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linear discriminant |
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| Nov17-07, 08:39 AM | #1 |
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linear discriminant
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 |
| Nov17-07, 01:52 PM | #2 |
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Recognitions:
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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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