Chapt 8 - Risk Models 2

Discussion in 'CT6' started by MindFull, Mar 20, 2010.

  1. MindFull

    MindFull Ton up Member

    The variability in claim numbers and claim amounts and parameter uncertainty...The example that was given, I'm really not too sure how the values were achieved. For E(X ij) = E (the upside down V ) = p... I'm not sure how that step goes. Could I have some help?

    Thanks.
     
  2. MindFull

    MindFull Ton up Member

    Any help with this??
     
  3. John Lee

    John Lee ActEd Tutor Staff Member

    Apologies your question slipped through the net.

    This part of the chapter shows how the examinable stuff (earlier in the chapter) can be applied to real life situations. As such exam questions don't test this stuff directly though could get you to work through similar things step by step.

    It is a bit odd.

    Xij is lognormal with mean Vi.
    But Vi is a RV with has mean p.

    So E(Xij) = E[E(Xij|Vi)] using the formula from the top of page 16 of the Tables.

    But E(Xij|Vi) is the mean of Xij when the mean (Vi) is fixed (say v):

    E(Xij|Vi = vi) = vi

    Now we let the vi vary and find the mean of it:

    E(Xij) = = E[E(Xij|Vi)] = E[Vi]

    Since Vi has mean p. Hence:

    E(Xij) = = E[E(Xij|Vi)] = E[Vi] = p
    This example was asked in September 2004 Q2.
     

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