Question

There are two types of claims that are made to an insurance company. Let \(N_{i}(t)\) denote the number of type \(i\) claims made by time \(t\), and suppose that \(\left\\{N_{1}(t), t \geqslant 0\right\\}\) and \(\left\\{N_{2}(t), t \geqslant 0\right\\}\) are independent Poisson processes with rates \(\lambda_{1}=10\) and \(\lambda_{2}=1 .\) The amounts of successive type 1 claims are independent exponential random variables with mean \(\$ 1000\) whereas the amounts from type 2 claims are independent exponential random variables with mean \(\$ 5000 .\) A claim for \(\$ 4000\) has just been received; what is the probability it is a type 1 claim?

Short answer

The probability that a claim of $4000 is a type 1 claim is approximately \(99.34\% \).

Step-by-step solution

Identify the given information

We are given the following information: - Type 1 claims: Poisson process with rate \(\lambda_1 = 10\), claim amounts are exponentially distributed with mean \(\$1000\) - Type 2 claims: Poisson process with rate \(\lambda_2 = 1\), claim amounts are exponentially distributed with mean \(\$5000\) - Received claim amount: \(\$4000\)

Define the random variables and probability distribution functions

Let \(X_1\) denote the amount of a type 1 claim and \(X_2\) denote the amount of a type 2 claim. Then, the probability density function (pdf) of \(X_1\) and \(X_2\) can be written as: \(f_{X_1}(x) = \frac{1}{1000}e^{-\frac{x}{1000}}\) for \(x \geq 0\) \(f_{X_2}(x) = \frac{1}{5000}e^{-\frac{x}{5000}}\) for \(x \geq 0\)

Write down the main goal using Bayes' theorem

Our goal is to find the probability that a claim of \$4000 is a type 1 claim. We can use Bayes' theorem to find this probability. Let \(A_1\) be the event that the claim is of type 1, and \(A_2\) be the event that the claim is of type 2. We want to find \(P(A_1 | X = 4000)\), where \(X = 4000\) is the event that the claim amount is \$4000. Bayes' theorem states: \(P(A_1 | X = 4000) = \frac{P(X = 4000 | A_1)P(A_1)}{P(X = 4000)}\)

Find the probabilities using the pdf's and Poisson rates

The hardest part will be to find the different probabilities in the equation and for that we use the pdf's and Poisson rates. - \(P(X = 4000 | A_1) = f_{X_1}(4000) = \frac{1}{1000}e^{-\frac{4000}{1000}}\) - \(P(X = 4000 | A_2) = f_{X_2}(4000) = \frac{1}{5000}e^{-\frac{4000}{5000}}\) We are also given the Poisson rates which are \(10\) and \(1\), then we can infer that: - \(P(A_1) = \frac{\lambda_1}{\lambda_1 + \lambda_2} = \frac{10}{10 + 1}\) - \(P(A_2) = \frac{\lambda_2}{\lambda_1 + \lambda_2} = \frac{1}{10 + 1}\) With this information, we can compute \(P(X= 4000)\): - \(P(X = 4000) = P(X = 4000 | A_1)P(A_1) + P(X = 4000 | A_2)P(A_2)\)

Compute \(P(A_1 | X = 4000)\) using Bayes' theorem

Now we can compute the probability using Bayes' theorem: \(P(A_1 | X = 4000) = \frac{P(X = 4000 | A_1)P(A_1)}{P(X = 4000)}\) \(= \frac{\frac{1}{1000}e^{-\frac{4000}{1000}} \times \frac{10}{11}}{\frac{1}{1000}e^{-\frac{4000}{1000}} \times \frac{10}{11} + \frac{1}{5000}e^{-\frac{4000}{5000}} \times \frac{1}{11}}\) \(= \frac{1}{1 + \left(\frac{1}{5}e^{\frac{4000}{5000} - \frac{4000}{1000}}\right)}\) Calculating this value, we get: \(P(A_1 | X = 4000) \approx 0.9934\) Hence, the probability that a claim of \$4000 is a type 1 claim is approximately \(99.34\% \).