Support Vector Machine Initial Idea

27 Nov 2017

Initial Idea

Now, we know the distance from $H$ to $H_1$ is $\frac{\left|w^t\cdot x-b\right|}{\left\|w\right\|}=\frac1{\left\|w\right\|}$; where $H_1:w^t\cdot x-b=1$. As to distance between $H_1$ and $H_2$, we must minimize the term $\left\|w\right\|=(w^t\cdot w)^{\textstyle\frac12}$ to satisfy the condition that no points between $H_1$ and $H_2$.

Formulate Our Problem By Lagrange Multiplier

More precisely, by minimizing $w^t\cdot w$, we expect to have:
➀for positive examples, $w^t\cdot x-b\geq1$, where $y=1$.
➁for negative examples, $w^t\cdot x-b\leq-1$, where $y=-1$.
Then, combine ➀ and ➁ we can have:
$y_i\cdot(w^t\cdot x_i-b)\geq1$, $\forall i\in\{1,2,...,n\}$
We can formulate our problem as:
$\underset w{min}\frac12w^t\cdot w$, subject to $y_i\cdot(w^t\cdot x_i-b)\geq1,\forall i$.
The first part is the target, the second part is the constraint.

We introduce $\alpha_1,\alpha_2,\dots\alpha_n$ to be the lagrange multiplier and express in below lagrangian to be our objective function:
$L(w,b,\alpha)=\frac12w^t\cdot w-\sum_{i=1}^n\alpha_i\cdot(y_i\cdot(w^t\cdot x_i-b)-1)$
;where $L$ is actually the maximum likelihood function for $w$, $\alpha$, $b$ to be estimated out the best optimal value at the corresponding largest possibility.

Regularize The Objective Function

Next to inspect inside $L(w,b,\alpha)$ to get the best optimal value of $w$, $\alpha$, $b$:
$\begin{array}{l}L(w,b,\alpha)\\=\frac12w^t\cdot w-\sum_{i=1}^n\alpha_i\cdot(y_i\cdot(w^t\cdot x_i-b)-1)\\=\frac12w^t\cdot w-\sum_{i=1}^n\alpha_i\cdot y_i\cdot(w^t\cdot x_i-b)+\sum_{i=1}^n\alpha_i\end{array}$
To get the optimal $\alpha$, we should take partial derivatives of $L$ on $w$, $b$ respectively and equate them to zero:
➀$\frac{\partial L}{\partial w}=w-\sum_{i=1}^n\alpha_i\cdot y_i\cdot x_i=0$, then, we have $w=\sum_{i=1}^n\alpha_i\cdot y_i\cdot x_i$
➁$\frac{\partial L}{\partial b}=\sum_{i=1}^n\alpha_i\cdot y_i$
➂use the deduction result of ➀ and ➁ in $L(w,b,\alpha)$, we obtain:
$\begin{array}{l}L(w,b,\alpha)\\=\frac12(\sum_{i=1}^n\alpha_i\cdot y_i\cdot x_i)^t\cdot(\sum_{i=1}^n\alpha_i\cdot y_i\cdot x_i)\\-\sum_{i=1}^n\alpha_i\cdot y_i\cdot((\sum_{i=1}^n\alpha_i\cdot y_i\cdot x_i)^t\cdot x_i-b)\\+\sum_{i=1}^n\alpha_i\end{array}$
➃reduce from the second term, we have below deduction:
$\begin{array}{l}\sum_{i=1}^n\alpha_i\cdot y_i\cdot((\sum_{i=1}^n\alpha_i\cdot y_i\cdot x_i)^t\cdot x_i-b)\\=\sum_{i=1}^n\alpha_i\cdot y_i\cdot(\sum_{i=1}^n\alpha_i\cdot y_i\cdot x_i)^t\cdot x_i\\-\sum_{i=1}^n\alpha_i\cdot y_i\cdot b\\=(\sum_{i=1}^n\alpha_i\cdot y_i\cdot x_i)^t\cdot(\sum_{i=1}^n\alpha_i\cdot y_i\cdot x_i)\\\dots\sum_{i=1}^n\alpha_i\cdot y_i\cdot b=0\end{array}$
➄take it back to $L(w,b,\alpha)$, we get:
$\begin{array}{l}L(w,b,\alpha)\\=-\frac12(\sum_{i=1}^n\alpha_i\cdot y_i\cdot x_i)^t\cdot(\sum_{i=1}^n\alpha_i\cdot y_i\cdot x_i)\\+\sum_{i=1}^n\alpha_i\\=-\frac12\sum_{i,j=1}^n\alpha_i\cdot\alpha_j\cdot y_i\cdot y_j\cdot(x_i^t\cdot x_j)\\+\sum_{i=1}^n\alpha_i\end{array}$

