# SVM hyperplane margin

so that $$H_0$$ is equidistant from $$H_1$$ and $$H_2$$.

However, here the variable $$\delta$$ is not necessary. So we can set $$\delta=1$$ to simplify the problem.

$$w\cdot x+b=1$$ and

$$w\cdot x+b=−1$$

Why is this assumption is taken? If it is taken, we can get the distance between two planes as $$2$$ directly because both are parallel and differ by $$2$$. how it is $$\frac{2}{\left\|w\right\|}$$ instead.

got equations from this https://www.svm-tutorial.com/2015/06/svm-understanding-math-part-3/

After we have

$$w^Tx + b = \pm \delta$$

We can always divide everything by $$\delta$$,

$$\left( \frac{w}{\delta}\right)^Tx + \left( \frac{b}{\delta}\right)=\pm1$$

Now, we can set $$\tilde{w}=\frac{w}{\delta}$$ and $$\tilde{b}=\frac{b}{\delta}$$.

$$\tilde{w}^Tx+\tilde{b}=\pm1$$

This is as if we have set $$\delta=1$$ from the beginning.

The derivation of the distance formula has been given in equation $$(19)$$ in the article that you linked to and you might like to be more specific if you can't understand it. The distance should be $$\frac{2\delta}{\|w\|}$$ if $$\delta$$ is not set to be $$1$$.

• Thanks, i have checked that equation (19). what am not able to get intuitively is if both hyper planes are parallel, after replacing W and X with scalar values what you get is two lines differing by 2 from -1 to 1 so the margin will be 2. Sep 16, 2019 at 3:52
• Referring equation (3) we get y-ax+b = ±1 so both differ by 2. Sep 16, 2019 at 3:59
• Note that the distance between $y=x+1$ and $y=x-1$ is $\sqrt2$ rather than $2$. The distance is measured along the normal direction. Sep 16, 2019 at 5:12