Question

Policyholders of a certain insurance company have accidents at times distributed according to a Poisson process with rate \(\lambda .\) The amount of time from when the accident occurs until a claim is made has distribution \(G\). (a) Find the probability there are exactly \(n\) incurred but as yet unreported claims at time \(t\). (b) Suppose that each claim amount has distribution \(F\), and that the claim amount is independent of the time that it takes to report the claim. Find the expected value of the sum of all incurred but as yet unreported claims at time \(t\).

Short answer

The probability of having exactly \(n\) unreported claims at time \(t\) is given by: \[P(Y_n(t) = n) = \sum_{k=n}^{\infty} P(N(t-x) = k-n) \cdot P(N(x) = k) \cdot \binom{k}{n} \cdot (1 - G(x))^n \cdot G(x)^{k-n} \] The expected value of the sum of all incurred but as yet unreported claims at time \(t\) is given by: \[E[\text{Unreported claims at time } t] = \mu \cdot \sum_{n=0}^{\infty} n \cdot P(Y_n(t) = n)\] Where \(\mu\) is the expected claim amount found using the distribution \(F\).

Step-by-step solution

Find the probability of having exactly \(n\) accidents at time \(t\).

Since the accident occurrence follows a Poisson process with rate \(\lambda\), the number of accidents \(N(t)\) occurring in the interval \([0, t]\) has a Poisson distribution with mean \(\lambda t\). The probability that there are exactly \(n\) accidents at time \(t\) is: \[P(N(t) = n) = \frac{(\lambda t)^n e^{-\lambda t}}{n!}\]

Calculate the probability of a claim being unreported.

For a claim to be unreported, it must've occurred in the time interval \([0, t-x]\), where \(x\) is the time it takes to report the claim. Using the distribution \(G\), we can find the probability that it takes more than \(x\) time to report a claim by calculating the complementary probability \(1 - G(x)\).

Find the probability of having \(n\) unreported claims at time \(t\).

To find the probability of having exactly \(n\) unreported claims at time \(t\), we need to consider the convolution of the Poisson distribution and the unreported claim probability. Let \(Y_n(t)\) be the number of unreported claims at time \(t\). The probability of having exactly \(n\) unreported claims is: \[P(Y_n(t) = n) = \sum_{k=n}^{\infty} P(N(t-x) = k-n) \cdot P(N(x) = k) \cdot \binom{k}{n} \cdot (1 - G(x))^n \cdot G(x)^{k-n} \] Where - \(P(N(t-x) = k-n)\) is the probability of having \((k-n)\) accidents in the time interval \([0, t-x]\) (from Step 1), - \(P(N(x) = k)\) is the probability of having \(k\) accidents in the time interval \([0, x]\) (from Step 1), - \(\binom{k}{n}\) is the binomial coefficient (combinations), - \((1 - G(x))^n\) is the probability of \(n\) unreported claims, - \(G(x)^{k-n}\) is the probability of \((k-n)\) reported claims.

Find the expected claim amount.

To calculate the expected claim amount, we will use the distribution \(F\). The expected claim amount, denoted by \(\mu\), can be found as: \[\mu = E[X] = \int_0^{\infty} x \cdot f(x) dx\] where \(f(x)\) is the probability density function (pdf) corresponding to the distribution \(F\).

Calculate the expected value of unreported claims at time \(t\).

Since the claim amount is independent of the time to report a claim, we can simply multiply the probability of having exactly \(n\) unreported claims (from Step 3) with the expected claim amount (from Step 4) to find the expected value of the sum of all incurred but yet unreported claims at time \(t\): \[E[\text{Unreported claims at time } t] = \mu \cdot \sum_{n=0}^{\infty} n \cdot P(Y_n(t) = n)\]