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Linear discriminant

  1. Nov 17, 2007 #1

    I am doing this exercise (2 class problem with 2-dimensional features) and I have solved the linear discriminant function which turns out be y1(x) - y2(x) = 2x1 +2x2

    I am having difficulty in finding the class posterior probabilities frm the linear discriminant function obtained.

    But I tried this way and got stuck.

    We know that yi(x) = ln{p(x|ci)p(Ci)}

    from the linear discriminant function obtained

    y1(x) = ln{p(x|C1)p(C1)} = 2x1 + 2x2 + y2(x)
    ln{p(x|C1)p(C1)} = 2x1 + 2x2 + ln{p(x|C2)p(C2)}
    ln[{p(x|C1)p(C1)} - {p(x|C2)p(C2)} = 2x1 + 2x2

    Using exponential for both side we get
    p(x|C1)p(C1)/p(x|C2)p(C2) = exp(2x1 + 2x2)

    p(C1) and p(C2) are constatnt so we can neglect us giving the following

    p(x|C1)/p(x|C2)= exp(2x1 + 2x2)

    From this point I am not sure how to separate both posterior probabilities.

    plz help...Thank you
    Last edited: Nov 17, 2007
  2. jcsd
  3. Nov 17, 2007 #2


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    As long as p(x|C1)/p(x|C2)= exp(2x1 + 2x2) is the only condition that needs to hold, p(x|C1) = exp(2x1), p(x|C2)= exp(-2x2) would satisfy it.
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