April 2019 Q6

Discussion in 'SP8' started by Jun Wu, Apr 12, 2022.

  1. Jun Wu

    Jun Wu Active Member

    Dear All
    Hope you are well

    I find this question very confusing at the set up

    Solution sets out N|Phi=Theta ~Binomial (3,theta) , why is Phi= Theta? are we using this interchangeably?

    I get N the number of claims are being told to be Binomial

    But i am confused with this set up : ie ' N with parameter Phi follows a binomial distribution with parameters n = 3 and p = Theta
    The parameter Phi has the pdf .... which involves Theta. Then why do we set Phi = Theta?

    A more intuitive set up is from Page 39 of CH18, where we have loss ratio depending on theta, loss ratio Xk itself is gamma dist (Vk, Vk/Theta), Theta itself has its own distribution of uniform (50,500). To me this means distribution of Xk depends on theta, so it makes sense to me when working out Beta, small Phi and Lambda and then Z.

    Thanks !!
    Jun
     
  2. Busy_Bee4422

    Busy_Bee4422 Ton up Member

    Hi Jun

    You need to be very careful conceptually because there at least three traps.

    The use of the dummy variables capital letter phi and theta is different from the notes.

    The exam question is dealing with claim numbers hence the use of the binomial distribution. If you reread the question you will see that the binomial distribution relates to the number of claims for an individual.

    Ordinarily you would be estimating the benchmark claims ratio and its moments. In this case you are estimating the benchmark claim numbers per individual and its moments. Your claim numbers per individual is essentially your 'claims ratio'.
     
    Jun Wu likes this.
  3. o.menary11

    o.menary11 Keen member

    Hi, i am struggling to understanding the following transition in the solution to this question, could i have some insight into any missing steps please?


    E[3 Φ(1- Φ)]= ∫ 3θ(1- θ)π(θ)dθ


    thanks
     
  4. Ian Senator

    Ian Senator ActEd Tutor Staff Member

    No missing steps, just stats of expected values, which says E[x]=int[xf(x)dx]
     

Share This Page