance policy payments to Company XYZ and $$X$$ be the number of annual warranty claims. Then, $$Y=5000(X-1) \;\; \text{if}\;\; X>1$$

and $$0$$ otherwise. \begin{align} E[Y] &= 5000 \cdot E[X-1\mid X>1] \\ &= 5000 \cdot \sum_{x=2}^\infty (x-1) \cdot P[X=x-1]\\ &=5000 \cdot \sum_{x=1}^\infty (x) \cdot P[X=x]\\ &=5000[E(X)-0\cdot P[X=0]]\\ &=5000[E(X)] \end{align}

Note that $$P(X=0) = 0.6 \implies c= -\ln 0.6$$. So, $$E[Y]=5000[-\ln 0.6].$$ This is not the correct answer, however. Can someone please point out where I went wrong?

• The first sum ought to be $5000\times \sum_{x=2}^{\infty} (x-1)P(X=x)$.
– lulu
Commented Jan 18, 2021 at 23:53
• @lulu Could you please what you mean in a bit more detail? Which summation expression are you referring to? The first one or the last one? Commented Jan 19, 2021 at 0:02
• I mean, what I wrote is $E[Y]$ (by definition). You incorrectly have $P(X=x-1)$. You get a payout of $x-1$ when $X=x$.
– lulu
Commented Jan 19, 2021 at 0:04
• Should say: there's nothing to do with conditional probabilities here. It's just the straight probability.
– lulu
Commented Jan 19, 2021 at 0:05
• @Lulu Thanks, makes sense now. I wonder what summation-related manipulation would lead to the correct answer in this case... Commented Jan 19, 2021 at 0:16

Number of claims is $$X \sim \mathsf{Pois}(c),$$ so $$P(X=0) = e^{-c} = 0.6,$$ which implies $$c = -\log_e(.6) = 0.511.$$ So $$E(X) = 0.5108.$$

c =-log(.6); c
[1] 0.5108256


Because of XYZ can't collect on the first claim, you can't multiply $$c$$ by \\$5000 to get the average annual payout as $$5000c.$$

There may be some years with only one claim so XYZ collects nothing, and there may be years with several valid claims so thqt XYZ collects from the insurance company.

Here is a strictly computational method. Technically, Poisson probabilities extend from $$0$$ to $$\infty.$$ But terms decay rapidly to 0, Let's use the total number of claims for illustration. You can come extremely close to $$E(X)$$ by summing only the first 100 terms of the infinite series,

k = 1:99;  s = sum(k*dpois(k, -log(.6)))
s
[1] 0.5108256  # exactly E(X) to many places


Now we find the expected number $$Y$$ of claims with payouts. We seek $$E(Y) = 0.1108$$ so that we can give the expected annual payout of $$5000\,E(Y)\approx \554$$.

We let the 100-vector j represent the number of paid claims and the 1000-vector i be the number of claims.

j = c(0,0,1:98);  i = 0:99
5000*s
[1] 0.1108256  # E(Y) = avg nr. payable claims
[1] 554.1281   # 5000E(Y) = avg annual payout


Note: in case you're wondering whether it is sufficient to sum the first 100 terms of the series, the first 50 would have been more than enough:

j = c(0,0.1:48); i = 0:48
sum(j*dpois(i, -log(.6)))
[1] 0.1508256


Almost all of the probability of the distribution $$\mathsf{Pois}(-log(.6))$$ is below 6.

qpois(.9999, -log(.6))
[1] 5