CT3 IFoA October 2012 Question 4

Discussion in 'CT3' started by trackr, Mar 26, 2016.

  1. trackr

    trackr Member

    Hi there, I understand that they used convolutions for this question.

    However when I solved it without seeing the answer, I used MGFs. Would I get full marks for that method?

    I assumed U to be continuous uniform dist. So Mz(t)=E (exp(t^z)) = E (exp(t^(U+X))).

    Now since U & X are independent. I separated them out and simplified it. In the end I got the distribution of Z as Uniform dist with a = t/2 and b = (2+t)/2.

    Can someone reject or confirm my method?

    Thanks in advance!
     
  2. suraj

    suraj Member

    Range of Z is from -∞ to +∞. So it can't be a Uniform Distribution.
     
  3. John Lee

    John Lee ActEd Tutor Staff Member

    Interesting, I get an MGF which is no standard distribution...
     
  4. trackr

    trackr Member

    Hi John,

    Could you please elaborate on that? So is my method right/wrong? If wrong why so?
     
  5. John Lee

    John Lee ActEd Tutor Staff Member

    \(M_Z (t) = M_U (t) M_X (t) = 1/t (e^t - 1) * e^{½t^2} \)
     
    trackr likes this.
  6. trackr

    trackr Member

    Yeah that is what I got too.

    Thank you for replying :)
     
    John Lee likes this.

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