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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 位 is equal to 位. 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 = 位 脳 Time interval = 5 脳 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 渭 is: Variance = 渭^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) 脳 (Mean amount per claim) = (20 claims) 脳 (\(2000) = \)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) 脳 (Variance of amount per claim) = (20 claims) 脳 (\(4,000,000) = \)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

Cars pass a certain street location according to a Poisson process with rate \(\lambda\). A woman who wants to cross the street at that location waits until she can see that no cars will come by in the next \(T\) time units. (a) Find the probability that her waiting time is \(0 .\) (b) Find her expected waiting time. Hint: Condition on the time of the first car.

An event independently occurs on each day with probability \(p .\) Let \(N(n)\) denote the total number of events that occur on the first \(n\) days, and let \(T_{r}\) denote the day on which the \(r\) th event occurs. (a) What is the distribution of \(N(n) ?\) (b) What is the distribution of \(T_{1}\) ? (c) What is the distribution of \(T_{r} ?\) (d) Given that \(N(n)=r\), show that the set of \(r\) days on which events occurred has the same distribution as a random selection (without replacement) of \(r\) of the values \(1,2, \ldots, n\).

Suppose that \(\left\\{N_{0}(t), t \geqslant 0\right\\}\) is a Poisson process with rate \(\lambda=1 .\) Let \(\lambda(t)\) denote a nonnegative function of \(t\), and let $$ m(t)=\int_{0}^{t} \lambda(s) d s $$ Define \(N(t)\) by $$ N(t)=N_{0}(m(t)) $$ Argue that \(\\{N(t), t \geqslant 0\\}\) is a nonhomogeneous Poisson process with intensity function \(\lambda(t), t \geqslant 0 .\) Hint: Make use of the identity $$ m(t+h)-m(t)=m^{\prime}(t) h+o(h) $$

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Let \(X\) and \(Y\) be independent exponential random variables with respective rates \(\lambda\) and \(\mu\). Let \(M=\min (X, Y)\). Find (a) \(E[M X \mid M=X]\) (b) \(E[M X \mid M=Y]\), (c) \(\operatorname{Cov}(X, M)\)
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