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individual risk model continued

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  • individual risk model continued

    I tried posting an attachment, however when I tried to view it I couldn't.

    I will restate the problem so that it can be read properly.

    A portfolio of independent policies has three classes of policies.

    class 1
    # in class = 100
    prob of claim/policy
    claim amount = 3

    class 2
    # in class = 200
    prob of claim/policy = 0.02
    claim amount = 1

    class 3
    # in class = 50
    prob of claim/policy = 0.04
    claim amount = 2


    if the relative security loading theta is set such that the probability of the aggregate claim amounts exceed the total premium is 0.05, determine theta using the normal approximation to the aggregate claims.

    Thank you

  • #2
    {(Theta)E[X]-E[X]}/Var(X)=1.645 Where X is the total claims.
    Last edited by Ken; December 4 2005, 01:18 PM.
    Whether you are the lion or the gazelle, when the sun comes up, you better be running.


    • #3
      Originally posted by Ken
      {(Theta)E[X]-E[X]}/Var(X)=1.645 Where X is the total claims.
      I'm not so sure. If we let L denote aggregate claims and P denote total premiums, we are asked to find the relative security loading theta such that Pr(L>P)=.05. This implies that Pr(L<P)=.95. Now, since aggregate losses are assumed to be normally distributed, we perform the normal approximation by subtracting E(L) and then dividing by sqrt(Var(L)) to transform to a standard normal distribution: Pr(L<P)=Pr((L-E(L))/sqrt(Var(L)) < (P-E(L))/sqrt(Var(L)))=Pr(Z<(P-E(L))/sqrt(Var(L)))=.95.

      Directly following from the last equality, (P-E(L))/sqrt(Var(L))=1.645. But by definition, (1+theta)=P/E(L), so we can write P as (1+theta)E(L). This allows us to write P-E(L) as (1+theta)E(L)-E(L)=theta*E(L). So, I believe our final equality would be theta*E(L)/sqrt(Var(L))=1.645 as opposed to (theta*E(L)-E(L))/Var(L)=1.645.

      Professor Broverman, Professor K., or anyone else, could you please confirm the correct equality, especially if neither of these is it? Thanks.
      act justly. walk humbly. .


      • #4

        Your analysis looks correct. It looks like Ken
        used theta instead of 1+theta , and also
        left out the square root in the denominator.
        Sam Broverman

        [email protected]


        • #5
          As always, thank you for taking the time to help with this, Prof. Broverman.
          act justly. walk humbly. .