Sept 2011 Q.7 (ii)

Discussion in 'CT6' started by madhurikumar, Apr 9, 2014.

  1. madhurikumar

    madhurikumar Member

    Type 1 risk - 80% prob - Normal
    Type 2 Risk - 20% prob - Normal

    While I attempted this question , I presumed the loss distribution of the sum of these two Normal distributions was Normal.

    Could someone throw more light on this part of the solution and explain why the cumulative pdf isn't normal ?

    Thanks in advance.
     
  2. suraj

    suraj Member

    X is not sum of two Normal variables. It's a mixture of two Normal RV X1 and X2.

    So

    \(f_X(x) = 0.8~f_{X_1}(x_1) + 0.2~f_{X_2}(x_2) \)

    not X = 0.8 X1 + 0.2 X2

    See previous question on this paper which is also on Mixture Distribution.
     
    John Lee likes this.
  3. madhurikumar

    madhurikumar Member

    Argh. Yes I get it. Thank you Suraj.
     
  4. Jammy

    Jammy Member

    Hi, I still have the same doubt as madhuri, can someone please elaborate and clear this up?
     
  5. Katherine Young

    Katherine Young ActEd Tutor Staff Member

    Suraj is perfectly right:

    We don't have \(X = 0.8 X1 + 0.2 X2\) (which would be Normal).

    Instead we have \(fx(x)=0.8fx_1(x_1)+0.2fx_2(x_2)\) (which isn't Normal).

    To check this isn't Normal, you can substitute in the PDFs of a normal distribution into this equation, (one PDF for \(fx_1(x_1)\) and another PDF for \(fx_2(x_2)\), then try to rearrange your equation to make the whole thing look like just one PDF of a Normal distribution. You won't be able to do it.

    Try it and see!

    If you still don't believe us, google it! eg https://en.wikipedia.org/wiki/Mixture_distribution
     
    John Lee likes this.

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