# Why can we do this in SVM optimization?

I posted this question in the datascience stackexchange but no one answered so I thought I would post it here aswell:

Ok, so I've been trying to read up on how SVM:s work and started with maximal margin classifiers. At page 132 in ESL (Elements of Statistical Learning) the authors "reformulates" the optimization problem but I can't seem to understand what they are doing from 4.47 to 4.48. Does anyone know?

Here is an excerpt: What I don't understand is why we can arbitrarly set the magnitude of beta to 1/M. What does a positively scaled multiple mean in this case? Just a multiple larger than 0?

For reference, the optimization problem in hand is $$\begin{gather}\tag{1} \max_{\beta, \beta_0} M\\ \text{subject to }y_i(x_i^T\beta + \beta_0)\geq M\|\beta\|, \forall i. \end{gather}$$
Assume the pair $$(\widehat{\beta}, \widehat{\beta}_0)$$ is a solution to this problem and let $$\widehat{M}$$ be the achieved objective value. Also assume that $$(\beta^*, \beta_0^*)$$ is a solution to $$(4.48)$$. We have $$\begin{equation*} y_i(x_i^T\widehat{\beta} + \widehat{\beta}_0) \geq \widehat{M}\| \widehat{\beta} \|, \forall i \end{equation*}$$ which is equivalent to $$\begin{equation*} y_i \left( x_i^T \frac{\widehat{\beta}}{\| \widehat{\beta} \|\widehat{M}} + \frac{\widehat{\beta}_0}{\| \widehat{\beta} \|\widehat{M}} \right) \geq 1,\forall i \end{equation*}$$ therefore, the pair $$\left( \frac{\widehat{\beta}}{\|\widehat{\beta}\|\widehat{M}}, \frac{\widehat{\beta}_0}{\| \widehat{\beta}\| \widehat{M}} \right)$$ satisfies the constraints of $$(4.48)$$. Since $$(\beta^*, \beta_0^*)$$ is a solution to $$(4.48)$$, we get $$\begin{equation*} \| \beta^*\| \leq \left\| \frac{\widehat{\beta}}{\| \widehat{\beta} \| \widehat{M}}\right\| = \frac{1}{\widehat{M}} \qquad\Rightarrow\qquad 1 \geq \widehat{M}\| \beta^* \|. \end{equation*}$$ But also $$\begin{equation*} y_i \left( x_i^T \beta^* + \beta_0^* \right) \geq 1 \geq \widehat{M}\| \beta^*\|,\forall i \end{equation*}$$ hence, $$(\beta^*, \beta_0^*)$$ satisfy the constraints of $$(1)$$.
• We assumed that $(\beta^*, \beta_0^*)$ is a solution to (4.48). This means that among all other pairs $(z, z_0)$ that also satisfy the constraints of (4.48), $\|\beta^*\|^2$ must be less than or equal to $\| z\|^2$ (since the norm squared was our minimization objective and $(\beta^*, \beta_0^*)$ was the solution that minimizes it). Dropping the squares and letting $(\widehat{\beta}/(\| \widehat{\beta} \|\widehat{M}), \widehat{\beta}_0/(\| \widehat{\beta} \|\widehat{M}))$ be our $(z, z_0)$, the result above follows. – EuxhenH Mar 24 at 23:29