Question

A company sells three policies. For policy A, all claim payments are 10,000 and a single policy has a Poisson number of claims with mean 0.01. For policy B, all claim payments are 20,000 and a single policy has a Poisson number of claims with mean 0.02. For policy C, all claim payments are 40,000 and a single policy has a Poisson number of claims with mean 0.03. All policies are independent. For the coming year there are 5,000, 3,000, and 1,000 of policies A, B, and C, respectively. Calculate the expected total payment, the standard deviation of total payment, and the probability that total payments will exceed 30,000.

Answer 1

For the three polices,

The means are :

10,000(0.01) = 100

20,000(0.02) = 400

40,000(0.03) = 1,200

Because the Poisson variance is equal to the mean, the variances are

`10,000^(2)(0.01) = 1,000,000\\20,000^(2)(0.02) = 8,000,000\\40,000^(2)(0.03) = 48,000,000.`

The overall mean is

`5,000(100)+3,000(400)+1,000(1,200)=2,900,000`

and the variance is
`5,000(1,000,000)+3,000(8,000,000)+1,000(48,000,000 )=7.7*10^(10)`

Total claims are the sum of 9,000 compound Poisson distributions which itself is a compound Poisson distribution with

`\lambda = 5,000(0.01)+3,000(0.02)+1,000(0.03)=140`

and the severity distribution places probability 50/140 on 10,000, 60/140 on 20,000, and 30/140 on 40,000. Using the recursive formula with units of 10,000,

P(S=0)=e

For the three polices, the means are

10,000(0.01) = 100, 20,000(0.02) = 400, and 40,000(0.03) = 1,200.

Because the Poisson variance is equal to the mean, the variances are

`\lambda = 5,000(0.01)+3,000(0.02)+1,000(0.03)=140`

The overall mean is
`5,000(100)+3,000(400)+1,000(1,200)=2,900,000` and the variance is
`5,000(1,000,000)+3,000(8,000,000)+1,000(48,000,000 )=7.7*10^(10) .`

Total claims are the sum of 9,000 compound Poisson distributions which itself is a compound Poisson distribution with
`\lambda = 5,000(0.01)+3,000(0.02)+1,000(0.03)=140` and the severity distribution places probability 50/140 on 10,000, 60/140 on 20,000, and 30/140 on 40,000. Using the recursive formula with units of 10,000

`P(S=0)=e^(-140)`

`P(S=1)=(140/1)*(5/14)*e^(-140)=50 e^(-140)`

`P(S=2)=140/2[(5/14)*50*e^(-140)+(2*6/14)e^(-140)]=1310e^(-140)`

`P(S=3)=140/3[(5/14)*1,310e^(-140)+(2*6/14)e^(-140)]=21,873.33e^(-140)`

P(S > 3) = 1 - P (S ≤ 3)

=
`1 - 23,233.33e^(-140)`