Answer:
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
[tex]10,000^{2}(0.01) = 1,000,000\\20,000^{2}(0.02) = 8,000,000\\40,000^{2}(0.03) = 48,000,000.[/tex]
The overall mean is
[tex]5,000(100)+3,000(400)+1,000(1,200)=2,900,000[/tex]
and the variance is [tex]5,000(1,000,000)+3,000(8,000,000)+1,000(48,000,000 )=7.7*10^{10}[/tex]
Total claims are the sum of 9,000 compound Poisson distributions which itself is a compound Poisson distribution with
[tex]\lambda = 5,000(0.01)+3,000(0.02)+1,000(0.03)=140[/tex]
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
[tex]\lambda = 5,000(0.01)+3,000(0.02)+1,000(0.03)=140[/tex]
The overall mean is [tex]5,000(100)+3,000(400)+1,000(1,200)=2,900,000[/tex] and the variance is [tex]5,000(1,000,000)+3,000(8,000,000)+1,000(48,000,000 )=7.7*10^{10} .[/tex]
Total claims are the sum of 9,000 compound Poisson distributions which itself is a compound Poisson distribution with [tex]\lambda = 5,000(0.01)+3,000(0.02)+1,000(0.03)=140[/tex] 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
[tex]P(S=0)=e^{-140}[/tex]
[tex]P(S=1)=(140/1)*(5/14)*e^{-140}=50 e^{-140}[/tex]
[tex]P(S=2)=140/2[(5/14)*50*e^{-140}+(2*6/14)e^{-140}]=1310e^{-140}[/tex]
[tex]P(S=3)=140/3[(5/14)*1,310e^{-140}+(2*6/14)e^{-140}]=21,873.33e^{-140}[/tex]
P(S > 3) = 1 - P (S ≤ 3)
= [tex]1 - 23,233.33e^{-140}[/tex]
