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Chapter 5: Problem 85

An insurance company pays out claims on its life insurance policies in accordance with a Poisson process having rate \(\lambda=5\) per week. If the amount of money paid on each policy is exponentially distributed with mean \(\$ 2000\), what is the mean and variance of the amount of money paid by the insurance company in a four-week span?

Short Answer

Expert verified
The mean and variance of the amount of money paid by the insurance company in a four-week span are \(40,000 and \)80,000,000, respectively.

Step by step solution

01

Find mean of Poisson distribution

The mean of a Poisson distribution with rate λ is equal to λ. In this scenario, the rate is 5 claims per week. The mean of the number of claims paid in a 4-week span is: Mean = λ × Time interval = 5 × 4 = 20 claims
02

Find the variance of Poisson distribution

The variance of a Poisson distribution is equal to the mean. Thus, the variance of the number of claims in the 4-week span is: Variance = Mean = 20 claims
03

Mean of Exponential distribution

In the given scenario, the amount paid on each policy is exponentially distributed with mean \(2000. So, the mean amount paid per claim is \)2000.
04

Variance of Exponential distribution

The variance of an exponential distribution with mean μ is: Variance = μ^2 In this case, the mean is $2000. So the variance of the amount paid per claim is: Variance = (2000)^2 = $4,000,000
05

Mean of total payout in 4-week span

Mean of the total amount paid in the 4-week span is equal to the product of the mean number of claims and the mean amount paid per claim: Mean payout = (Mean number of claims) × (Mean amount per claim) = (20 claims) × (\(2000) = \)40,000
06

Variance of total payout in 4-week span

To find the variance of the total payout, we use the formula for the variance of the sum of independent random variables: Variance of total payout = (Mean number of claims) × (Variance of amount per claim) = (20 claims) × (\(4,000,000) = \)80,000,000 So, the mean and variance of the amount of money paid by the insurance company in a four-week span are \(40,000 and \)80,000,000, respectively.

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Most popular questions from this chapter

The water level of a certain reservoir is depleted at a constant rate of 1000 units daily. The reservoir is refilled by randomly occurring rainfalls. Rainfalls occur according to a Poisson process with rate \(0.2\) per day. The amount of water added to the reservoir by a rainfall is 5000 units with probability \(0.8\) or 8000 units with probability \(0.2 .\) The present water level is just slightly below 5000 units. (a) What is the probability the reservoir will be empty after five days? (b) What is the probability the reservoir will be empty sometime within the next ten days?

Let \(X\) and \(Y\) be independent exponential random variables with respective rates \(\lambda\) and \(\mu\). (a) Argue that, conditional on \(X>Y\), the random variables \(\min (X, Y)\) and \(X-Y\) are independent. (b) Use part (a) to conclude that for any positive constant \(c\) $$ \begin{aligned} E[\min (X, Y) \mid X>Y+c] &=E[\min (X, Y) \mid X>Y] \\ &=E[\min (X, Y)]=\frac{1}{\lambda+\mu} \end{aligned} $$ (c) Give a verbal explanation of why \(\min (X, Y)\) and \(X-Y\) are (unconditionally) independent.

If \(X_{i}, i=1,2,3\), are independent exponential random variables with rates \(\lambda_{i}\), \(i=1,2,3\), find (a) \(\left.P \mid X_{1}

Two individuals, \(A\) and \(B\), both require kidney transplants. If she does not receive a new kidney, then \(A\) will die after an exponential time with rate \(\mu_{A}\), and \(B\) after an exponential time with rate \(\mu_{B} .\) New kidneys arrive in accordance with a Poisson process having rate \(\lambda\). It has been decided that the first kidney will go to \(A\) (or to \(B\) if \(B\) is alive and \(A\) is not at that time) and the next one to \(B\) (if still living). (a) What is the probability that \(A\) obtains a new kidney? (b) What is the probability that \(B\) obtains a new kidney?

(a) Let \(\\{N(t), t \geqslant 0\\}\) be a nonhomogeneous Poisson process with mean value function \(m(t) .\) Given \(N(t)=n\), show that the unordered set of arrival times has the same distribution as \(n\) independent and identically distributed random variables having distribution function $$ F(x)=\left\\{\begin{array}{ll} \frac{m(x)}{m(t)}, & x \leqslant t \\ 1, & x \geqslant t \end{array}\right. $$ (b) Suppose that workmen incur accidents in accordance with a nonhomogeneous Poisson process with mean value function \(m(t) .\) Suppose further that each injured man is out of work for a random amount of time having distribution F. Let \(X(t)\) be the number of workers who are out of work at time \(t\). By using part (a), find \(E[X(t)]\)
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