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Chapter 4: Problem 53

Find the average premium received per policyholder of the insurance company of Example \(4.27\) if \(\lambda=1 / 4\) for one-third of its clients, and \(\lambda=1 / 2\) for two-thirds of its clients.

Short Answer

Expert verified
The average premium received per policyholder of the insurance company is \(3/4\).

Step by step solution

01

Identify Premium Values and Weights

We have 2 different premiums received per policyholder: - λ = 1/4 for one-third of its clients (weight 1/3) - λ = 1/2 for two-thirds of its clients (weight 2/3)
02

Calculate Weighted Sum of Premiums

To find the weighted sum, multiply each premium value by its corresponding weight, and then add the results together. Weighted Sum = (1/4 * 1/3) + (1/2 * 2/3)
03

Simplify Weighted Sum

Now, we will simplify the weighted sum by performing the multiplication operations. Weighted Sum = (1/12) + (2/3)
04

Calculate Common Denominator

To add the two fractions, we need to find the common denominator, which is the smallest common multiple of 12 and 3. Common Denominator = 12
05

Add Fractions

Now, we can add the two fractions using the common denominator of 12. Weighted Sum = (1/12) + (8/12) = (1+8)/12 = 9/12
06

Simplify Fraction

Finally, we can simplify the fraction to get the final result. Average Premium = 9/12 = 3/4 The average premium received per policyholder of the insurance company is \(3/4\).

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Key Concepts

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

Weighted Average
Understanding the concept of a weighted average is crucial when dealing with diverse sets of data that have different levels of importance or frequency. In the context of insurance, for example, the premium received per policyholder can vary and these variations are not equally distributed. We can think of a weighted average as a means of balancing out these discrepancies to find a single representative value.

In the exercise given, to compute the average premium, premium values are given different 'weights' based on the proportion of clients they apply to. For instance, a premium paid by one-third of clients has a lower impact on the overall average than that paid by two-thirds of the clients. To perform this calculation, each premium value is multiplied by its respective weight, and then all of these products are summed up to find the total weighted sum. Subsequently, the sum is simplified to find the representative average premium which, in this case, is calculated as 3/4.
Probability Models
Probability models are mathematical representations that help us understand and predict the likelihood of different outcomes. In insurance, for instance, these models can forecast the probability of certain events, such as claims or policy cancellations. The premise is based on defining a set of all possible outcomes (the sample space) and assigning probabilities to these that reflect their likelihood of occurrence.

For the calculation of an average premium in the textbook problem, \( \lambda \) symbolizes the premium rate. It reflects a model where different premium rates apply to different client groups, each with a specific probability—symbolized as one-third and two-thirds in the exercise. This indicates the model's inclusion of varying likelihoods which are critical for the calculation of the weighted average premium.
Mathematical Expectation
The term mathematical expectation, often used interchangeably with 'expected value', is fundamental in statistical and financial analyses. The expectation represents the long-term average outcome of a random variable if its experiment were to be repeated numerous times.

From the viewpoint of an insurance company, mathematical expectation helps in estimating the average amount that should be charged as a premium to not only cover all claims but also generate profit. The process involves multiplying each potential outcome by its probability and summing these products. In our exercise scenario, the expected value would be the average premium, which is determined by taking into account that different premiums have different chances of occurrence based on the client distribution. This weighted approach aligns closely with the mathematical expectation of revenue the insurance company would have from its policyholders.

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Most popular questions from this chapter

A professor continually gives exams to her students. She can give three possible types of exams, and her class is graded as either having done well or badly. Let \(p_{i}\) denote the probability that the class does well on a type \(i\) exam, and suppose that \(p_{1}=0.3, p_{2}=0.6\), and \(p_{3}=0.9 .\) If the class does well on an exam, then the next exam is equally likely to be any of the three types. If the class does badly, then the next exam is always type \(1 .\) What proportion of exams are type \(i, i=1,2,3 ?\)

Consider the Ehrenfest urn model in which \(M\) molecules are distributed between two urns, and at each time point one of the molecules is chosen at random and is then removed from its urn and placed in the other one. Let \(X_{n}\) denote the number of molecules in urn 1 after the \(n\) th switch and let \(\mu_{n}=E\left[X_{n}\right]\). Show that (a) \(\mu_{n+1}=1+(1-2 / M) \mu_{n}\). (b) Use (a) to prove that $$ \mu_{n}=\frac{M}{2}+\left(\frac{M-2}{M}\right)^{n}\left(E\left[X_{0}\right]-\frac{M}{2}\right) $$

A Markov chain is said to be a tree process if (i) \(\quad P_{i j}>0\) whenever \(P_{j i}>0\), (ii) for every pair of states \(i\) and \(j, i \neq j\), there is a unique sequence of distinct states \(i=i_{0}, i_{1}, \ldots, i_{n-1}, i_{n}=j\) such that $$ P_{i_{k}, i_{k+1}}>0, \quad k=0,1, \ldots, n-1 $$ In other words, a Markov chain is a tree process if for every pair of distinct states \(i\) and \(j\) there is a unique way for the process to go from \(i\) to \(j\) without reentering a state (and this path is the reverse of the unique path from \(j\) to \(i\) ). Argue that an ergodic tree process is time reversible.

Consider three urns, one colored red, one white, and one blue. The red urn contains 1 red and 4 blue balls; the white urn contains 3 white balls, 2 red balls, and 2 blue balls; the blue urn contains 4 white balls, 3 red balls, and 2 blue balls. At the initial stage, a ball is randomly selected from the red urn and then returned to that urn. At every subsequent stage, a ball is randomly selected from the urn whose color is the same as that of the ball previously selected and is then returned to that urn. In the long run, what proportion of the selected balls are red? What proportion are white? What proportion are blue?

Consider the following approach to shuffling a deck of \(n\) cards. Starting with any initial ordering of the cards, one of the numbers \(1,2, \ldots, n\) is randomly chosen in such a manner that each one is equally likely to be selected. If number \(i\) is chosen, then we take the card that is in position \(i\) and put it on top of the deck-that is, we put that card in position 1 . We then repeatedly perform the same operation. Show that, in the limit, the deck is perfectly shuffled in the sense that the resultant ordering is equally likely to be any of the \(n !\) possible orderings.
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