Ch9, section 4.7 concluding remarks.

Discussion in 'CT6' started by Ark raw, Jul 30, 2017.

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  1. Ark raw

    Ark raw Member

    I've been stuck here for a while now. The things I couldn't understand are:
    1. how ultimate ruin is certain when loading factor=0 and F(x) has any form?
    2. how ψ(U) when F(x)=1-e^(-αx) is the same as ψ(αU) when F(x)=1-e^(-x) ?


     

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    Ark raw, Jul 30, 2017
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  2. John Lee

    John Lee ActEd Tutor Staff Member

    1. I believe the proof is quite messy - but intuitively if you only collect premium equal to the mean of the claims - then you've got a horizontal random walk and so eventually you WILL get a negative value.
    2. You're just changing the units. So if \(X \sim Exp(\alpha)\) then using functions of a random variable from CT3 chapter 3, when \(Y=\alpha X\) we will have \(F_Y (y) = F_X(y/\alpha) = 1-e^{-y}\)
     
    Last edited: Aug 11, 2017
    John Lee, Aug 2, 2017
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  3. Ark raw

    Ark raw Member

    But I still can't understand how it relates to ψ(U) and ψ(αU)
     
    Ark raw, Aug 10, 2017
    #3
  4. John Lee

    John Lee ActEd Tutor Staff Member

    If we're working in pounds then U is in pounds, claims X are in pounds and the probability of ruin is \(\psi(U)\).

    if we're now working in pence then Y=100X and we'll now have surplus 100U and the probability of ruin is \(\psi(100U)\)
     
    John Lee, Aug 11, 2017
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