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I am struggling with this point (since my Math is abysmal). I have looked at various YouTube videos (most of them just gloss over this or the Math is beyond me) and read various blog posts (same problem).

Basically, all I want to understand is: Why do the equations for the hyperplanes that the support vectors are on look like this?

$\vec{w}^T\vec{x}+b=1$

$\vec{w}^T\vec{x}+b=-1$

Why 1/-1?

I understand that for points on the hyperplane, this is true:

$\vec{w}^T\vec{x}+b=0$

For points "above" or in the direction of $\vec{w}$ it's $ > 0 $, for the other case it's $ < 0 $

I just don't really understand where the 1/-1 come from. Form what I have read, it has to do with scaling and ultimately choosing 1 is arbitrary, but I don't get it.

What is the problem with the scaling and I kind of makes the optimization problem easier, but the penny just won't drop.

I understand that by doing $c\vec{w}^T\vec{x}+cb=0$, I get the same hyperplane, but I don't understand why because of that I can just go: $\vec{w}^T\vec{x}+b=1$.

So in my head I am picturing this:

The equation for the support vectors are $\vec{w}^T\vec{x}+b= +d / -d $. I basically make two copies of my hyperplane and move them upwards/downwards until each copy hits the first data point and then I place my hyperplane in the middle, so it is equidistant from the other two hyperplanes. Then my hyperplane is the one with the maximum margin.

But I still don't understand why it is okay to just say that d = 1.

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1 Answer 1

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By definition $y$ is labeled as $\{-1, 1\}$. A separating hyperplane is one such that

$\vec{w}^T\vec{x} + b - y > 0$ when $y = 1$

and

$\vec{w}^T\vec{x} + b - y < 0$ when $y = -1$

The above inequalities become equalities at the margins of a maximal margin classifier "band", which is what you have shown above. Which class is labeled as $-1$ and which as $1$ is chosen arbitrary here.

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  • $\begingroup$ hang on a second... why -y? It though it was just wx+b > 0?! $\endgroup$ Commented Oct 21, 2018 at 14:55

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