Var[S(2)] Chapter 18 Page 16

Discussion in 'CM2' started by Thabo Motaung, Feb 22, 2024.

  1. Thabo Motaung

    Thabo Motaung Member

    Good day

    In the Course Notes on page 16 there is a Question that wants us to calculate the probability of the insurer being insolvent based on the following information/assumptions:

    N(2) ~ Poi(60), X ~ logN(3, 1.1), c = 1200, U = 1000
    Now when it comes to calculating Var[S(2)] i struggle to understand how they come up with the solution:

    Since S(2) = X1 + X2 + ... + XN(60). And so E[S(2)] = 60*exp{mu + 0.5*sigma^2} ( This I understand how they got here)

    But now how did they get to calculate Var[S(2)] = 60*exp{6 + 2.2} ?

    I would have thought that Var[S(2)] = 60^2*exp(6 + 1.1)*(exp(1.1) - 1)
     
  2. Thabo Motaung

    Thabo Motaung Member

    Edit: Saw my mistake, it should be Var[S(t)] = lambda*E[X^2] = lambda*[Var(X) + [E(x)]^2]
     
    CapitalActuary likes this.

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