/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 59 There are two types of claims th... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 59

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

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

Step by step solution

01

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: \(\$4000\)
02

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 \(x \geq 0\)
03

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: \(P(A_1 | X = 4000) = \frac{P(X = 4000 | A_1)P(A_1)}{P(X = 4000)}\)
04

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 \(P(X= 4000)\): - \(P(X = 4000) = P(X = 4000 | A_1)P(A_1) + P(X = 4000 | A_2)P(A_2)\)
05

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 \(99.34\% \).

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Most popular questions from this chapter

A cable car starts off with \(n\) riders. The times between successive stops of the car are independent exponential random variables with rate \(\lambda .\) At each stop one rider gets off. This takes no time, and no additional riders get on. After a rider gets off the car, he or she walks home. Independently of all else, the walk takes an exponential time with rate \(\mu\). (a) What is the distribution of the time at which the last rider departs the car? (b) Suppose the last rider departs the car at time \(t .\) What is the probability that all the other riders are home at that time?

Determine $$ \operatorname{Cov}[X(t), X(t+s)] $$ when \(\\{X(t), t \geqslant 0\\}\) is a compound Poisson process.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be independent and identically distributed exponential random variables. Show that the probability that the largest of them is greater than the sum of the others is \(n / 2^{n-1} .\) That is, if $$ M=\max _{j} X_{j} $$ then show $$ P\left\\{M>\sum_{i=1}^{n} X_{i}-M\right\\}=\frac{n}{2^{n-1}} $$ Hint: What is \(P\left\\{X_{1}>\sum_{i=2}^{n} X_{i}\right\\} ?\)

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 \(t\).

Prove that (a) \(\max \left(X_{1}, X_{2}\right)=X_{1}+X_{2}-\min \left(X_{1}, X_{2}\right)\) and, in general, (b) \(\max \left(X_{1}, \ldots, X_{n}\right)=\sum_{1}^{n} X_{i}-\sum_{i
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