Poisson process, claims arrival

Question

In some insurance contract the claims arrival process is the Poisson process with $\lambda = 3$. The single claim values are (independent from each other and from arrival process) continuously distributed on $(0,1)$ with density function $f(y)=1$.

The policyholder has the following strategy:

• doesn't report the claim if its value is smaller than $1/3$,
• reports first claim which value is bigger than $1/3$ and then reports all claims regardless of its size.

What is the expected value of not-reported claims in that insurance contract?

I was trying to that in the following way:

Let $K$ be the sum of the value of not-reported claims. Then $\mathbb{E}K = \mathbb{E}[\mathbb{E}(K|N)]$ where $N$ is the number of all claims and we know that $N$~$ Poiss(3)$. Now I want to determine $\mathbb{E}(K|N=n)$ for each $n$ because then I would be able to finish this task. But I find it hard:

• $\mathbb{E}(K|N=0) = 0$ it's obvious.
• $\mathbb{E}(K|N=1) = ?$
• $\mathbb{E}(K|N=2) = ?$

$\dots$

• $\mathbb{E}(K|N=n) = ?$

I know also that

• $\mathbb{P}($the single claim is reported$) = \frac{2}{3}$ and
• $\mathbb{P}($the single claim is not-reported$) = \frac{1}{3}$
• $ \mathbb{E}($the single claim$)=\frac{1}{2} $

EDIT* I want to calculate the expected value of not-reported (equivalently - reported) claims in that insurance contract during first (one) year.

Answer 1

First of all, it should be obvious that the arrival times of the claims are irrelevant; what matters is that they occur. So the Poisson process is an unnecessary detail. This is something you have missed; in particular, it is the fact that we are never asked for the number of claims, reported or otherwise, in a fixed period of time, nor are we asked for when the claims occur. We are interested in the number of unreported claims until the first reported claim.

Once you understand this, the rest is straightforward, since all we are doing is considering independent Bernoulli trials with parameter $p = 2/3$ as the probability of reporting a claim. Then it follows that the number of unreported claims is a geometric random variable $X$ with $$\Pr[X = x] = (1-p)^x p, \quad x = 0, 1, 2, \ldots.$$ Since unreported claims $Y^* = Y \mid Y < 1/3$ are independent and uniform on $[0, 1/3)$, the expectation is calculated as $$\operatorname{E}[Y^*] = \operatorname{E}[\operatorname{E}[Y^* \mid X]].$$ This computation is left as an exercise.

Answer 2

Can anyone look at my solution, I'm still not sure of it:

• $\mathbb{E}(K|N=0) =0$
• $\mathbb{E}(K|N=1) = \mathbb{P}($the single claim is not-reported$) \cdot \mathbb{E}($the single claim which is not reported$) = \frac{1}{3} \cdot \frac{1}{6}$
• $\mathbb{E}(K|N=2) = \frac{1}{3} \cdot \frac{1}{6} + \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{6}$

$\dots$

• $\mathbb{E}(K|N=n) = \sum_\limits{k=1}^{n}\frac{1}{6} \cdot ( \frac{1}{3} )^k = \frac{1}{12}(1-\frac{1}{3^n} ) $

Now I can calculate the expected value of non-reported claims:

$\mathbb{E}(K)=\mathbb{E}(\mathbb{E}(K|N=n)) = \mathbb{E}(\frac{1}{12}(1-\frac{1}{3^N} )) = \frac{1}{12}(1-\mathbb{E}(\frac{1}{3^N}) ) = \frac{1}{12}(1-\sum_\limits{n=0}^{\infty}(\frac{1}{3^n} \cdot \frac{3^n}{n!}e^{-3}) ) = \frac{1}{12}(1-e^{-2} )$