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9 October, 20:58

Two insureds own delivery vans. Insured A had two vans in year 1 and one claim, two vans in year 2 and one claim, and one van in year 3 with no claims. Insured B has no vans in year 1, three vans in year 2 and two claims, and two vans in year 3 and three claims. The number of claims for insured each year has a Poisson distribution. Use semiparametric empirical Bayes estimation to obtain the estimated number of claims for each insured in year 4.

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  1. 9 October, 21:14
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    For A: 2.208

    For B: 2.9024

    Step-by-step explanation:

    To solve we will do the following:

    For A:

    no insured (x) claims (y) xy Var=total / (y-1)

    2 1 2 0.08

    2 1 2 0.08

    1 0 0 0.64

    Weighted average 0.8 0.16

    a var - X - 0.64

    k mean/a - 1.25

    Z (credibility) 1 / (1+k) 0.444

    no of claims for next year (y=4)

    = 0.44*4 + 0.56*0.8

    = 2.208

    For B:

    no insured (x) claims (y) xy Var=total / (y-1)

    0 0 0 0

    3 6 12 12

    2 3 6 2

    Weighted average 2.4 3.5

    a var - X 1.1

    k mean/a 2.18

    Z (credibility) 1 / (1+k) 0.314

    no of claims for next year (y=4)

    0.314*4 + 0.686*2.4

    = 2.9024
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