I just have a question regarding the answer of the 17th question.
As I've understood the question, the large C value condition imposes to w1 to be 0 and w2 a small value around 0. So after training the classifier should be determined only by signe(x2). Thus, the boundary line should be the horizontal line of equation x2=0 but the correction gives the opposite : x1=0 that for me is a vertical line that doesn't fit with the dataset.
Am I wrong and if so what am I missing?
Thanks by advance for answering!
Edit : I got it, I fell in the trap reading norm L2 of w instead of w2 square, that explains the answer.
I'm still a bit confused by the same question.
Even if w2 is close to 0, it will still be a (small) non-zero real number. Given this and the fact that w1 will not affect the regularization term, if we choose w1 = w2 we will have the line x1 + x2 = 0, which does not misclassify any samples and therefore makes the loss lower. What am I missing?
Exam 2021 question 17
Hello all !
I just have a question regarding the answer of the 17th question.
As I've understood the question, the large C value condition imposes to w1 to be 0 and w2 a small value around 0. So after training the classifier should be determined only by signe(x2). Thus, the boundary line should be the horizontal line of equation x2=0 but the correction gives the opposite : x1=0 that for me is a vertical line that doesn't fit with the dataset.
Am I wrong and if so what am I missing?
Thanks by advance for answering!
Edit : I got it, I fell in the trap reading norm L2 of w instead of w2 square, that explains the answer.
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I'm still a bit confused by the same question.
Even if w2 is close to 0, it will still be a (small) non-zero real number. Given this and the fact that w1 will not affect the regularization term, if we choose w1 = w2 we will have the line x1 + x2 = 0, which does not misclassify any samples and therefore makes the loss lower. What am I missing?
Boundary line is x1=0 meaning that you basically only use x1 for classification. Data points with x1<0 and those with x1>0 will be distinguished.
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