Can anyone explain in detail why $wx^++b=1$ for positive support vectors in SVM and $wx^-+b=-1$ for negative support vectors in SVM?
1 Answer
The decision boundary of SVM is $wx+b=0$. When you shift this line to positive and negative sides by some amount, you'll get parallel lines of the form: $wx+b=\pm d$. $wx+b=d$ will hit positive SVs and $wx+b=-d$ will hit negative SVs. Instead of using a third variable ($d$) here, we could scale $w$ and $b$ such that our line does not change. For example, $2x+3y=0$ and $4x+6y=0$ are still the same. Similarly, one could search for $w$ and $b$ such that $wx+b=\pm1$ at SVs without loss of generality.
-
$\begingroup$ Hi, so according to what you wrote, 2x+2 and 4x+4 are the same line? And could you given an example of expressing the same line in the 3 forms you showed? $\endgroup$ Commented Oct 21, 2018 at 14:15