/*! 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] | 91Ó°ÊÓ

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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Distribution
The exponential distribution is commonly used to model the time between events in a Poisson process. It is defined by its probability density function (PDF), which describes the likelihood of a particular value occurring. In our insurance example, the claim amounts for both types of events follow an exponential distribution.
This means, for a type 1 claim, given that the mean is \(1000, the probability density function is:\[ f_{X_1}(x) = \frac{1}{1000} e^{-\frac{x}{1000}} \text{ for } x \geq 0 \]Similarly, for a type 2 claim with a mean of \)5000, the PDF becomes:\[ f_{X_2}(x) = \frac{1}{5000} e^{-\frac{x}{5000}} \text{ for } x \geq 0 \]
Understanding these distributions is crucial because it helps us compute the probability of different claim amounts, which is key for insurance companies in risk assessment and financial planning. The exponential distribution is memoryless, meaning the probability of an event occurring in the future is independent of the past.
Bayes' Theorem
Bayes' theorem is a fundamental concept in probability theory and statistics that helps update the probability of a hypothesis based on new evidence. It is particularly useful in situations where we need to determine conditional probabilities.
In the context of our insurance claim problem, we want to determine the probability that a received \(4000 claim is of type 1. Bayes' theorem allows us to do so by using the known conditional probabilities:\[ P(A_1 | X = 4000) = \frac{P(X = 4000 | A_1) P(A_1)}{P(X = 4000)} \]Here, \(P(X = 4000 | A_1)\) is the probability that the amount is \)4000 given it is a type 1 claim, which is derived from the exponential distribution. \(P(A_1)\) represents the prior probability of a claim being type 1, based on the Poisson rates.
Using Bayes' theorem provides a structured method to incorporate both the likelihood of the observed claim amount and the prior knowledge about the event types, resulting in a more accurate assessment of the situation.
Probability Density Function
The probability density function (PDF) is a core component of continuous probability distributions. It describes the likelihood that a random variable takes on a given value. For continuous variables, the PDF is crucial in calculating probabilities over intervals.
In our example, the PDFs for claim types 1 and 2 were derived from their respective exponential distributions:- For type 1 claims, with mean \(1000:\[ f_{X_1}(x) = \frac{1}{1000} e^{-\frac{x}{1000}} \text{ for } x \geq 0 \]- For type 2 claims, with mean \)5000:\[ f_{X_2}(x) = \frac{1}{5000} e^{-\frac{x}{5000}} \text{ for } x \geq 0 \]
The PDF helps compute the likelihood of receiving a claim of a specific amount, like $4000, under each claim type. Moreover, when integrated over a range, the area under the PDF curve represents the probability that the variable falls within that interval.
Understanding the PDF is vital for evaluating how various random variables behave and for making precise statistical inferences.
Random Variables
Random variables are essential building blocks in probability theory. They are variables that take on values resulting from the outcome of a random event and are crucial in modeling real-world phenomena.
In this exercise, we assigned random variables to claim amounts for type 1 and type 2 claims: \(X_1\) and \(X_2\). These random variables are distributed according to the exponential distribution due to the nature of the claims process.
- \(X_1\) represents the amount of a type 1 claim.- \(X_2\) represents the amount of a type 2 claim.
These random variables help describe the uncertainty in claim amounts, which is foundational for predicting future events and assessing risk. Moreover, understanding how random variables operate allows for evaluating probabilities using tools like probability density functions and Bayes' theorem, ultimately supporting better decision-making based on statistical data.

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Most popular questions from this chapter

Consider a two-server parallel queuing system where customers arrive according to a Poisson process with rate \(\lambda\), and where the service times are exponential with rate \mu. Moreover, suppose that arrivals finding both servers busy immediately depart without receiving any service (such a customer is said to be lost), whereas those finding at least one free server immediately enter service and then depart when their service is completed. (a) If both servers are presently busy, find the expected time until the next customer enters the system. (b) Starting empty, find the expected time until both servers are busy. (c) Find the expected time between two successive lost customers.

Let \(X_{1}\) and \(X_{2}\) be independent exponential random variables, each having rate \(\mu .\) Let $$ X_{(1)}=\operatorname{minimum}\left(X_{1}, X_{2}\right) \text { and } X_{(2)}=\operatorname{maximum}\left(X_{1}, X_{2}\right) $$ Find (a) \(E\left[X_{(1)}\right]\) (b) \(\operatorname{Var}\left[X_{(1)}\right]\) (c) \(E\left[X_{(2)}\right]\) (d) \(\operatorname{Var}\left[X_{(2)}\right]\)

Consider a conditional Poisson process in which the rate \(L\) is, as in Example \(5.29\), gamma distributed with parameters \(m\) and \(p\). Find the conditional density function of \(L\) given that \(N(t)=n\).

Teams 1 and 2 are playing a match. The teams score points according to independent Poisson processes with respective rates \(\lambda_{1}\) and \(\lambda_{2} .\) If the match ends when one of the teams has scored \(k\) more points than the other, find the probability that team 1 wins. Hint: Relate this to the gambler's ruin problem.

A doctor has scheduled two appointments, one at \(1 \mathrm{P} . \mathrm{M}\). and the other at \(1: 30 \mathrm{P.M}\). The amounts of time that appointments last are independent exponential random variables with mean 30 minutes. Assuming that both patients are on time, find the expected amount of time that the \(1: 30\) appointment spends at the doctor's office.

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