# Linear discriminant

1. Nov 17, 2007

### sensitive

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. Nov 17, 2007

### EnumaElish

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.