iai oct 2014

Discussion in 'CT6' started by djjain80, Aug 11, 2017.

  1. djjain80

    djjain80 Member

    Que.4 part 1
    Could Mx(t) given in d solution b simplified as
    (.9+.1exp5t)^5
     
  2. John Lee

    John Lee ActEd Tutor Staff Member

    I'd love to help - but without a photo/scan of the question and the line in the solution it makes it a lot more effort. ;)
     
  3. vgarg

    vgarg Member

    Let \(N\) be the number of car accidents, \(X_i\) be the cost of replacement of tyres on \(i^{th}\) car involved in accident and \(Y_i\) be the number of tyres that require replacement on \(i^{th}\) car involved in an accident. Then we have the following

    • \(N \sim Bin(500, p), M_N(t) = (1-p + pe^t)^{500}\)
    • \(Y_i \overset{i.i.d}{\sim} Bin(4, 0.1), M_{Y_i}(t) = (0.9 + 0.1e^t)^4 \)
    • \(X_i = 5Y_i \quad \forall i \)
    • \(S = \sum_{i=1}^{N}X_i = \sum_{i=1}^{N}5Y_i\)
    The moment generating function of \(S\) is then

    \begin{align} M_S(t) &= \mathbb E\left [e^{t\sum_{i=1}^{N}5Y_i} \right ]\\ &= \mathbb E\left [(M_{Y_i}(5t))^N \right ]\\ &= M_N(\ln(M_{Y_i}(5t))) \\ &= (1-p + p((0.9 + 0.1e^{5t})^4))^{500} \end{align}
     
    John Lee and djjain80 like this.

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