Why $L(w,b,\alpha)$ = the target plus the constraint?

But, why do we design $L(w,b,\alpha)$ in the expression of the target plus the constraint? Recall that in the lagrange multiplier article, I have shown you by taking $L(x,\lambda)=f(x)+\lambda\cdot g(x)$, finally leads to $\lambda<0$.

➀if we choose $L(x,\lambda)=f(x)-\lambda\cdot g(x)$, then $\lambda>0$. This is by artificial design. Whether plus or minus sign is used in the expression, could we have the similar achievement.
➁now, back to the formulate of our problem:
$\underset{w,b}{min}\frac12w^t\cdot w$, subject to $y_i\cdot(w^t\cdot x_i-b)\geq1$
➂suppose we take the constraint function function to be $g(x)=y_i\cdot(w^t\cdot x_i-b)\geq1$, if $\lambda$ would be negative by prior lagrangian proof, to get positive $\lambda$, we should use the minus operator in $L(x,\lambda)$.
➃Therefore, almost all SVM articles design $L(w,b,\alpha)$; where $\alpha$ is the lagrange multiplier given in the expression by using the minus operator:
$L(w,b,\alpha)=\frac12w^t\cdot w-\sum_{i=1}^n\alpha_i\cdot(y_i\cdot(w^t\cdot x_i-b)-1)$
➄such design guarantees that we could have $\forall\alpha_i>0$, which would be one major condition that must be satisfied in the following SVM article.

After solving the $\alpha_i$, we can get $w=\sum_{i=1}^n\alpha_i\cdot y_i\cdot x_i$, then, we can classify a new object x by:
$\begin{array}{l}f(x)\\=sign(w^t\cdot x-b)\\=sign((\sum_{i=1}^n\alpha_i\cdot y_i\cdot xi)^t\cdot x-b)\\=sign(\sum_{i=1}^n\alpha_i\cdot y_i\cdot(xi)^t\cdot x-b)\end{array}$

Non-Linear Training Set

Here, we have a question, what, if the separating surface of the two classes is not linear?
This might imply that the training set is distributed in a non-linear pattern. Suggestion would be made to transform the training set into another high-dimensional(maybe) space, such that they could be linearly separable.
➀let $\Phi(\cdot)$ to be the transformation function to some high-dimensional space, then:
$\begin{array}{l}L(\alpha)\\=-\frac12\sum_{i,j=1}^n\alpha_i\cdot\alpha_j\cdot y_i\cdot y_j\cdot(\Phi(x_i).\times\Phi(x_j))\\+\sum_{i=1}^n\alpha_i\end{array}$
, where $k(x_i,x_j)=\Phi(x_i).\times\Phi(x_j)$, $k(x_i,x_j)$ is the kernel function, the element-wise dot product $\Phi^t(x_i).\times\Phi(x_j)$ in the high-dimensional space is equivalent to the kernel function of the space of the output(or maybe input) parameters.
➁we can take $k(x_i,x_j)=e^{-(\frac{\left|x_i-x_j\right|^2}{2\cdot\delta^2})}$,
where, $k(x_i,x_j)\in R^{n2},\Phi(x_i).\times\Phi(x_j)\in R^{n2}$
$x_i,x_j\in R^{n1}$ and $n1\leq n2$.