Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Linear discriminant

  1. Nov 17, 2007 #1
    Hi

    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

    EnumaElish

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Linear discriminant
  1. Linear Correlation (Replies: 2)

  2. Linear estimation (Replies: 2)

  3. Linear regression (Replies: 7)

  4. Linear MMSE (Replies: 1)

Loading...