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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) Show that the limiting probabilities of the reversed Markov chain are the same as for the forward chain by showing that they satisfy the equations $$ \pi_{j}=\sum_{i} \pi_{i} Q_{i j} $$ (b) Give an intuitive explanation for the result of part (a).

On a chessboard compute the expected number of plays it takes a knight, starting in one of the four corners of the chessboard, to return to its initial position if we assume that at each play it is equally likely to choose any of its legal moves. (No other pieces are on the board.) Hint: Make use of Example 4.36.

A taxi driver provides service in two zones of a city. Fares picked up in zone \(A\) will have destinations in zone \(A\) with probability \(0.6\) or in zone \(B\) with probability \(0.4\). Fares picked up in zone \(B\) will have destinations in zone \(A\) with probability \(0.3\) or in zone \(B\) with probability \(0.7 .\) The driver's expected profit for a trip entirely in zone \(A\) is 6 ; for a trip entirely in zone \(B\) is \(8 ;\) and for a trip that involves both zones is 12 . Find the taxi driver's average profit per trip.

For a branching process, calculate \(\pi_{0}\) when (a) \(P_{0}=\frac{1}{4}, P_{2}=\frac{3}{4}\). (b) \(P_{0}=\frac{1}{4}, P_{1}=\frac{1}{2}, P_{2}=\frac{1}{4}\). (c) \(P_{0}=\frac{1}{6}, P_{1}=\frac{1}{2}, P_{3}=\frac{1}{3}\).

Suppose in the gambler's ruin problem that the probability of winning a bet depends on the gambler's present fortune. Specifically, suppose that \(\alpha_{i}\) is the probability that the gambler wins a bet when his or her fortune is \(i\). Given that the gambler's initial fortune is \(i\), let \(P(i)\) denote the probability that the gambler's fortune reaches \(N\) before 0 . (a) Derive a formula that relates \(P(i)\) to \(P(i-1)\) and \(P(i+1)\). (b) Using the same approach as in the gambler's ruin problem, solve the equation of part (a) for \(P(i)\) (c) Suppose that \(i\) balls are initially in urn 1 and \(N-i\) are in urn 2, and suppose that at each stage one of the \(N\) balls is randomly chosen, taken from whichever urn it is in, and placed in the other urn. Find the probability that the first urn becomes empty before the second.

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