Finding Class Posterior Probabilities from Linear Discriminant Function

  • Context: Graduate 
  • Thread starter Thread starter sensitive
  • Start date Start date
  • Tags Tags
    Linear
Click For Summary
SUMMARY

The discussion focuses on deriving class posterior probabilities from a linear discriminant function for a two-class problem with two-dimensional features. The linear discriminant function is defined as y1(x) - y2(x) = 2x1 + 2x2. The user successfully manipulates the equation to express the ratio of class conditional probabilities, leading to the equation p(x|C1)/p(x|C2) = exp(2x1 + 2x2). The conclusion indicates that specific forms for p(x|C1) and p(x|C2) can be proposed, such as p(x|C1) = exp(2x1) and p(x|C2) = exp(-2x2), to satisfy the derived condition.

PREREQUISITES
  • Understanding of Linear Discriminant Analysis (LDA)
  • Familiarity with class conditional probabilities
  • Knowledge of logarithmic and exponential functions
  • Basic concepts of Bayesian inference
NEXT STEPS
  • Research the derivation of posterior probabilities in Bayesian classification
  • Study the properties and applications of Linear Discriminant Analysis (LDA)
  • Explore the relationship between likelihood functions and posterior probabilities
  • Learn about the implications of class prior probabilities in classification tasks
USEFUL FOR

Data scientists, machine learning practitioners, and statisticians interested in classification techniques and the application of Linear Discriminant Analysis in probabilistic models.

sensitive
Messages
33
Reaction score
0
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.


please help...Thank you
 
Last edited:
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 2 ·
Replies
2
Views
9K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 0 ·
Replies
0
Views
1K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
9K