Poisson distribution insurance policies

Question

The number of claims for a single policy follows a Poisson distribution with a mean claim count of 0.24.

1. Determine the distribution of the number of claims registered on the total portfolio, when you know that the portfolio consists of 1500 independent policies.

2. Using the normal approximation,

a. calculate the probability of 320 or more claims on the total portfolio.

b. determine the 95% quantile of the distribution of the number of claims registered on the total portfolio.

Answer 1

The poisson distribution has only one parameter: lambda. Lambda is both the expected value and the variance of the distribution. For the case of 1500 policies and a claim rate of 0.24, the expected number of claims per portfolio is:

0.24 * 1500 = λ = 360

So the expected distribution is:

P(x) = λ^xexp(-λ) / x! = (360)^xexp(-360)/x!

Now we switch to the normal approximation. For large lambda, we can use a normal distribution with mean and variance = lambda.

So the probability of 320 or more claims is equal to the 1 - CDF(320) for a N(360, Sqrt(360)) distribution (the latter being a shorthand for a normal distribution with mean = 360 and variance = 360).

You should calculate this yourself. I get 0.0.9824925

To get the 95% quantile, we are asking what value of x gives F(x) = 0.95 where F is CDF of the N(λ,λ) distribution. I get 391.2089. Meaning that in 95% of cases, you would expect 391.2089 or fewer claims.